Mathematics Pune University MCQs

 This set of Mathematics Multiple Choice Questions & Answers  focuses on “Properties of Rational Numbers”.

1. Which of the following type of numbers are closed under only multiplication?

a) Rational Numbers

b) Integers

c) Whole Numbers

d) Natural Numbers

Answer: c

Explanation:

Numbers Closed Under

Addition Subtraction Multiplication Division

a) Rational Numbers Yes Yes … No

b) Integers … Yes … No

c) Whole Numbers … … Yes …

d) Natural Numbers … No … …

2. By using the properties of rational numbers solve the following equation  + .

a) 26

b) 24

c) 28

d) 0

Answer: a

Explanation: By using the properties of rational numbers like addition and multiplication, we can solve the equation and get to the correct answer.

 + 

=  + 

= 26.

3. Solve: \

 

 * 

 

\).

a) \

 

 \

 

 \

 

 \(\frac{5}{2}\)

Answer: c

Explanation: Using the property of multiplication, we can solve the question.

\

 

 * 

 

\)

= \

 

 * 

 

\)

= 5/3.

4. Which number should be subtracted from \

 

 

\)?

a) \

 

 \

 

 \

 

 0

Answer: b

Explanation: Using the properties of addition and subtraction we can solve this problem.

\(\frac{16}{15} – x = \frac{-16}{15}\)

Therefore \(\frac{16}{15} + \frac{16}{15}\) = x

Therefore x = \(\frac{32}{15}\)

Option \(\frac{-32}{15}\) cannot be the correct answer, as the question itself asks to subtract the number in order to obtain \(\frac{-16}{15}\).

5. Simplify: \

 

 

 

 

 \

 

 \

 

 \

 

 \(\frac{121}{16}\)

Answer: d

Explanation: In order to simplify the given equation we have to equate the denominators.

We will do this by taking the least common multiple.

Therefore \(\frac{11}{2} + \frac{11}{4} – \frac{11}{8} + \frac{11}{16} = \frac{88}{16} + \frac{44}{16} – \frac{22}{16} + \frac{11}{16}\)

= \(\frac{121}{16}\).

6. Which of the following rational number is in the standard form?

a) \

 

 \

 

 \

 

 \(\frac{-21}{56}\)

Answer: a

Explanation: A rational number is said to be in it’s standard form when the numbers cannot be further reduced.

Here, we can see that the options other than the option \(\frac{16}{21}\) can be reduced.

When \(\frac{16}{24}\) is reduced, we get \(\frac{2}{3}\)… Standard Form.

When \(\frac{21}{15}\) is reduced, we get \(\frac{7}{5}\)… Standard Form.

When \(\frac{-21}{56}\) is reduced, we get \(\frac{-3}{8}\)… Standard Form.

7. Fill in the blanks \Missing open brace for subscript \

 

 \

 

 \

 

 \(\frac{-5}{4}\)

Answer: d

Explanation: Using the properties of addition and subtraction we can solve this problem.

\(\frac{-15}{32} / x = \frac{12}{32}\)

Therefore x = \(\frac{-15}{32} / \frac{12}{32}\)

Therefore x = \(\frac{-15}{32} * \frac{32}{12}\)

Therefore x = \(\frac{-15}{12}\)

Therefore x = \(\frac{-5}{4}\).

8. What is the multiplicative inverse of \

 

 \

 

 \

 

 1

d) 9

Answer: d

Explanation: \(\frac{15}{135}\) * x = 1

Therefore \(\frac{1}{9}\) * x = 1

Therefore x = 9

Multiplicative inverse of a particular number, is the number which gives 1 on multiplication with the number. There is one and only one multiplicative inverse for all numbers.

9. Which of the following lies between 2 and 3?

a) \

 

 \

 

 \

 

 \(\frac{12}{3}\)

Answer: a

Explanation: The correct answer to the question is \(\frac{5}{2}\), since it is the only number that lies between 2 and 3. While the others don’t lie in the given range.

Options \(\frac{6}{2}\) and \(\frac{6}{3}\) are not correct because the question states “between” and not “equal to”.

10. What is the reciprocal of \

 

 x

b) -x

c) \

 

 \(\frac{1}{-x}\)

Answer: b

Explanation: The number obtained by dividing a number by 1 is its reciprocal. All the numbers have one and only one reciprocal number.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Representation of Numbers on Number Line”.

1. On a number line, the arrangement of numbers is as follows _________

a) Negative -> 0 -> positive

b) Positive -> 0 -> negative

c) Positive -> 0 -> positive

d) Negative -> 0 -> negative

Answer: a

Explanation: When we refer to any number line, we observe that considering 0 as the center, the numbers to the left are negative and to the right are positive. The correct option would be “negative -> 0 -> positive”.

2. The number \

 

 Right side

b) Left side

c) On the center

d) Can be on either of the sides

Answer: b

Explanation: As we know that on a number line the negative numbers are placed on the left side of the 0 and positive numbers on the right side. Here \(\frac{-16}{7}\) is a negative number and hence has to be placed on the left side. Other options are wrong as negative numbers cannot be place on right side or at the center.

3. The number line for natural numbers is ____________

a) the line that extends indefinitely on both sides

b) the line that extends indefinitely to the right, but from 0

c) the line that extends indefinitely only to the right side of 1

d) the line that extends indefinitely on both sides, but you can see numbers only between –1, 0 and 0, 1 etc

Answer: c

Explanation: The set of natural numbers is [1, 2, 3, 4, 5…..] i.e. natural numbers start from 1 and extends to infinity on the positive side. The option which states that ‘The line that extends indefinitely to the right, but from 0′ shows the number line for whole numbers.

4. When the numbers \

 

 

 

 

 \

 

 \

 

 \

 

 \(\frac{5}{2}\)

Answer: b

Explanation: All the numbers given to us are positive and according to the sign convention they would be placed on the right side of the 0. Here we have 4 numbers which are

\(\frac{5}{2}\) = 2.5

\(\frac{7}{6}\) = 1.67

\(\frac{5}{4}\) = 1.25

\(\frac{12}{11}\) = 1.09

Among these four decimal numbers the smallest is 1.09 which is decimal representation of \(\frac{12}{11}\), this will be nearest to the center i.e. 0.

5. Which of the following form an equivalent pair of rational numbers?

a) \

 

 

 \

 

 

 \

 

 

 \(\frac{-230}{300} \,and \,\frac{-46}{60}\)

Answer: d

Explanation: When the rational numbers in their reduced form give equal fractions those numbers are called equivalent pair of rational numbers.

Here, we have 4 pairs of rational numbers in fractional form.

\

 

 

 

 

 

 

/13 and 8/13

\(\frac{-230}{300} \,and \,\frac{-46}{60}\) when reduced we get \(\frac{-23}{30} \,and \,\frac{-23}{30}\). Here we see that both the numbers are equal and hence this pair is equivalent pair.

6. A number line which consists only integers will consider all the numbers like 1, -1, 2.5, etc.

a) True

b) False

Answer: b

Explanation: The number line which consists of only integers, the set of integers is [……,-3, -2, -1, 0, 1, 2, 3, ……] Hence the number line won’t consist of decimal like 2.5. Hence the given statement is false.

7. -8 is ____ than 8 and -12 is ____ than -9.

a) greater and smaller

b) smaller and greater

c) greater and greater

d) smaller and smaller

Answer: d

Explanation: When the numbers -8 and 8 placed on the number line -8 would be on the left side of 0 and 8 would be on the right side of 0. Hence -8 is smaller than 8. Similarly, for -12 and -9, -12 is smaller than -9. This is because as we go towards left on a number line the magnitude of the number increases but due to its negative sign it is smaller than the rest of the number to the right of that number.

This set of Mathematics Quiz for Class 8 focuses on “Rational Numbers Between Two Rational Numbers”.

1. Which number lies between the numbers \

 

 

 \

 

 \

 

 \

 

 \(\frac{1}{2}\)

Answer: a

Explanation: The given fractional numbers represents \(\frac{1}{4}\) = 0.25 and \(\frac{1}{6}\) = 0.17

Our options are \(\frac{1}{5}\) = 0.20

\(\frac{1}{6}\) = 0.17

\(\frac{1}{3}\) = 0.33

\(\frac{1}{2}\) = 0.50

The only option between 0.17 and 0.25 is \(\frac{1}{5}\) i.e. 0.20.

2. The rational number which is not lying between 6 and 7 is ________

a) \

 

 \

 

 \

 

 \(\frac{13}{2}\)

Answer: a

Explanation: The only number which is not between 6 and 7 is \(\frac{36}{7}\)

\(\frac{36}{7}\) = 5.14

\(\frac{45}{7}\) = 6.42

\(\frac{46}{7}\) = 6.57

\(\frac{13}{2}\) = 6.5.

3. Between any consecutive two integers, there are always infinite integers.

a) True

b) False

Answer: b

Explanation: The set of integers is […., -1, 0, 1, ….] there cannot be any integers between two integers.

4. The maximum number of integers between two consecutive natural numbers is ________

a) zero

b) 2

c) 3

d) infinite

Answer: d

Explanation: There can be an infinite number of rational numbers between any two consecutive natural numbers. But there cannot be any integers between two consecutive natural numbers.

5. Which on the following number lies between \

 

 

 \

 

 \

 

 \

 

 \(\frac{3}{1000000}\)

Answer: a

Explanation: The given fractional numbers represents \(\frac{3}{10}\) = 0.3 and \(\frac{3}{10000}\) = 0.00003

Our options are in fractions, converting it into decimal form we get,

\(\frac{3}{1000}\) = 0.0003

\(\frac{3}{10000}\) = 0.00003

\(\frac{3}{100000}\) = 0.000003

\(\frac{3}{1000000}\) = 0.0000003.

The only option between 0.3 and 0.00003 is \(\frac{3}{1000}\) i.e. 0.0003.

6. \

 

 \

 

 

 \

 

 

 \

 

 

 \(\frac{7}{2}\) and \(\frac{6}{5}\)

Answer: c

Explanation: When we place the 4 given pairs and the number on the number line we observe that the number \(\frac{32}{9}\) lies between the pair \(\frac{7}{2}\) and \(\frac{10}{2}\), while other options fail to capture the given number between them. Hence the correct pair is \(\frac{7}{2}\) and \(\frac{10}{2}\).

7. Which of the following is the greatest number among the given rational numbers?

a) \

 

 \

 

 \

 

 \(\frac{101}{20}\)

Answer: a

Explanation: The numbers when placed on the number line, the number which is placed farthest from 0 is the greatest. Here the numbers are, \(\frac{139}{26}\) = 5.34

\(\frac{72}{21}\) = 3.42

\(\frac{99}{36}\) = 2.75

\(\frac{101}{20}\) = 5.05

So, we conclude that the greatest number is \(\frac{139}{26}\) and hence would be placed farthest from 0.

8. [….., -1, 0, 1, …….] the given set shows which type of numbers?

a) Rational numbers

b) Integers

c) Natural numbers

d) Whole numbers

Answer: b

Explanation: The given set is of integers.

The sets of other numbers are,

Whole number -> [0, 1, 2, 3, 4, ……]

Natural number -> [1, 2, 3, 4, …….]

Rational number -> [……, 0, ……..] This set consists of all numbers on the number line.

9. Pick the smallest number of the following.

a) –\

 

 –\

 

 –\

 

 –\(\frac{101}{20}\)

Answer: a

Explanation: The numbers when placed on the number line, the number which is placed nearest from 0 is the smallest. Here the numbers are, –\(\frac{139}{26}\) = -5.34

–\(\frac{72}{21}\) = -3.42

–\(\frac{99}{36}\) = -2.75

–\(\frac{101}{20}\) = -5.05

So, we conclude that the smallest number is –\(\frac{139}{26}\) and hence would be placed smallest from 0. Thing to remember: When the numbers are placed on the left side of the number line, the number farthest from the center is the smallest number. This is due to the negative sign.

10. Pick the option which has the ordered pair of BIG and SMALL number.

a) \

 

 

 \

 

 

 \

 

 

 \(\frac{12}{4}\) and \(\frac{13}{5}\)

Answer: d

Explanation: Here we have to find the ordered pair of numbers and the order should be big following by small. This condition is only satisfied by one pair of numbers which is \(\frac{12}{4}\) and \(\frac{13}{5}\).

This set of Mathematics MCQs for Class 8 focuses on “Solving Linear Equations with Variables on One Side and Number on the Other”.

1. Find the solution for the equation 3x + 3 = 4.

a) 3

b) \

 

 4

d) \(\frac{1}{4}\)

Answer: b

Explanation: 3x + 3 = 4

Step 1: Subtract 3 from both the sides,

3x + 3 – 3 = 4 – 3

3x = 1

Step 2: Divided both the sides by 3,

x = \(\frac{1}{3}\)

So, we have solution for the equation x = \(\frac{1}{3}\)

2. In equation 3x + 4 = 10, by transposing the variable on RHS we get ________

a) -4 = 10 – 3x

b) 4 = 3x + 10

c) 4 = -3x + 10

d) -4 = – 3x – 10

Answer: c

Explanation: While shifting the variable from LHS to RHS the sign changes from positive to negative and vice versa. Here, the given equation is 3x + 4 = 10.

3. Solve equation 7x + 14 = 21 to find value of x.

a) x = 1

b) x =-1

c) x = 2

d) x = -2

Answer: a

Explanation: 7x + 14 = 21

Step 1: Solving the equation, we take constant 14 on the RHS making 7x = 7

Step 2: Dividing LHS and RHS with 7 we get x = 1.

4. Solve: \

 

 

 

 x = 1

b) x = 2

c) x = 3

d) x = 0

Answer: d

Explanation: \(\frac{5}{2}x + \frac{3}{2} = \frac{6}{4}\)

Step 1: Reducing the fraction we get,

\(\frac{5}{2}x + \frac{3}{2} = \frac{3}{2}\)

Step 2: Multiplying LHS and RHS by 2 we get,

5x + 3 = 3

i.e. 5x = 0

Step 3: Dividing both sides by 5 we get,

x = 0.

5. Solve: 7x – 2 = 3.

a) x = \

 

 x = \

 

 x = \

 

 x = \(\frac{5}{3}\)

Answer: a

Explanation: 7x – 2 = 3

Step 1: Transposing -2 on RHS we get,

7x = 5

Step 2: Dividing both sides by 7 we get,

x = \(\frac{5}{7}\).

6. Pick the equation from the given one’s which have solution as z = 2.

a) 2z -2 = 3

b) 3z -2 = -2

c) 3z -3 = 3

d) 4z + 3 = 3

Answer: c

Explanation: Solving the 3z – 3 = 3 we get,

Step 1: Transposing -3 on RHS we get,

3z = 6

Step 2: Dividing both sides by 3 we get,

z = 2

Hence, this is the solution of our question.

7. Pick the equation which has the solution in the form of prime number.

a) 2x = 3

b) 3z = -6

c) 4y – 3 = 2

d) 2z – 2 = 2

Answer: d

Explanation: A prime number is the number which can be formed by multiplying only two numbers which are 1 and the number itself.

Solving equation, 2z – 2 = 2

Step 1: Transposing – 2 on RHS we get,

2z = 4

Step 2: Dividing equation by 2 we get,

z = 2

Since 2 is greater than 0 and can be formed only by multiplying 1 and 2 it is a prime number.

8. Solve: 16 = 3m – 2.

a) m = -5

b) m = 5

c) m = 6

d) m = -6

Answer: c

Explanation: Step 1: Transposing – 2 to LHS we get,

18 = 3m

Step 2: Dividing both the sides by 3 we get,

m = 6.

9. What is the solution of the equation, 7 + 2x = 0?

a) \

 

 \

 

 \

 

 \(\frac{-7}{2}\)

Answer: d

Explanation: 7 + 2x = 0

Step 1: Transposing 7 to RHS we get,

2x = -7

Step 2: Dividing both sides by 2 we get,

x = \(\frac{-7}{2}\).

10. The solution of equation 3x + 6 = -3 is x = -3.

a) True

b) False

Answer: a

Explanation: 3x + 6 = -3

Step 1: Transposing 6 to RHS we get,

3x = -9

Step 2: Dividing both sides by 3 we get

x = -3.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Applications of Linear Equation ”.

1. What should be added to twice the rational number \

 

 

 \

 

 \

 

 \

 

 \(\frac{13}{4}\)

Answer: b

Explanation: Twice the rational number is  × \(\frac{13}{3} = \frac{26}{3}\)

Let x be added to \(\frac{26}{3}\),

x + \(\frac{26}{3} = \frac{13}{6}\)

x = \(\frac{13}{6} – \frac{26}{3}\)

x = \(\frac{13}{6} – \frac{52}{6}\)

x = \(\frac{-39}{6}\)

x = \(\frac{-13}{2}\).

2. Perimeter of a square is 48 m. What is the length of one side?

a) 12 m

b) 192 m

c) 13 m

d) 14 m

Answer: a

Explanation: Formula for perimeter is square,

Perimeter = 4 × 

48 = 4 × 13

Dividing throughout by 4 we get,

Length of one side = 12.

3. At Tanay’s birth, he was 34 years younger than his father. What will be his age after 20 years?

a) 20

b) 54

c) 0

d) 14

Answer: a

Explanation: At birth Tanay’s age would be 0, at that time difference in ages was 34 years.

As the time passes Tanay and his father’s age increases but the difference remains same.

Hence, Tanay’s age would be

i.e. Age at his birth + 20 years

i.e. 0 + 20 years

i.e. 20 years.

4. If a side of a regular hexagon is 20 cm then what will be the perimeter of that regular hexagon?

a) 120 cm

b) 120 m

c) 60 cm

d) 60 m

Answer: a

Explanation: A regular hexagon has six equal sides.

Perimeter of a regular hexagon = 6 × 

Perimeter of a regular hexagon = 6 × 

Perimeter of a regular hexagon = 120 cm.

5. Sum of two distinct numbers is positive then which of the following statements is correct?

a) Both numbers are equal and negative

b) Greater one positive and smaller one negative

c) Both positive

d) Both negative

Answer: d

Explanation: Here we are given that there are two distinct numbers, so the option stating equal and negative is ruled out. The second information we have is the sum is negative, so if the greater number is positive than the sum is positive and hence the option which states greater one positive and smaller one negative also is ruled out. Now we have two options left they are both positive or negative. We know that sum two positive number is positive. So the option left is both negative which is the correct options.

6. There are two parties in an election the ratio of their votes is 2:3. The total number of voters who took part in that election are 1200. How many votes did the winning party get?

a) 480

b) 600

c) 1200

d) 720

Answer: d

Explanation: Let the common ratio be x.

∴ 2x + 3x = 1200

∴ 5x = 1200

∴ x = 240

The party which gets higher number of votes wins the election so, the party which has 3x votes is the winning party.

Number of votes = 3x

= 3 × 240

= 720.

7. If circumference of a circle is 6.28 cm then what is the radius of the circle?

a) 2 cm

b) 1 cm

c) 4 cm

d) 3 cm

Answer: b

Explanation: Circumference of a circle = 2 × π × Radius

6.28 = 2 × 3.14 × Radius

Radius = 1.

8. The degree of linear equation in one variable is _______

a) 1

b) 2

c) 3

d) 4

Answer: a

Explanation: Degree is the highest power of any equation.

In any linear equation the highest power is 1

Hence the degree of linear equation in one variable is 1.

9. The process of shifting any constant or variable in an equation from one side to other is ___________

a) assosiativity

b) distributivity

c) commutativity

d) transposition

Answer: d

Explanation: This is the property used while we solve equations. In this property when we take a component of equation from LHS to RHS or vice versa the signs change, the change of signs is in order to balance the equation.

10. Linear equation in one variable has ________ number of solution/s.

a) one

b) one and only one

c) two

d) infinite

Answer: b

Explanation: The number of solution of any equation depends on the degree of the equation (degree is the highest power of the equation. Example x 2 +x-2=0 here the degree of the equation is 2 as the highest power in this equation is 2). Since the linear equation in one variable has it’s degree as 1 always, therefore the answer would be ‘one and only one solution’. The option ‘one’ is not accurate and hence one should select ‘one and only one’.

This set of Mathematics Multiple Choice Questions and Answers for Class 8 focuses on “Solving Linear Equations with Variables on Both the Sides”.

1. If Malik has 3 coins and his sister has 4 coins each of 2 rupees, what will be the sum of amount both have?

a) 14 rupees

b) 13 rupees

c) 12 rupees

d) 7 rupees

Answer: a

Explanation: There are 7 coins in total. When we multiply 2 × 7 we get 14. Hence, the total amount with them is 14 rupees.

2. The perimeter of a rectangle is 12 cm and it’s breadth is 2 cm. What will be it’s length?

a) 2 cm

b) 3 cm

c) 4 cm

d) 5 cm

Answer: b

Explanation: For any rectangle the formula for perimeter is 2 × . Here we already know the breadth of the rectangle. When we substitute the known values in the formula for perimeter of a rectangle we get,

12 = 2 × 

∴ 6 = Length + 2

∴ Length = 4 cm.

3. At present Disha’s mother is three times the age of Disha. After 5 years their ages will sum up to 70 years. Find the present age.

a) Disha 10 and her mother 30

b) Disha 12 and her mother 36

c) Disha 13 and her mother 39

d) Disha 15 and her mother 45

Answer: d

Explanation: Let Disha’s age be x years then her mother would be 3x years old. After 5 years, their ages are x + 5  and 3x + 5  respectively.

 +  = 70

∴ x + 5 + 3x + 5 = 70

∴ 4x + 10 = 70

∴ 4x = 60

∴ x = 15

Since x = 15 we know that Disha’s present age is 15 years, to find her mother’s age we have  i.e. 45 years.

4. Bansal has 7 times as many two-rupee coins as he has five-rupee coins. If he has in all a sum of rupees 95, how many coins of each denomination does he have?

a) 5 two-rupee and 2 five-rupee

b) 15 two-rupee and 5 five-rupee

c) 5 two-rupee and 15 five-rupee

d) 15 two-rupee and 5 five-rupee

Answer: b

Explanation: Let the number of five-rupee coins with Bansal be x. Then the number of two-rupee coins will be 7x. The amount from

 Five-rupee coins = 5x

 Two-rupee coins = 2x × 7 = 14x

Hence the total amount with Bansal is 19x.

∴ 19x = 95

∴ x = 5

Thus, the number of five-rupee coin = 5

and number of two-rupees coin = 15.

5. The sum of three consecutive numbers is 789, what are those consecutive numbers?

a) 262, 263 and 264

b) 263, 264 and 265

c) 264, 265 and 266

d) 265, 266 and 267

Answer: a

Explanation: Let the first number be x. Since they are consecutive numbers the next number would be  and the third number would be . Now as the sum of the numbers is given as 789 we solve the equation by adding the three consecutive variables.

x +  +  = 789

∴ 3x + 3 = 789

∴ 3x = 789 – 3

∴ 3x = 786

∴ x = 262

Hence the three consecutive numbers are 262, 263 and 264.

6. The sum of two natural numbers is 5, the numbers are in a ratio 2:3. Find the numbers.

a) 2 and 3

b) 1 and 4

c) 0 and 5

d) 2 and 4

Answer: a

Explanation: Let the common multiple be x.

∴ 2x + 3x = 5

∴ 5x = 5

∴ x = 1

Since, x = 1 the natural numbers are 2 and 3.

7. If the perimeter of square is 28 m. Find the length of the side.

a) 7 cm

b) 7 m

c) 14 m

d) 11 m

Answer: b

Explanation: The formula for perimeter of a square is 4 × 

Perimeter = 4 × 

∴ 28 = 4 × 

∴ side = 7

Hence, the length of the side of the square is 7 m.

8. Sum of consecutive multiples of 23 is 1656. Find the multiples.

a) 529, 552, 575

b) 629, 662, 675

c) 189, 222, 275

d) 389, 332, 375

Answer: a

Explanation: Let , x and  be the three consecutive multiples. Now, the sum of these numbers is 1656.

 + x +  = 1656

∴ x – 23 + x + x + 23 = 1656

∴ 3x = 1656

∴ x = 552

Hence, x – 23 = 529 and x + 23 = 575

The three consecutive multiples of 23 which sum up to 1656 are 529, 552 and 575.

9. When two integers are added the sum is -52. If the integers are in the ratio 6:7, then find the integers.

a) -24 and 28

b) 24 and -28

c) -24 and -28

d) 24 and 28

Answer: c

Explanation: Let the common multiple be x. We know that the integers sum up to -52.

6x + 7x = -52

∴ 13x = -52

∴ x = -4

Since, x = -4 the integers are -24 and -28.

10. If Raj scores 27 marks less than the highest scorer and the highest scorer has 2 marks less than the maximum achievable score, then find the score that Raj scored, if the maximum achievable score is 100.

a) 70

b) 71

c) 72

d) 73

Answer: b

Explanation: The highest scorer scores 2 marks less then maximum achievable score.

Score of highest scorer = Maximum achievable score – 2

∴ Score of highest scorer = 100 -2

∴ Score of highest scorer = 98

Now, Raj’s score can be calculated by,

Raj’s score= Score of highest scorer – 27

∴ Raj’s score = 98 – 27

∴ Raj’s score = 71.

This set of Mathematics Online Test for Class 8 focuses on “Applications of Linear Equation ”.

1. The digits of a two-digit number differ by 4. If the digits are interchanged, and the resulting number is added to the original number, we get 152. What can be the original number?

a) 95

b) 40

c) 73

d) 59

Answer: a

Explanation: Let us take the two digit number such that the digit in the units place is x. The digit in tens place differs by 4 ∴ the digit in tens place is .

∴the two digit number obtained = [10 * ] + [x]

∴the two digit number obtained = 10x + 40 + x

∴the two digit number obtained = 11x + 40

When the digits are interchanged we obtain the number = [10 * x] + []

The new number obtained = 11x + 4

As we know the original number and the new number add up to 152.

∴ [11x + 40] + [11x + 4] = 152

∴ 22x + 44 = 152

∴ 22x = 110

∴ x = 5

The original number = [10 * ] + [x]

The original number = 11x + 40

The original number = 11 *  + 40

The original number = 55 + 40

The original number = 95.

2. Jon is thrice as old as Kavya. Five years ago his age was two times Kavya’s age. Find their present age.

a) Kavya’s age = 15 years; Jon’s age = 5 years

b) Kavya’s age = 5 years; Jon’s age = 15 years

c) Kavya’s age = 5 years; Jon’s age = 5 years

d) Kavya’s age = 15 years; Jon’s age = 15 years

Answer: b

Explanation: Let us consider Kavya’s age as x years.

Then Jon’s age would be 3x years.

Kavya’s age five years ago was  years.

Jon’s age five years ago was  years.

It is given that Jon’s age five years ago was two times Kavya’s age.

Thus,  = 2 * 

Or 3x -5 = 2x – 10

Or x = 5

∴ Jon’s age = 3x

∴ Jon’s age = 3 * 5

∴ Jon’s age = 15 years

∴ Kavya’s age = x

∴ Kavya’s age = 5 years.

3. Arya takes a number adds \

 

 \

 

 \

 

 23

d) 3

Answer: a

Explanation: Let the number chosen by Arya be x.

The first operation carried out by Arya was adding \(\frac{13}{3}\) to the number.

Hence, the equation formed is x + \(\frac{13}{3}\) = 12

The second operation is to divide throughout by 3.

Hence, the equation is modified to \(\frac{x}{3} + \frac{13}{9}\) = 4

\(\frac{x}{3} + \frac{13}{9}\) = 4

∴ \(\frac{3x}{9} + \frac{13}{9} = \frac{36}{9}\)

∴ 3x + 13 = 36

∴ 3x = 23

∴ x = \(\frac{23}{3}\).

4. When a number is subtracted from 484, we get 459. The number subtracted is square of?

a) 5

b) 4

c) 3

d) 2

Answer: a

Explanation: Let the number subtracted be x.

484 – x = 459

∴ 484 – 459 = x

∴ x = 25

∴ the subtracted number is 25 and 25 is square of number 5.

5. Raj buys books worth rupees four hundred, he has coins of denomination two-rupees. How many coins does he need to pay the bill?

a) 200

b) 100

c) 400

d) 150

Answer: a

Explanation: Let the number of coins required be x.

The amount to be paid is rupees 400.

 = 400

∴ 2x = 400

∴ x = 200

Raj needs 200 coins each of two rupees in order to pay the bill amount.

6. Form an equation for all multiples of 12.

a) 3x

b) 12x

c) 4x

d) 3x

Answer: b

Explanation: If a set is formed consisting all the multiples of 12 it would be like [12,24,36,48,60,…..] The set of multiples of 12 is formed by multiplying the set of natural numbers with 12.

Let the set of natural number be represented by x

∴ the general equation of multiples of 12 = 12x

7. Sita wants to buy books of five hundred-rupees and she has 12 fifty-rupees notes. How many notes will she have after the payment?

a) 1

b) 2

c) 3

d) 4

Answer: b

Explanation: If Sita wants to pay five hundred-rupees in fifty-rupees notes then she has to give the shopkeeper 10 notes each of fifty-rupees. After giving 10 notes, she will be left with 2 notes. Hence, the correct answer to this question is 2.

8. If Ram’s present age is 3 years and Shyam is twice Ram’s present age. What will be Shyam’s age after 10 years?

a) 16

b) 17

c) 18

d) 19

Answer: a

Explanation: Let Ram’s present age be x years.

∴ Shyam’s present age will be 2x years.

After 10 years Shyam’s age would be,  years.

∴ Shyam’s age after 10 years = 2 * 3 + 10

∴ Shyam’s age after 10 years = 6 + 10

∴ Shyam’s age after 10 years = 16 years.

9. If the perimeter of a regular hexagon is 192 m then find the Length of each side of the regular hexagon.

a) 32 cm

b) 32 m

c) 23 m

d) 23 cm

Answer: b

Explanation: Perimeter of a regular hexagon = 6 * 

∴ 192 = 6 * 

∴ side = 32 m.

10. If the perimeter of a scalene triangle is 23 cm, with side 1 with Length 12 cm and side 2 with Length 3 cm. Find the Length of third side.

a) 23 cm

b) 12 cm

c) 2 cm

d) 8 cm

Answer: d

Explanation: Perimeter of a scalene triangle = Length of side 1 + Length of side 2 + Length of side 3

∴ 23 = 12 + 3 + Length of side 3

∴ 23 = 15 + Length of side 3

∴ Length of side 3 = 8 cm.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Reduce the Linear Equation and Find the Value of the Variable”.

1. Solve: \

 

 

 x = \

 

 x = \

 

 x = \

 

 x = \(\frac{167}{14}\)

Answer: a

Explanation: \(\frac{2x-11}{2} + \frac{x}{7}\) = 5

∴\(\frac{7*}{7*2} + \frac{2*x}{2*7}\) = 5

∴\(\frac{14x-77+2x}{14}\) = 5

∴16x – 77 = 5 * 14

∴16x – 77 = 70

∴16x = 147

∴x = \(\frac{147}{16}\).

2. Solve: \

 

 

 

 \

 

 \

 

 \

 

 \(\frac{1}{10}\)

Answer: c

Explanation: \(\frac{x}{5} + \frac{x}{7} = \frac{1}{35}\)

∴\(\frac{7*x+5*x}{5*7} = \frac{1}{35}\)

∴\(\frac{7x + 5x}{35} = \frac{1}{35}\)

∴\(\frac{12x}{35} = \frac{1}{35}\)

∴12x = 1

∴x = \(\frac{1}{12}\).

3. Solve: 3 = 5.

a) \

 

 \

 

 2

d) -2

Answer: a

Explanation: 3 = 5

∴3x – 9 = 5x + 10

∴3x – 5x = 10 – 9

∴-2x = 1

∴x = \(\frac{-1}{2}\).

4. Solve: \

 

 

 \

 

 \

 

 3

d) 2

Answer: a

Explanation: \(\frac{m-1}{2} + \frac{2m+3}{3}\) = 2

∴\(\frac{3*}{3*2} + \frac{3*}{2*3}\) = 2

∴\(\frac{3m-3+6m+9}{6}\) = 2

∴\(\frac{9m+6}{6}\) = 2

∴9m + 6 = 12

∴9m = 6

∴m = \(\frac{2}{3}\).

4. Solve: 12x – 3 = 9.

a) 1

b) -1

c) 12

d) -12

Answer:

Explanation: 12x – 3 = 9

∴12x = 12

∴x = 1.

5. Solve: \

 

 

 \

 

 \

 

 \

 

 \(\frac{7}{59}\)

Answer: c

Explanation: \(\frac{2v-3}{2} + \frac{3v}{4}\) = 12

∴\(\frac{4*}{4*2} + \frac{2*3v}{4*2}\) = 12

∴\(\frac{8v-12+6v}{8}\) = 12

∴8v – 12 + 6v = 12 * 8

∴14v – 12 = 96

∴14v = 108

∴v = \(\frac{108}{14} = \frac{59}{7}\).

6. Solve: 0.25 = 1.

a) \

 

 \

 

 \

 

 \(\frac{11}{12}\)

Answer: a

Explanation: 0.25 = 1

∴3t – 1 = 1

∴3t = 2

∴t = \(\frac{2}{3}\).

7. Solve: 2y – 2 = 7y + 1.

a) \

 

 \

 

 \

 

 \(\frac{3}{2}\)

Answer: a

Explanation: 2y – 2 = 7y + 1

∴2y – 7y = 1 + 2

∴-5y = 3

∴y = \(\frac{-3}{5}\).

8. Solve: 3n – 13 = 13n + 31.

a) \

 

 \

 

 \

 

 \(\frac{44}{10}\)

Answer: b

Explanation: 3n – 13 = 13n +31

∴3n – 13n = 31 + 13

∴-10n = 44

∴n = \(\frac{-44}{10} = \frac{-22}{5}\).

9. Solve: 2x – 3 = 0.

a) \

 

 \

 

 \

 

 \(\frac{5}{3}\)

Answer: a

Explanation: 2x – 3 = 0

∴2x = 3

∴x = \(\frac{3}{2}\).

10. Solve: \

 

 

 \

 

 \

 

 \

 

 \(\frac{16}{10}\)

Answer: c

Explanation: \

 

 

 = 1 * 

∴28y – 24 = 12y – 14

∴28y – 12y = -14 + 24

∴16y = 10

∴y = \(\frac{10}{16} = \frac{5}{8}\).

This set of Mathematics Question Bank for Class 8 focuses on “Reduce the Given Linear Equations to a Simpler Form”.

1. Solve: \

 

 

 \

 

 \

 

 \

 

 \(\frac{-9}{91}\)

Answer: b

Explanation: \(\frac{3x+7}{12} + \frac{22x-1}{3}\) = 1

∴ \(\frac{3*+12*}{12*3}\) = 1

∴ 9x + 21 + 264x – 12 = 36

∴ 273x + 9 = 36

∴ 273x = 27

∴ x = \(\frac{27}{273} = \frac{9}{91}\).

2. Simplify: \

 

 

 10x = 5

b) 5x = 10

c) 10x + 5 = 0

d) 5x + 10 = 0

Answer: a

Explanation: \(\frac{2x+1}{2} + \frac{2x-1}{3}\) = 1

∴ \(\frac{3*+2*}{6}\) = 1

∴ \(\frac{6x+3+4x-2}{6}\) = 1

∴ 6x + 3 + 4x – 2 = 6

∴ 10x + 1 = 6

∴ 10x = 5

3. Solve: 11x – 2 = 7x + 12.

a) –\

 

 \

 

 \

 

 \(\frac{14}{6}\)

Answer: b

Explanation: 11x – 2 = 7x + 12

∴ 11x – 7x = 12 + 2

∴ 4x = 14

∴ x = \(\frac{14}{4} = \frac{7}{2}\).

4. When the equation \

 

 

 x=0

b) x=1

c) x-1=0

d) x+1=0

Answer: a

Explanation: \(\frac{12x}{2} + \frac{6x}{5}\) = 0

∴ \(\frac{5*12x}{2*5} + \frac{2*6x}{5*2)}\) = 0

∴ 60x+12x=0

∴ 72x=0

∴ x=0.

5. Solve: 12x-13=2x*.

a) x+\

 

 x-\

 

 x=\

 

 x+\(\frac{14}{13}\)=0

Answer: c

Explanation: 12x-13=2x*

∴ 12x-13=22x-24x

∴ 12x-13= -2x

∴ 14x=13

∴ x=\(\frac{13}{14}\).

6. Simplify: \

 

 x=\

 

 13x+145=0

c) 13x=145

d) 145x+13=0

Answer: c

Explanation: \(\frac{13x-1}{12}\)=12

∴ 13x-1=144

∴ 13x=145

∴ x=\(\frac{145}{13}\).

7. Simplify: t – 12 = 12t – 1.

a) 1+t=0

b) 1-t=0

c) t+1=0

d) t=0

Answer: b

Explanation: t – 12 = 12t – 1.

∴ -11t=11

∴ t=1.

8. Solve: 13x-2=21x+2.

a) 2x+1=0

b) 2x-1= 0

c) 1 + 2x = 0

d) 1 – 2x = 0

Answer: a

Explanation: 13x-2=21x+2

∴ 13x-21x=2+2

∴ -8x=4

∴ x=\(\frac{-1}{2}\).

9. What will be the solution for equation 3j+2=1-3j.

a) \

 

 \

 

 \

 

 \(\frac{-1}{12}\)

Answer: a

Explanation: 3j+2=1-3j

∴ 3j+3j=1-2

∴ 6j=-1

∴ j=\(\frac{-1}{6}\).

10. Solve: 12x+2=13x-1.

a) 1

b) 2

c) 3

d) 4

Answer: c

Explanation: 12x+2=13x-1

∴ 12x-13x=-1-2

∴ –x=-3

∴ x=3.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Age/Money/Denomination of Currency Word Problems”.

1. Sara is twice the age of Marry. The sum total of Sara’s age and Marry’s age after five years is 52. What is Marry’s present age?

a) 14 years

b) 19 years

c) 28 years

d) 33 years

Answer: c

Explanation: Let Sara’s age be x years. ∴ Marry’s age is 2x years.

Now after five years Sara’s age =  years and Marry’s age =  years

We know that, +=52

∴ x+5+2x+5=52

∴ 3x+10=52

∴ 3x=42

∴ x=14 years

∴ Sara’s present age is 14 years. As we know that, Marry’s age is twice that of Sara’s age

∴ Marry’s age 28 years.

2. Mohan has to pay two hundred rupees for a book but has only a note of two thousands rupees, what amount will he get back?

a) 2000 rupees

b) 200 rupees

c) 1800 rupees

d) 1400 years

Answer: c

Explanation: Mohan has to pay two hundred rupees. He has two thousand rupees with him. Let amount he gets back be x rupees.

∴ 2000-x=200

∴ 2000-200=x

∴ x=1800 rupees.

3. If Akshat has twelve two rupees coins and two five rupees coins. What is the total amount with him?

a) 43 rupees

b) 34 rupees

c) 23 rupees

d) 32 rupees

Answer: b

Explanation: Akshat has 12 two rupees coins and 2 five rupees = +

∴ Total Amount = 24+10

∴ Total Amount = 34 rupees.

4. A boy has all five rupees coins and he need to pay one hundred and thirty five rupees to a shopkeeper for his grocery. How coins does he need to pay the total amount?

a) 27

b) 28

c) 29

d) 30

Answer: a

Explanation: The total amount to be paid is 135 rupees.

The boy has only 5 rupee coins. Let the number of coins required be x.

∴ \(\frac{135}{5}\)=x

∴ 27=x

∴ the boy requires 27 coins in order to pay his billed amount.

5. If a girl buys an ice cream worth twenty seven rupees and pays the shopkeeper with a note worth fifty rupees. What will be the change she received from the shopkeeper?

a) 27 rupees

b) 23 rupees

c) 22 rupees

d) 21 rupees

Answer: b

Explanation: Let the amount received by the girl from the shopkeeper be x rupees.

∴ 27+x=50

∴ x=50-27

∴ x=23.

6. At present Rahul’s age is 27 years and Rajiv’s age is 19 years. What is the sum of their ages after five years?

a) 56 years

b) 65 years

c) 46 years

d) 64 years

Answer: a

Explanation: The present age of Rahul and Rajiv is 27 and 19 respectively, after five years their age will be Rahul’s age =years and Rajiv’s age = years.

Sum of Rahul’s age and Rajiv’s age = +

∴ Sum of Rahul’s age and Rajiv’s age = 32+24

∴ Sum of Rahul’s age and Rajiv’s age = 56 years.

7. The notebook costs thirty rupees. How many ten rupees notes will be required to pay the whole amount?

a) 3 notes

b) 4 notes

c) 5 notes

d) 6 notes

Answer: a

Explanation: If a notebook costs thirty rupees and one has ten rupees notes. Let the number of ten rupees note required be x.

∴ 10x=30

∴ x=\(\frac{30}{10}\)

∴ x=3.

8. Present ages of Anu and Raj are in the ratio 4:5. Eight years from now the ratio of their ages will be 5:6. Find their present ages.

a) Raj’s age is 32 years and Anu’s age is 40 years

b) Raj’s age is 40 years and Anu’s age is 48 years

c) Raj’s age is 32 years and Anu’s age is 32 years

d) Raj’s age is 40 years and Anu’s age is 32 years

Answer: d

Explanation: Let the present ages of Anu and Raj be 4x years and 5x years respectively.

After eight years. Anu’s age =  years;

After eight years, Raj’s age =  years.

∴ the ratio of their ages after eight years = \

 

 

 

=5

∴ 24x+48=25x+40

∴ –x=-8

∴ x=8

∴ Anu’s age = 4×8 = 32years

∴ Raj’s age = 5×8 = 40years.

9. Two brothers have their age in the ratio of 2:3. After five years what will be the ratio of their ages. The sum of their ages after 5 years is 30.

a) \

 

 \

 

 \

 

 \(\frac{3}{2}\)

Answer: a

Explanation: Let the common ratio be x. The age of two brothers after five years would be  years and years.

Therefore +=30

Therefore 2x+5+3x+5=30

Therefore 5x=20

Therefore x=4

Therefore the brothers age would be 10 years and 15 years respectively.

Now, after five years their ages would be 15 years and 20 years respectively.

The ratio of the brothers ages = \(\frac{15}{20} = \frac{3}{4}\).

10. A boy has two two rupees coins, one five rupees coins and two ten rupees coins. In what combination should the boy select three coins so the amount is maximum?

a) two five rupees and one ten rupees

b) three ten rupees and one five rupees

c) four two rupees and one ten rupees

d) two ten rupees and one five rupees

Answer: d

Explanation: If the boy selects two ten rupees and one five rupees coins then he’ll get the maximum amount. While in any other combination the amount would be less.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Polygon”.

1. A simple closed curve made up of only line segments is called a ________

a) polygon

b) square

c) rectangle

d) triangle

Answer: a

Explanation: The options other than polygon are special cases of polygons. Polygon can have ‘n’ number of sides but all the sides should be joined by a straight line.

2. Which of the following is a polygon?

a) 

b) mathematics-questions-answers-polygon-q2b

c) mathematics-questions-answers-polygon-q2c

d) mathematics-questions-answers-polygon-q2d

Answer: d

Explanation: Incorrect options are not polygons because they have their side as curved lines. As we know that the polygon is a figure which consists of ‘n’ numbers of side but all of them should be straight lines and not curved.

3. Which of the following is not a polygon?

a) mathematics-questions-answers-polygon-q3a

b) mathematics-questions-answers-polygon-q3b

c) mathematics-questions-answers-polygon-q3c

d) mathematics-questions-answers-polygon-q3d

Answer: a

Explanation: It cannot be considered as a polygon as all the sides aren’t connected. Polygons form a closed figure and hence this figure won’t be considered as a polygon.

4. The maximum number of sides that a polygon can have is ________

a) three

b) four

c) five

d) infinite

Answer: d

Explanation: The definition of a polygon states that ‘A simple closed curve made up of only line segments is called a polygon.’ Here the basic condition for a polygon is that the figure should be a closed figure and there can be ‘n’ numbers for sides.

5. Any triangle is a ______ sided polygon.

a) 1

b) 2

c) 3

d) 4

Answer: c

Explanation: A triangle is three sided polygon. The triangles are classified into three types  Equilateral Triangle  Isosceles Triangle  Scalene Triangle. This classification is done on the basis of length of each side and the angle between these sides.

 Equilateral Triangle: In this triangle all sides are of equal length and all the angles measure=60°

mathematics-questions-answers-polygon-q5a

 Isosceles Triangle: In this triangle, two sides are of equal length and the base angles are equal.

mathematics-questions-answers-polygon-q5b

 Scalene Triangle: In this triangle, all the sides are of different and all the angles are different in measure.

mathematics-questions-answers-polygon-q5c

6. A polygon having ten sides is called decagon.

a) True

b) False

Answer: a

Explanation: The word polygon is a Latin word . This is derived by the number of sides possessed by the figure. Here we take an example of six sided figure in Latin six is hex therefore the name of the figure is ‘hexagon’. Similarly, ten sided figure is known as a decagon.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Classification of Polygons”.

1. Which of the following is correct?

a) A polygon is a figure which comprises of straight lines

b) A polygon is a figure which comprises of all types of curves

c) A polygon is a closed figure made up of straight lines

d) A polygon is a figure which comprises of straight lines and curves

Answer: c

Explanation: The definition of a polygon states that ‘A simple closed curve made up of only line segments is called a polygon’. The other options which does not state the a polygon is a closed figure and made of straight lines are wrong.

2. Which of the following is a triangle?

a) mathematics-questions-answers-classification-polygons-q2a

b) mathematics-questions-answers-classification-polygons-q2b

c) mathematics-questions-answers-classification-polygons-q2c

d) mathematics-questions-answers-classification-polygons-q2d

Answer: a

Explanation: There are four types of figures in the options, here we need to find which of the following is a triangle. The triangle is a polygon which have three sides which also form three angles with each other.

3. Which of the following can be classified as pentagon?

a) mathematics-questions-answers-classification-polygons-q3a

b) mathematics-questions-answers-classification-polygons-q3b

c) mathematics-questions-answers-classification-polygons-q3c

d) mathematics-questions-answers-classification-polygons-q3d

Answer: d

Explanation: A polygon which has five sides and five angles is called a pentagon. One can find this by a small trick, a student should learn to break the word. For example, Hexagon: Hexa+Gon . As we know that hexa is six, it would be easy for the students to learn the names of the polygons.

4. All parallelograms are square.

a) True

b) False

Answer: b

Explanation: All the squares are special cases of parallelogram. This is because parallelogram is a four-sided polygon with opposite sides parallel to each other. Moreover, a square is four-sided closed figure with all angles measuring 90° and all sides equal. So, the given statement is false.

5. If a polygon has its opposite sides equal and parallel to each other and all angles measuring 90°, identify the polygon.

a) Parallelogram

b) Square

c) Rectangle

d) Rhombus

Answer: b

Explanation: If a rhombus has it’s opposites sides equal and parallel then it can be square or a rhombus but the question also states that the angle between the sides is 90°. Square is a special type of rhombus which has all the angles as 90°.

6. When a square is pushed from one corner keeping the base as it is, which polygon is formed?

a) Square

b) Parallelogram

c) Rhombus

d) Rectangle

Answer: c

Explanation: If we select the upper corner of a square and push it the square will bend in the direction of the force so, the vertical edges would tilt in the direction of the force and this would lead to the formation of new polygon named as rhombus.

7. A seven sided figure is called _______

a) hexagon

b) seven sided polygon

c) heptagon

d) decagon

Answer: c

Explanation: One can find this by a small trick, a student should learn to break the word. For example, Hexagon: Hexa+Gon. As we know that Hexa is six. Similarly, Seven is known as Hepta. ∴Heptagon.

8. A polygon with minimum number of sides is ________

a) triangle

b) line

c) square

d) angle

Answer: a

Explanation: A polygon with minimum numbers of sides is a triangle. A single line does not have one side, an angle has two sides, but it is not closed. A triangle has three sides and its closed. Square has four sides and its closed. So, triangle is the polygon with minimum number of sides.

9. A parallelogram each of whose angles measures 90° is ___________

a) square

b) rectangle

c) parallelogram

d) square as well as Rectangle

Answer: b

Explanation: If a parallelogram has all it’s angle as 90° then it’s a special case of parallelogram and it can be either square or a rectangle.

10. If in a triangle two of it’s angles are 35° and 45° respectively. What is the measure of it’s third angle?

a) 35°

b) 45°

c) 90°

d) 110°

Answer: d

Explanation: In a triangle the sum of all angles is 180°. So, now we can calculate the third angle.

Angle 1 + Angle 2 + Angle 3 = 180°

∴ 35° + 45° + Angle 3 = 180°

∴ Angle 3 = 110°.

11. What is the general form of naming a polygon?

a) Polygon

b) Hexagon

c) n-gon 

d) Pentagon

Answer: c

Explanation: One can find this by a small trick, a student should learn to break the word. For example Hexagon: Hexa+Gon now as we know that hexa is six. Similarly, if a polygon has ‘n’ sides then the polygon would be named as n-gon.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Quadrilaterals – Diagonals”.

1. A/An __________ is a line segment connecting two non-consecutive vertices of a polygon.

a) diagonal

b) side

c) median

d) altitude

Answer: a

Explanation: When a line segment joins two vertices which are not placed besides each other, it is known as a diagonal. There should always be a vertex between the two vertices which join the diagonal. For example, in the below figure, the dotted line represents the diagonal for the square, whereas the solid lines cannot be considered as the diagonals because there is no vertex between them.

mathematics-questions-answers-diagonals-q1

2. If two diagonals of a square intersect, what would be the angle of intersection?

a) 45°

b) 180°

c) 90°

d) 120°

Answer: c

Explanation: The diagonals of any square are perpendicular bisectors of each other. So, the angle between the diagonals would be 90°. The diagram of any square with two diagonals would be like this.

mathematics-questions-answers-diagonals-q2

3. How many diagonals can be drawn to a closed curve?

a) ∞

b) 1

c) 2

d) 0

Answer: d

Explanation: In a closed curve there , there are no vertex. Since there are no vertex in any closed curves, there won’t be any diagonals. Hence, zero diagonals.

mathematics-questions-answers-diagonals-q3

4. Which of the following quadrilaterals has two pairs of adjacent sides equal and diagonals intersecting at right angles?

a) Rhombus

b) Parallelogram

c) Square

d) Kite

Answer: d

Explanation: The quadrilateral which has it’s adjacent sides equal, one can think of two such polygons, they are square and kite. The second condition given is the diagonals intersect at right angles which is true for both square and kite. Among kite and square, kite has adjacent sides equal whereas square has all 4 sides equal, the correct answer would be kite.

5. How many diagonals does a triangle have?

a) ∞

b) 1

c) 3

d) 0

Answer: d

Explanation: Triangle has only three sides, it cannot form any diagonals. As diagonals are formed by figure which have four or more sides. Since it is not possible to draw a diagonal in a triangle ∴ the correct option would be 0.

6. Which of the following is correct?

a) The diagonals of kite bisect each other at 90°

b) The diagonals of kite bisect each other

c) The diagonals of kite make a right angle with each other

d) The diagonals of kite make an angle of 45° with each other

Answer: c

Explanation: A kite is a quadrilateral which has its two adjacent sides equal and its diagonals are perpendicular to each other (i.e. Angle =90°).

7. What type of triangle is formed by the two diagonals of any square and any side of that square?

a) Isosceles Triangle

b) Right Angled Triangle

c) Equilateral Triangle

d) Isosceles Right Angles Triangle

Answer: d

Explanation: The figure shown below is the triangle formed by the diagonals and the side of any square. As we know that the diagonals of any square are perpendicular bisectors of each other. Hence the triangle formed would be isosceles right-angled triangle.

mathematics-questions-answers-diagonals-q7

8. Diameters of a circle can be considered as diagonals.

a) True

b) False

Answer: b

Explanation: Diameters cannot be considered as diagonals because circle does not have any side, hence according to the definition the given statement is false.

9. The side of a square is 6 cm, what would be the square of it’s diagonal’s length?

a) √72

b) 36

c) √36

d) 72

Answer: d

Explanation: As the diagonals of a square are perpendicular bisectors of each other, the form a right angles triangle. Here to find the length of the diagonal could be found by using Pythagoras Theorem.

∴ By Using Pythagoras Theorem we get ,

∴  2 +  2 = x 2

∴ 36+36=x 2

∴ 72=x 2

10. Raj runs in a park which is in a shape of rectangle, he completes one lap and then he runs across the diagonal of the park. The dimensions of the park are 100 m × 120 m. What is the total distance travelled by him diagonally?

a) 156.2

b) 200

c) 1000

d) 75

Answer: a

Explanation: We know that all the angles subtended by the sides of a rectangle are 90° in measure. Therefore we can use Pythagoras theorem to find the distance travelled by Raj diagonally.

 2 +  2 = x 2

X = 156.2 m.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Quadrilaterals – Sum Angle Property”.

1. What is the general formula to count the sum of all the interior angles of a polygon with ‘n’ number of sides?

a) 180°

b) 360°

c) 90°

d) 720°

Answer: a

Explanation: The general formula for the sum total of all the interior angles of any polygon is 180°. For example: As we know that the sum of all the interior angles of any triangle is 180°.

So, now we can recheck it with the above formula. Sum of interior angles =180°

Any triangles have three sides so n=3.

∴ Sum of interior angles =180°

∴ Sum of interior angle =180°.

2. What is the sum of all the angles in any triangle?

a) 90°

b) 180°

c) 270°

d) 240°

Answer: b

Explanation: : As we know that the sum of all the interior angles of any triangle is 180°. Sum of interior angles =180°

Any triangles have three sides so n=3.

∴ Sum of interior angles =180°

∴ Sum of interior angle =180°.

3. If the angles of a four sided polygon are in the ratio 7:8:9:12, then the angles would be?

a) 70, 80, 90, 100

b) 70, 70, 80, 80

c) 120, 70, 80, 100

d) 70, 80, 90, 120

Answer: d

Explanation: Let us assume x to be the common multiple.

Now, the sides of the four sided polygon would be 7x, 8x, 9x and 12x.

The sum total of any four sided polygon is 360°.

Therefore 7x+8x+9x+12x=360

Therefore 36x=360

Therefore x=10°

Therefore the angles are, 7x=70, 8x=80, 9x=90 and 12x=120.

One can recheck their answer by, verifying the total as One can recheck their answer by, verifying the total as 360°.

4. In a triangle the sides are 4 cm, 5 cm and 4 cm. One of the base angle is 80°. Find the measure of the apex angel.

a) 30°

b) 40°

c) 20°

d) 10°

Answer: c

Explanation: One can get the hint that the given triangle is isosceles triangle by observing the length of all the sides. As in an isosceles triangle two sides are of equal length and also the base angles are equal in measure. Let the apex angle for denoted by the variable x.

Therefore 80+80+x=180

Therefore 160+x=180

Therefore x=20.

5. A four sided polygon has all sides equal in length, what would be the angle between two adjacent sides?

a) Exactly 90°

b) 90° or any acute angle

c) 90° or any obtuse angle

d) Can be any angle

Answer: a

Explanation: When a four sided polygon has its all sides equal in length it is a square. Any square has it’s angles =90°. Hence the answer would be 90°. Here the other options won’t be accurate. For example if a student selects the option 90° or any acute angle it would be partially correct because in a square the angles are always right angles.

6. In a quadrilateral the angles are 40°,120°,10° and x. Find x.

a) 40°

b) 120°

c) 190°

d) 180°

Answer: c

Explanation: A quadrilateral is a four sided polygon. Therefore the sum of all the interior angles is 360°.

Therefore 40+120+10+x=360

Therefore 170+x=360

Therefore x=190.

7. What angle is subtended by a semicircle?

a) 360°

b) 300°

c) 180°

d) 90°

Answer: c

Explanation: A circle subtends a whole of 360°, this is the measure of one complete rotation over the circumference of the circle. Now, the semicircle would subtend an angle of 180°.

8. What would be the measure of all the angles in a scalene triangle?

a) 60°, 60° and 60°

b) 120°, 30° and 30°

c) 90°, 45° and 45°

d) 60°, 30° and 90°

Answer: d

Explanation: In a scalene triangle the measure of all the angles and the sides is different. The options other then 60°, 30° and 90° are either equilateral triangles or isosceles triangles.

9. Any right angled triangle must have how may angles in total?

a) Two angles

b) Four angles

c) One angle

d) Three angles

Answer: d

Explanation: In any triangle irrespective of any type of triangle it has three angles in total. Here if the question would ask the ‘how many angles other than right angle’ then the answer would be two angles.

10. The formula 180° is the formula for the sum of exterior angle of ‘n’ sided polygons.

a) True

b) False

Answer: b

Explanation: The formula 180° is the formula for the interior angles of the polygon.

As we know the sum angle property of the square is 360°.

Therefore 180°

Therefore 180°

Therefore 360°.

This set of Mathematics Exam Questions and Answers for Class 8 focuses on “Sum of Measures of the Exterior Angles of a Polygon”.

1. What are exterior angles of any polygon?

a) All angles within the polygon

b) The angles with measure as 90°

c) The angles forming a pair of supplementary angles

d) The angles forming a pair of complementary angles

Answer: c

Explanation: If an extended ray along a side of a polygon form a linear pair with the same side of the polygon then the angle made by the polygon and the extended ray is said to be exterior angle of that polygon. Any polygon having ‘n’ number of sides the sum of all external angels is 360°.

2. The sum total of all the external angles is always ________

a) 180°

b) 180°

c) 360°

d) 360°

Answer: d

Explanation: As we know that the external angles for a supplementary pair of angles with the internal angles. We take any square of any length, as we know that all the angles of a square measure 90° and we also know that they are supplementary.

Therefore 90°+x = 180°

Therefore x = 90°

Similarly the other three external angles also measure the same. Therefore the sum total of all the external angles is 360°.

3. If perimeter of a rectangle is 24 m, then what would be the measure of each of the external angle of that rectangle?

a) 90°

b) 180°

c) 360°

d) 45°

Answer: a

Explanation: We know that the rectangle has all it’s angle as 90°. All the external angles make straight angle with the internal angles. Therefore all the angles would be 90°. The information provided about the perimeter is of no use.

4. Find the value of x.

mathematics-questions-answers-sum-measures-exterior-angles-polygon-q4

a) 160°

b) 140°

c) 120°

d) 150°

Answer: a

Explanation: As we know that sum of all the internal angles of the triangle is 180°.

Therefore x+y+z=180°

Therefore 30°+130°+z=180°

Therefore z=120°

Now we know that the external angle is always a linear angle.

Therefore x+y=180°

Therefore 20°+y=180°

Therefore y=160°.

5. In the given figure find the values of x & y respectively.

mathematics-questions-answers-sum-measures-exterior-angles-polygon-q5

a) 14° & 166°

b) 166° & 14°

c) 170° & 10°

d) 10° & 170°

Answer: b

Explanation: The polygon in the adjoining figure is a hexagon. In order to find the value of x we ought to know the sum of all the internal angles of any hexagon. The formula used is 180°. Therefore we get the sum of all the angles as 720°.

Therefore 84°+126°+149°+x°+104°+91°=720°

Therefore x=166°

Now we use the property of external angles and find the value of y.

Therefore x+y=180°

Therefore y=14°.

6. In any equilateral triangle, find the sum of all the external angles.

a) 180°

b) 120°

c) 60°

d) 360°

Answer: d

Explanation: As we know from our prior knowledge that all the angles in any equilateral triangle the angles measure 60°. Therefore the external angle would be measuring 120° as the pair of internal and external angles is always linear. Since there are three identical external angles, therefore our answer would be 120 ×3=360°.

7. Find the value of x.

mathematics-questions-answers-sum-measures-exterior-angles-polygon-q7

a) 3/140

b) 150/7

c) 140/3

d) 7/150

Answer: c

Explanation: Here we are given a triangle with all it’s internal angles marked in terms of x.

Now we will use the property of triangle.

Therefore °+°+°=180°

Therefore x+20+x+30+x-10=180

Therefore 3x=140

Therefore x=140/3.

8. Find the value of x.

mathematics-questions-answers-sum-measures-exterior-angles-polygon-q8

a) 170/2°

b) 2/170°

c) 170°

d) 70°

Answer:

Explanation: Here we will use the property of external angles.

Therefore °+x=180°

Therefore x-10+x=180°

Therefore 2x=170

Therefore x=170/2.

9. External angle and internal angle of a same polygon form a pair of linear angle together.

a) True

b) False

Answer: a

Explanation: External angle of any polygon can be seen when the side of the polygon is extended and hence the external angle and the internal angle form a pair of linear angle. The sum of these two angles is always equal to 180°. They are also known as straight angles.

10. A polygon can have each of its exterior angles as 70°.

a) True

b) False

Answer: b

Explanation: Each exterior angle of any n-sided polygon=/n

If the exterior angle is 70°,

Then /n = 70

Therefore n=5.1. Here ‘n’ is the number of sides and since the number of sides cannot be any decimal number the answer to the statement is false.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Kinds of Polygons”.

1. How many parallel sides does a trapezium have?

a) 1

b) 2

c) 3

d) All the sides

Answer: b

Explanation: A polygon which has it’s one pair of opposite sides parallel to each other is known as trapezium. So the number of sides parallel would be 2. The option which states that all the sides are parallel can be straight away ruled out because if all the sides are parallel it won’t be forming a closed figure.

2. Which of the following figure is a trapezium?

a) mathematics-questions-answers-kinds-polygons-q2a

b) mathematics-questions-answers-kinds-polygons-q2b

c) mathematics-questions-answers-kinds-polygons-q2c

d) mathematics-questions-answers-kinds-polygons-q2d

Answer: c

Explanation: A polygon which has it’s one pair of opposite sides parallel to each other is known as trapezium. In the given options only one satisfy the conditions for a polygon, others have some or the other condition that is not being satisfied.

3. Raj has A trapezoidal shaped paper. He wants to give equal sized triangles to his friends, how many friends will get the triangle shaped cutting?

a) 1

b) 2

c) 3

d) 4

Answer: c

Explanation: If a student has three identical cutting of a right angled triangle and places it in a specific manner then the formation formed with these triangles is known as trapezium. So, here Raj can give three triangles to his friends.

4. If the non-parallel sides of a trapezium are equal in length then the trapezium is known as _______

a) equal sided trapezium

b) isosceles trapezium

c) equilateral trapezium

d) parallel trapezium

Answer: b

Explanation: If the two non-parallel sides of the trapezium are equal in length then the trapezium is known as Isosceles Trapezium. This is a special type of trapezium.

5. A kite has ________ sides equal.

a) opposite

b) parallel

c) adjacent

d) intersecting

Answer: c

Explanation: Kite is a special type of polygon in which the consecutive  sides of the figure are equal in length and the diagonals are perpendicular but only one of them bisects the other. Therefore here the correct answer would be adjacent.

6. ________ is a special case of parallelogram.

a) Rectangle

b) Square

c) Trapezium

d) Rhombus

Answer: b

Explanation: In a parallelogram all the sides are of equal length and the opposite sides are parallel to each other. This all conditions are satisfied by square and since all the angles measure 90°, therefore it is considered as the special case of parallelogram.

7. Take two set square of 30°-60°-90° measure, if they are placed along their hypotenuse then which type of a polygon is formed?

a) Rectangle

b) Square

c) Parallelogram

d) Trapezium

Answer: a

Explanation: When the set square are placed along their hypotenuse they form a rectangle. The set squares would look like the below figure. On the contrary if the set squares are of measure 45°-45°-90°, then the polygon formed is square.

mathematics-questions-answers-kinds-polygons-q7

8. Which of the following is not a polygon?

a) Square

b) Rectangle

c) Parallelogram

d) Circle

Answer: d

Explanation: Circle is not considered to be a polygon because it is a closed curve whereas the definition says that a ‘n’ sided closed figured containing straight lines is known as polygon. Since the circle does not satisfy all the conditions for a polygon it is not considered as a polygon.

9. __________ type of triangles has two angles equal.

a) Isosceles

b) Scalene

c) Equilateral

d) Any type of triangle

Answer: a

Explanation: As we know that there are three types of triangle

 Equilateral Triangle: All sides are equal in length and all the angles measure 60°

 Isosceles Triangle: Two sides are of equal length and the corresponding base angles would be of equal measure.

 Scalene Triangle: All three sides of the different lengths and all the angles are of different measures.

10. If perimeter of any rectangle is 26 cm and one of the sides measures 7 cm, then the measure of the adjacent side is?

a) 6 m

b) 6 cm

c) 12 m

d) 12 cm

Answer: b

Explanation: Formula for perimeter of is, Perimeter = 2 × 

Therefore 26=2 × 

Therefore 26=14+2Breadth

Therefore 12=2×Breadth

Therefore Breadth=6.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Properties of Polygons-1”.

1. What will be the sum of internal angles of a septagon?

a) 650

b) 700

c) 450

d) 900

Answer: d

Explanation: The sum of the interior angles of a polygon can be calculated using the formula = ( number of sides – 2) × 180

Here, number of sides is 7.

So, the sum of interior angles of a septagon =  × 180 = 5 × 180 = 900

2. What parameter is used to find the area and perimeter of a polygon?

a) Curve

b) Size

c) Length

d) Angle

Answer: c

Explanation: The length of the polygon can be used to find the perimeter and area of a polygon. Since, the perimeter and area are in relation with the sides of the polygon, therefore they play a major role in finding the two parameters.

3. What will be the area of the given figure if the length of one side is 10 cm and the height of the parallelogram is 15 cm?

a) 225 cm 2

b) 1225 cm 2

c) 150 cm 2

d) 75 cm 2

Answer: c

Explanation: The given figure can be divided into two triangles. The total area of the parallelogram will be the sum of the areas of two triangles.

Area of triangle = \(\frac{1}{2}\) × base × height = \(\frac{1}{2}\) × 10 × 15 = 75 cm 2

Area of the parallelogram = 2 × area of triangle = 2 × 75 = 150 cm 2

4. A polygon is 3D shape with straight sides.

a) True

b) False

Answer: b

Explanation: A polygon is a shape constructed using straight sides. It is encompassed within the boundaries of lines in a regular or irregular manner. But a circle or any other curve does not fall under polygon. It is two dimensional in nature.

5. What will be the exterior angles of a hexagon?

a) 7.2°

b) 60°

c) 54°

d) 108°

Answer: b

Explanation: The exterior angles of a polygon can be found using the formula = \(\frac{360}{number \, of \,sides}\).

Since, the number of sides of a hexagon is 6.

Therefore, the exterior angle of a hexagon is = \(\frac{360}{6}\) = 60°

6. What is the circle which connects all the vertices of a polygon called?

a) Semi-circle

b) Incircle

c) Outer circle

d) Circumcircle

Answer: d

Explanation: A circle which connects all the vertices of the triangle is called as a circumcircle. Whereas, if touches the sides of the polygon internally then it is called as incircle.

7. What is a polygon with 50 sides called as?

a) Pentahexagon

b) Pentagon

c) Pentacontagon

d) Pentacontagion

Answer: c

Explanation: A polygon with fifty sides is called as pentacontagon. The sum of the interior angles of a pentacontagon is 8640° and the exterior angle of a pentacontagon is 7.2°.

8. What will be the sum of exterior angle of a regular polygon?

a) 360°

b) 180°

c) 540°

d) 120°

Answer: a

Explanation: The sum of the external angles of a regular polygon is always 360°. When we push all the angles of a convex polygon the angle converges and form a total of 360°. This angle is also called as perigon angle.

9. The circle inscribed inside a polygon is called a circumcircle.

a) False

b) True

Answer: a

Explanation: A circle inscribed in a polygon is called an incircle . A circle circumscribing the polygon, or just touching the vertices of the polygon is called as circumcircle.

10. The radius of the incircle is the apothem of the polygon.

a) True

b) False

Answer: a

Explanation: An apothem is a line from the center of the regular polygon at right angles to any of its sides. This line is also the radius of the incircle. Hence, an apothem is the radius of the incircle of a regular polygon. In the given figure, the line AB is the apothem of the polygon as well as the radius of the incircle.

This set of Mathematics Question Papers for Class 8 focuses on “Properties of Polygons-2”.

1. Circles or any other curve are polygons.

a) True

b) False

Answer: a

Explanation: A polygon is a two-dimensional figure with straight sides. Circles and other curve cannot be termed as polygon as they have no straight sides. Here, in the given figure the trapezium has two parallel sides and two straight lines.

2. Which of the following is not a property of a polygon?

a) Number of sides of shape

b) Angles between the sides of the shape

c) Length of the sides of a shape

d) Curve of the shape

Answer: d

Explanation: The number of sides, its angle between the sides and the length of the sides decide the type and shape of the polygon. For, a pentagon has five sides and internal angle of 108°.

3. A three-sided polygon with all three sides and internal angles different is an isosceles triangle.

a) True

b) False

Answer: b

Explanation: A three-sided polygon is called a triangle. If all the three sides are equal then it is called an equilateral triangle. If two sides of a triangle are equal then it is an isosceles triangle.

4. A triangle with acute internal angles is called an equilateral triangle.

a) True

b) False

Answer: b

Explanation: A triangle with internal acute angle is not always an equilateral triangle. But, an equilateral triangle has equal acute angle. In the given figure, the sides of the triangle are not equal, but it has acute interior angles.

5. Quadrangles are five-sided polygons.

a) False

b) True

Answer: a

Explanation: Quadrangles as the word suggest are four sided polygon. The word quadrangle is split as ‘Quadra’ meaning four, so a quadrangle has four angles and four sides.

6. The internal angle of a three-sided polygon is 360° and that of quadrangle is 180°.

a) True

b) False

Answer: b

Explanation: The sum of internal angle of a three-sided polygon is 180° and that of a quadrangle is 360°; because a quadrangle can be split into two triangles and we know that the sum of interior angles of the triangle is 180. So, two times 180 is 360.

7. Squares, rectangles, rhombuses are all types of parallelogram.

a) False

b) True

Answer: b

Explanation: Parallelogram has two pairs of parallel sides. In case of squares, rectangles and rhombuses the sides are parallel to each other. Hence, they are all types of parallelogram.

8. A ten-sided polygon is called a decagon.

a) True

b) False

Answer: a

Explanation: Deca means ten. So, a 10-sided polygon is called a decagon. The sum of the interior angles of a decagon is 1440° and the exterior angle of a decagon is 44°.

9. A regular polygon has unequal length and unequal angles between them.

a) False

b) True

Answer: b

Explanation: Regular polygons have equal length and equal angles between them. Whereas, irregular polygon has unequal sides and unequal angles between them.

10. Which of the following does not classify under polygon?

a)

b)

c)

d)

Answer: a

Explanation: A polygon is a two dimensional plane figure with straight lines. Figures b, c and d have straight lines and figure ‘a’ has curve. Since curves do not fall under polygon hence figure a is not a polygon.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Special Cases in Polygons”.

1. All the sides of rhombus are _______

a) equal

b) not equal

c) half of the side opposite

d) half of that of adjacent

Answer: a

Explanation: As we know that rhombus is a special type of polygon where all the angles measure acute angle and all the sides of the polygon are equal in length.

2. The diagonals of a rhombus are ______

a) perpendicular to each other

b) perpendicular bisector of each other

c) bisector of each other

d) perpendicular trisector of each other

Answer: b

Explanation: As we know that rhombus is a special type of polygon where all the angles measure acute angle and all the sides of the polygon are equal in length. The diagonals of rhombus have the property that they are perpendicular bisectors of each other i.e. they meet each other at right angle and bisect each other.

3. Rhombus has all the properties of a parallelogram and that of any kite.

a) True

b) False

Answer: a

Explanation: The given statement is true. Rhombus has all the properties of a parallelogram and that of any kite.

4. The diagonals of a rhombus are perpendicular bisectors of one another.

a) True

b) False

Answer: a

Explanation: As we know that rhombus is a special type of polygon where all the angles measure acute angle and all the sides of the polygon are equal in length. The diagonals of rhombus have the property that they are perpendicular bisectors of each other i.e. they meet each other at right angle and bisect each other.

5. What would be the measure of all the angles formed at the center of any rhombus due to its diagonals?

a) 90°

b) 180°

c) 45°

d) Can’t say

Answer: a

Explanation: As we know that rhombus is a special type of polygon, its diagonals are perpendicular bisectors of each other. Hence the measure of all the angles would be 90°.

6. A square is a ______ with equal sides.

a) rectangle

b) kite

c) hexagon

d) trapezium

Answer: a

Explanation: A rectangle has it’s opposite sides equal and all angles are right angles, if all the sides are equal then this figure fulfills all the properties of a square. Hence a square is a rectangle with all sides equal.

7. A square has all properties of which two polygons?

a) Rhombus and Parallelogram

b) Rectangle and Rhombus

c) Parallelogram and Rectangle

d) Circle and Trapezium

Answer: A square has all the properties of parallelogram and rhombus but not rectangle because it’s all the sides aren’t equal. Hence the correct option would be Rhombus and Parallelogram.

8. The triangle formed by the diagonals of a rectangle and its sides are similar by which test?

a) SAS

b) SSS

c) AA

d) Not similar

Answer: a

Explanation: The triangle formed by a side of a rectangle and the diagonals have,

 Side 1 

 Angle 

 Side 2 

Hence the triangles are similar by SAS test of similarity.

9. What is a diagonal?

a) A segment that connects any two nonconsecutive vertices in a polygon.

b) A 90 degree angle.

c) A polygon with four sides.

d) A segment that cuts through a polygon at 45 degrees.

Answer: a

Explanation: A diagonal is a segment which connects alternate sides, here the correct option is ‘A segment that connects any two nonconsecutive vertices in a polygon.’

10. All the squares are rectangles, but all rectangles are not squares.

a) True

b) False

Answer: a

Explanation: Rectangles are four sided closed polygons with opposite sides equal and parallel to each other, but square is a polygon with all sides equal and opposite sides parallel. Therefore the given statement is true and square are considered to be special cases i.e. rectangles with all sides equal.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Geometry – Constructing a Quadrilateral”.

1. Which of the following conditions does not fulfill the condition for constructing a quadrilateral?

a) When four sides and one diagonal are given

b) When two diagonals and three sides are given

c) When two adjacent sides and three angles are given

d) When two adjacent sides and two angles are given

Answer: d

Explanation: If we want to construct quadrilateral we need at least five information about the quadrilateral, but the answer states that two adjacent sides and two angles here we have only four information. So here we cannot construct a quadrilateral.

2. A quadrilateral can be constructed _______ if the lengths of its four sides and a diagonal is given.

a) uniquely

b) complete

c) incomplete

d) can’t be constructed

Answer: a

Explanation: If we want to construct quadrilateral we need at least five information about the quadrilateral. If we have four sides and a diagonal then we can construct a unique quadrilateral.

3. One can construct a unique quadrilateral by the knowledge of any four quantities.

a) True

b) False

Answer: b

Explanation: The basic condition for constructing a quadrilateral is, knowing five or more details about the polygon. So, here the statement would not be correct for constructing a unique quadrilateral.

4. If one desires to construct a unique triangle , how many details are required?

a) 1

b) 2

c) 3

d) 4

Answer: d

Explanation: The basic condition to construct any triangle is having at least four information about the triangle, one cannot construct a unique triangle without that. So, the options other than 4 would be incorrect. Hence 4 is the correct answer.

5. What would be the general formula of number of details to be known, for constructing a quadrilateral?

a) … where n is the number of sides

b) … where n is the number of sides

c) … where n is the number of sides

d) … where n is the number of sides

Answer: b

Explanation: If one wishes to construct a unique quadrilateral, the requirement is  number of details. If any student has n number of details, the constructed quadrilateral would not be a unique quadrilateral.

6. Which of the following is a valid quadrilateral?

a) ▀ABCD ∠A=60,∠B=120,∠C=120,∠D=90

b) ▀PQRS ∠A=90,∠B=90,∠C=90,∠D=90

c) ▀LMNO ∠A=90,∠B=150,∠C=90,∠D=60

d) ▀EFGH ∠A=120,∠B=120,∠C=120,∠D=120

Answer: b

Explanation: The basic property of any quadrilateral is that the sum of all the internal angles is 360°. Here except ▀PQRS the sum of all the internal angles of other quadrilaterals is greater than 360°. Hence all the other quadrilateral other then ▀PQRS are incorrect.

7. Which of the following is an invalid quadrilateral?

a) ▀ABCD ∠A=60,∠B=120,∠C=120,∠D=90

b) ▀PQRS ∠A=90,∠B=90,∠C=90,∠D=90

c) ▀LMNO ∠A=90,∠B=150,∠C=60,∠D=60

d) ▀EFGH ∠A=100,∠B=20,∠C=120,∠D=120

Answer: a

Explanation: The basic property of any quadrilateral is that the sum of all the internal angles is 360°. Here except ▀ABCD the sum of all the internal angles of other quadrilaterals is equal to 360°. Hence all the other quadrilaterals except ▀ABCD are incorrect.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Data Handling – Organising Data”.

1. What is frequency of object?

a) Frequency gives the number of times that an entry occurs

b) Frequency gives the number of times that an entry occurs per unit time

c) Frequency gives the percentage of times that an entry occurs

d) Frequency gives the fraction of times that an entry occurs

Answer: a

Explanation: Frequency gives the number of times that an entry occurs. Raw unstructured data can be difficult to interpret. If the data is organized and frequency determined it can help obtain meaningful information from it.

2. In the below dataset which represents subjects liked by students from class 6, determine the frequency of English subject.

English, Hindi, Hindi, Marathi, Science, English, Science, English, Mathematics, English, English, History, History, Geography, Mathematics

a) 3

b) 4

c) 2

d) 5

Answer: d

Explanation: Frequency gives the number of times that an entry occurs. English subject appears 5 times in the list. Hindi, History, Mathematics and Science appear 2 times. Marathi and Geography appear only 1 time.

3. Which is the highest consumed beverage from tea, coffee and milk in the list given below?

tea, tea, coffee, milk, tea, milk, tea, coffee, tea, coffee, milk, tea

a) milk

b) coffee

c) tea

d) tea and coffee

Answer: c

Explanation: Frequency gives the number of times that an entry occurs. Frequency of tea is 6, frequency of coffee is 3, frequency of milk is 2. Hence, the highest frequency of beverage consumed in the given list if tea.

4. Below is the list of answers given by students as their favorite pet. Determine the frequency of dog.

dog, cat, fish, turtle, cat, rabbit, turtle, dog, cat, dog, cat, dog, dog, turtle, dog, cat, cow, fish, rabbit, dog, cat, dog, cat, cat, dog, turtle, rabbit, cat, fish, dog, turtle

a) 5

b) 7

c) 8

d) 10

Answer: d

Explanation: Frequency gives the number of times that an entry occurs. In the given list, frequency of dog is 10, cat is 9, turtle is 4 and fish is 3.

5. Below is the list of cities nominated as the favorite city by tourists. Determine the city which has frequency 4.

Mumbai, Delhi, Mumbai, Agra, Agra, Chennai, Bengaluru, Mumbai, Agra, Bengaluru, Chennai, Mumbai, Delhi

a) Mumbai

b) Delhi

c) Agra

d) Bengaluru

Answer: a

Explanation: Frequency gives the number of times that an entry occurs. The frequency of Mumbai is 4, frequency of Agra is 3, frequency of Chennai, Bengaluru and Delhi is 2.

6. What is the frequency of dog being the favorite pet based on below frequency distribution table?

Animal Tally Marks

Dog |||| |||

Cat ||

Turtle ||||

Fish |||

a) 4

b) 2

c) 5

d) 8

Answer: d

Explanation: The frequency distribution table gives the number of times an entry occurs in the form of tally marks. The tally marks are grouped in multiples of 5 for easier counting. As seen from the above table, tally marks for dog is 8 i.e. a group of 5 and 3, their sum is 8.

7. Which of the below shows frequency distribution of number of visitors of India, Singapore, Malaysia and Australia?

India, Australia, Singapore, Malaysia, Malaysia, Australia, India, Malaysia, Australia, Singapore, Australia, Australia, Singapore, Singapore, Australia, India, Malaysia, Singapore, Singapore, Malaysia, India

a)

Country Tally Marks

Australia |||| |

India ||||

Malaysia ||||

Singapore |||| |

b)

Country Tally Marks

Australia |||| |||

India ||

Malaysia |||| |

Singapore |||

c)

Country Tally Marks

Australia ||||

India |||

Malaysia ||||

Singapore |

d)

Country Tally Marks

Australia |||| ||||

India ||||

Malaysia |||| ||

Singapore ||||

Answer: a

Explanation: The frequency distribution table gives the number of times an entry occurs in the form of tally marks. The frequency of Australia is 6, India is 4, Malaysia is 5 and Singapore is 6.

8. Below is the list of colors nominated as the favorite colors by students of class I. Determine the color which has frequency 9.

red, blue, red, yellow, green, yellow, green, yellow, yellow, red, green, red, blue, red, yellow, red, blue, yellow, red, blue, blue, yellow, yellow, green, green

a) red

b) blue

c) green

d) yellow

Answer: d

Explanation: Frequency gives the number of times that an entry occurs. The frequency of red is 7, frequency of blue is 5, frequency of yellow is 9 and green is 5.

9. Below is the list of cities nominated as the favorite city by tourists. Which is the correct frequency distribution diagram for favorite city?

Mumbai, Delhi, Agra, Mumbai, Agra, Agra, Chennai, Mumbai, Agra, Delhi, Delhi, Agra, Chennai, Mumbai, Delhi, Delhi, Agra, Delhi, Agra

a)

City Tally Marks

Mumbai |||

Delhi |||| |

Agra |||| ||||

Chennai |||

b)

City Tally Marks

Mumbai ||||

Delhi |||| |

Agra |||| ||

Chennai ||

c)

City Tally Marks

Mumbai ||||

Delhi |||| ||

Agra ||||

Chennai |

d)

City Tally Marks

Mumbai ||||

Delhi ||||

Agra |||| |||

Chennai ||||

Answer: b

Explanation: Frequency gives the number of times that an entry occurs. The frequency of Mumbai is 4, frequency of Agra is 7, Delhi is 6 and Chennai is 2.

10. Below is the inventory of beverages in the store. Which beverage is the highest in quantity in the store?

Coca-Cola, Sprite, Fanta, Pepsi, Fanta, Sprite Fanta, Pepsi, Coca-Cola, Fanta, Coca-Cola, Pepsi, Pepsi, Pepsi, Coca-Cola, Fanta, Pepsi, Sprite

a) Pepsi

b) Fanta

c) Coca-Cola

d) Sprite

Answer: a

Explanation: Frequency gives the number of times that an entry occurs. The frequency of Coca-Cola is 4, frequency of Fanta is 5, frequency of Pepsi is 6 and frequency of Sprite is 3. Hence, quantity of Pepsi is the highest in store.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Data Handling – Grouping Data”.

1. Below table represents marks obtained by students of class I in Mathematics. What is size of class intervals?

mathematics-questions-answers-grouping-data-q1

a) 10

b) 20

c) 5

d) 22

Answer: a

Explanation: Each group of 0-10, 10-20, etc. is called Class Interval. The size of class interval in the given example is 10.

2. Below table represents marks obtained by students of class I in Mathematics. Which class has highest frequency?

mathematics-questions-answers-grouping-data-q1

a) 0-10

b) 10-20

c) 20-30

d) 30-40

Answer: c

Explanation: Frequency gives the number of times that an entry occurs. Maximum number of students are in the class interval 20-30 i.e. 7.

3. Below table represents marks obtained by students of class I in Mathematics. Which class has lowest frequency?

mathematics-questions-answers-grouping-data-q1

a) 0-10

b) 10-20

c) 20-30

d) 40-50

Answer: d

Explanation: Frequency gives the number of times that an entry occurs. Maximum number of students are in the class interval 40-50 i.e. 2.

4. Below table represents marks obtained by students of class I in Mathematics. What is the upper limit of class interval 30-40?

mathematics-questions-answers-grouping-data-q1

a) 20

b) 30

c) 40

d) 10

Answer: c

Explanation: Upper class limit is the highest value in the given class interval. If a value is repeated in lower and upper-class interval, then the value is considered to belong to upper class interval. Here, the upper-class limit of class interval 30-40 is 40.

5. Below table represents marks obtained by students of class I in Mathematics. Which two classes have same frequency?

mathematics-questions-answers-grouping-data-q1

a) 0-10 and 10-20

b) 0-10 and 30-40

c) 10-20 and 30-40

d) 30-40 and 40-50

Answer: b

Explanation: Frequency gives the number of times that an entry occurs. The frequency of 0-10 and 30-40 is same and equal to 4.

6. Below table represents daily income of families from town A. What is size of class intervals?

mathematics-questions-answers-grouping-data-q6

a) 500

b) 1000

c) 1500

d) 2500

Answer: a

Explanation: Each group of 0-500, 500-1000, etc. is called Class Interval. The size of class interval in the given example is 500.

7. Below table represents daily income of families from town A. Which class has highest frequency?

mathematics-questions-answers-grouping-data-q6

a) 0-500

b) 500-1000

c) 1000-1500

d) 1500-2000

Answer: b

Explanation: Frequency gives the number of times that an entry occurs. Maximum number of students are in the class interval 500-1000 i.e. 10.

8. Below table represents daily income of families from town A. Which class has lowest frequency?

mathematics-questions-answers-grouping-data-q6

a) 0-500

b) 500-1000

c) 1000-1500

d) 1500-2000

Answer: d

Explanation: Frequency gives the number of times that an entry occurs. Maximum number of students are in the class interval 1500-2000 i.e. 5.

9. Below table represents daily income of families from town A. What is the upper limit of class interval 1500-2000?

mathematics-questions-answers-grouping-data-q6

a) 500

b) 1500

c) 2500

d) 2000

Answer: d

Explanation: Upper class limit is the highest value in the given class interval. If a value is repeated in lower and upper-class interval, then the value is considered to belong to upper class interval. Here, the upper-class limit of class interval 1500-2000 is 2000.

10. Below table represents daily income of families from town A. Which frequency is repeated?

mathematics-questions-answers-grouping-data-q6

a) 7

b) 5

c) 8

d) 10

Answer: a

Explanation: Frequency gives the number of times that an entry occurs. The frequency of 1000-1500 and 2000-2500 is same and equal to 7.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Data Handling – Bars with a Difference”.

1. Which of the following is not a property of histogram?

a) The bars are of equal width

b) The adjacent bars have no gap

c) The adjacent bars have gap

d) The height of column gives number of data items in the given data range

Answer: c

Explanation: Histogram is represented with bars of equal width with no gap between adjacent bars. The height of the bar represents the number of data items in the given data range called frequency.

2. Below is the histogram of number of people in the age range from 1 – 25 years. How many people are older than 5 and younger than 9 years?

mathematics-questions-answers-bars-difference-q2

a) 11

b) 5

c) 23

d) 24

Answer: a

Explanation: The height of histogram represents the number of items in the given range called as frequency. Here, the number of people older than 5 and younger than 9 years are 11.

3. Below is the histogram of number of people in the age range from 1 – 25 years. How many people are younger than 17 years?

mathematics-questions-answers-bars-difference-q2

a) 16

b) 23

c) 47

d) 63

Answer: d

Explanation: The height of the bar represents the number of data items in the given data range called frequency. The number of people younger than 17 years is sum of number of people between 1-5, 5-9, 9-13 and 13-17 which gives us 63.

4. Below is the histogram of number of people in the age range from 1 – 25 years. How many people are older than 13 years and younger than 25 years?

mathematics-questions-answers-bars-difference-q2

a) 37

b) 30

c) 63

d) 20

Answer: a

Explanation: The height of the bar represents the number of data items in the given data range called frequency. The number of people older than 13 and younger than 25 years is sum of number of people between 13-17, 17-21, 21-25 which gives us 37.

5. Below is the histogram of number of people in the age range from 1 – 25 years. Which group has maximum people?

mathematics-questions-answers-bars-difference-q2

a) 13-17

b) 9-13

c) 21-25

d) 17-21

Answer: a

Explanation: The height of the bar represents the number of data items in the given data range called frequency. Maximum frequency in the given histogram is 24 which is for the range 13-17. So maximum people are older than 13 and younger than 17 years.

6. Below is the histogram of number of people in the age range from 1 – 25 years. How many people are represented by the given histogram?

mathematics-questions-answers-bars-difference-q2

a) 37

b) 63

c) 72

d) 76

Answer: d

Explanation: The height of the bar represents the number of data items in the given data range called frequency. The number of people represented by the histogram is sum of frequency of bars in the given histogram. Hence, we get number of people represented by histogram as 76.

7. Below is the histogram of number of hours spent by students on mobile phones. Which group has maximum students?

mathematics-questions-answers-bars-difference-q7

a) 1-2

b) 2-3

c) 3-4

d) 4-5

Answer: a

Explanation: The height of the bar represents the number of data items in the given data range called frequency. Maximum students lie in the category of spending greater than 1 hour and less than 2 hours on mobile phones.

8. Below is the histogram of number of hours spent by students on mobile phones. How many students spend more than 4 hours on mobile phone?

mathematics-questions-answers-bars-difference-q7

a) 9

b) 1

c) 31

d) 12

Answer: b

Explanation: The height of the bar represents the number of data items in the given data range called frequency. The number of students spending more than 4 hours on mobile phone is 1.

9. Below is the histogram of number of hours spent by students on mobile phones. How many students spend more than 2 hours and less than 5 hours on mobile phone?

mathematics-questions-answers-bars-difference-q7

a) 31

b) 43

c) 22

d) 21

Answer: c

Explanation: The height of the bar represents the number of data items in the given data range called frequency. The number of students spending more than 2 hours and less than 5 hours on mobile phone is sum of frequency of 2-3, 3-4 and 4-5 which is 22.

10. Below is the histogram of number of hours spent by students on mobile phones. How many students are represented in the given histogram?

mathematics-questions-answers-bars-difference-q7

a) 43

b) 53

c) 10

d) 21

Answer: b

Explanation: The height of the bar represents the number of data items in the given data range called frequency. The total students represented by given histogram is sum of frequency of 1-2, 2-3, 3-4 and 4-5 which gives us 53.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Data Handling – Circle Graph or Pie Chart”.

1. Which graph represents relation between whole and its parts?

a) Histogram

b) Pie graph

c) Line graph

d) Stacked bar graph

Answer: b

Explanation: The correct answer is pie graph as pie graph represents relationship between whole and its parts. Histogram is represented with bars of equal width with no gap between adjacent bars. The height of the bar represents the number of data items in the given data range called frequency. Line graph is a line joining all the data points. A stacked bar graph is a chart that uses bars to show comparisons between categories of data with ability to break down and compare parts of a whole.

2. Below is the pie graph showing sales of store in each quarter. Which quarter has the highest sales?

mathematics-questions-answers-circle-graph-pie-chart-q2

a) 1 st quarter

b) 2 nd quarter

c) 3 rd quarter

d) 4 th quarter

Answer: b

Explanation: Pie graph shows relation between whole and its parts. Sales of 2 nd quarter is 40% of total sales of store. Sales of 2 nd quarter is higher than sales of other three quarters.

3. Below is the pie graph showing sales of store in each quarter. Which quarter has the lowest sales?

mathematics-questions-answers-circle-graph-pie-chart-q2

a) 1 st quarter

b) 2 nd quarter

c) 3 rd quarter

d) 4 th quarter

Answer: c

Explanation: Pie graph shows relation between whole and its parts. Sales of 3 rd quarter is 16% of total sales of store. Sales of 3 rd quarter is lower than sales of other three quarters.

4. Below is the pie graph showing mode of transport used by students of a class. Which is the most heavily used mode of transport?

mathematics-questions-answers-circle-graph-pie-chart-q4

a) Bike

b) Train

c) Bus

d) Walk

Answer: b

Explanation: Pie graph shows relation between whole and its parts. 37% of class uses train as a mode of transport. Percentages of students using other mode of transport is lower than those using train.

5. Below is the pie graph showing mode of transport used by students of a class. Which is the least used mode of transport?

mathematics-questions-answers-circle-graph-pie-chart-q4

a) Bike

b) Train

c) Bus

d) Walk

Answer: a

Explanation: Pie graph shows relation between whole and its parts. 11% of class uses bike as a mode of transport. Percentages of students using other mode of transport is higher than those using bike.

6. Below is the pie graph showing mode of transport used by students of a class. 22% of class use which mode of transport?

mathematics-questions-answers-circle-graph-pie-chart-q4

a) Bike

b) Train

c) Bus

d) Walk

Answer: c

Explanation: Pie graph shows relation between whole and its parts. According to the pie chart, 22% of class uses bus as a mode of transport. 37% use train, 30% use walk and 11% use bike as a mode of transport.

7. Below is the pie graph showing mode of transport used by students of a class. Which mode of transport is used by more than 25% of class?

mathematics-questions-answers-circle-graph-pie-chart-q4

a) Train and bike

b) Bus and walk

c) Bike and walk

d) Walk and train

Answer: d

Explanation: Pie graph shows relation between whole and its parts. According to the pie chart, 22% uses bus, 37% use train, 30% use walk and 11% use bike as a mode of transport. Thus, walk and train is used by more than 25% of class as mode of transport.

8. Below is the pie graph showing mother tongue of people in a society. Which is the most spoken mother tongue of that society?

mathematics-questions-answers-circle-graph-pie-chart-q8

a) Hindi

b) Marathi

c) Gujarati

d) Telugu

Answer: c

Explanation: Pie graph shows relation between whole and its parts. According to the pie chart, 38% of people in the society speak gujarati as their mother tongue. Percentage of people speaking other languages are lesser than 38%. Hence, gujarati is the most spoken mother tongue of the society.

9. Below is the pie graph showing mother tongue of people in a society. Which is the least spoken mother tongue of that society?

mathematics-questions-answers-circle-graph-pie-chart-q8

a) Hindi

b) Marathi

c) Gujarati

d) Telugu

Answer: d

Explanation: Pie graph shows relation between whole and its parts. According to the pie chart, 13% of people in the society speak telugu as their mother tongue. Percentage of people speaking other languages are greater than 13%. Hence, telugu is the least spoken mother tongue of the society.

10. Below is the pie graph showing mother tongue of people in a society. Which language is spoken as mother tongue by 20% of people of that society?

mathematics-questions-answers-circle-graph-pie-chart-q8

a) Hindi

b) Marathi

c) Gujarati

d) Telugu

Answer: c

Explanation: Pie graph shows relation between whole and its parts. According to the pie chart, 20% of people in the society speak hindi as their mother tongue. 13% speak telugu, 29% speak marathi and 38% speak gujarati as mother tongue.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Data Handling – Drawing a Pie Chart”.

1. What is the total angle at the center of pie chart?

a) 360°

b) 320°

c) 180°

d) 90°

Answer: a

Explanation: Pie chart is basically a circle. Total angle at the center of a circle is 360°. The central angle of the sectors will be a fraction of 360°.

2. Below is the pie graph showing sales of different items in first quarter of the year. Which item has the highest sales?

mathematics-questions-answers-drawing-pie-chart-q2

a) Food

b) Clothing

c) Electronics

d) Jewellery

Answer: b

Explanation: Pie graph shows relation between whole and its parts. Sales of clothing is 40% of total sales of store. Sales of clothing is higher than sales of other three items.

3. Below is the pie graph showing sales of different items in first quarter of the year. Which item has the lowest sales?

mathematics-questions-answers-drawing-pie-chart-q2

a) Food

b) Clothing

c) Electronics

d) Jewellery

Answer: c

Explanation: Pie graph shows relation between whole and its parts. Sales of electronics is 16% of total sales of store. Sales of electronics is lower than sales of other three items.

4. Below is the pie graph showing sales of different items in first quarter of the year. If total sales of company are Rs 600000, how much is the sales of food in Rs.?

mathematics-questions-answers-drawing-pie-chart-q2

a) 96000

b) 120000

c) 240000

d) 144000

Answer: b

Explanation: Pie graph shows relation between whole and its parts. 20% of sales of company is from food department. Total sales of company is Rs. 600000. Thus, sales of food department is Rs. 120000.

5. Below is the pie graph showing sales of different items in first quarter of the year. If total sales of company are Rs 1000000, how much is the sales of electronics in Rs.?

mathematics-questions-answers-drawing-pie-chart-q2

a) 160000

b) 200000

c) 400000

d) 240000

Answer: a

Explanation: Pie graph shows relation between whole and its parts. 16% of sales of company is from electronics department. Total sales of company is Rs. 1000000. Thus, sales of electronics department is Rs. 160000.

6. Below is the pie graph showing colors picked by students of a class. 22% of students picked which color?

mathematics-questions-answers-drawing-pie-chart-q6

a) Red

b) Blue

c) Green

d) Yellow

Answer: c

Explanation: Pie graph shows relation between whole and its parts. According to the pie chart, 22% of students picked green color. 37% picked blue, 30% picked yellow and 11% picked red color.

7. Below is the pie graph showing colors picked by students of a class. If total students of class are 200, how many students picked red color?

mathematics-questions-answers-drawing-pie-chart-q6

a) Red

b) Blue

c) Green

d) Yellow

Answer: c

Explanation: Pie graph shows relation between whole and its parts. According to the pie chart, 11% of students picked red color. Total number of students of a class is 200. Thus, number of students who picked red color is 22.

8. Below is the pie graph showing colors picked by students of a class. Number of students who picked blue color are 111. How many students were there in that class?

mathematics-questions-answers-drawing-pie-chart-q6

a) 200

b) 300

c) 100

d) 400

Answer: b

Explanation: Pie graph shows relation between whole and its parts. According to the pie chart, 37% of students picked blue color. Total number of students who picked blue color is 111. Thus, total number of students of that class is 300.

9. Below is the pie graph showing mother tongue of people in a society. What percent of people in society speak Hindi?

mathematics-questions-answers-drawing-pie-chart-q9

a) 18%

b) 17%

c) 26%

d) 39%

Answer: b

Explanation: Pie graph shows relation between whole and its parts. According to the pie chart, 17% of people in the society speak hindi as their mother tongue.

10 Below is the pie graph showing mother tongue of people in a society. Total number of people in society is 1500. How many people speak Marathi?

mathematics-questions-answers-drawing-pie-chart-q9

a) 255

b) 270

c) 390

d) 585

Answer: d

Explanation: Pie graph shows relation between whole and its parts. According to the pie chart, 39% of people in the society speak marathi as their mother tongue. Total number of people in the society is 1500. Thus, 585 people speak marathi as their mother tongue in the society.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Data Handling – Chance and Probability”.

1. How many outcomes can be obtained by tossing a coin?

a) 1

b) 2

c) 3

d) 4

Answer: b

Explanation: A coin has two sides i.e. head and tail. When a coin is tossed, we can get either head or tail. Thus, there are 2 outcomes when a coin is tossed.

2. How many outcomes can be obtained by tossing 2 coins?

a) 1

b) 2

c) 3

d) 4

Answer: d

Explanation: A coin has two sides i.e. head  and tail . When a coin is tossed, we can get either head or tail. When 2 coins are tossed, the possible outcomes are HH, HT, TH, TT. Thus, there are 4 outcomes when 2 coins are tossed.

3. How many outcomes can be obtained by tossing 3 coins?

a) 1

b) 2

c) 8

d) 4

Answer: c

Explanation: A coin has two sides i.e. head  and tail . When a coin is tossed, we can get either head or tail. When 3 coins are tossed, the possible outcomes are HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Thus, there are 8 outcomes when 3 coins are tossed.

4. How many outcomes can be obtained by rolling a die?

a) 6

b) 2

c) 3

d) 4

Answer: a

Explanation: A die has 6 faces. The faces are numbered 1, 2, 3, 4, 5 and 6. When a die is rolled, there are 6 outcomes.

5. How many outcomes can be obtained by rolling 2 dice?

a) 6

b) 2

c) 3

d) 4

Answer: a

Explanation: A dice has 6 faces. The faces are numbered 1, 2, 3, 4, 5 and 6. When a dice is rolled, there are 6 outcomes. When 2 dice are rolled, possible outcomes are 36 as below:

     

     

     

     

     

     .

6. A bag has 5 balls. One of each color red, blue, green, yellow and pink. What are the possible outcomes of a ball picked from bag?

a) 6

b) 2

c) 3

d) 5

Answer: d

Explanation: A bag has 5 balls. One of each color red, blue, green, yellow and pink. When a ball is picked from bag, it can be either of the 5 balls. So possible outcome of picking ball from bag is 5.

7. Which of the below represents possible outcomes when 2 coins are tossed? Here, head is represented by H and tail is represented by T.

a) HH, HT, TH, TT

b) HH, HH, HT, HH

c) HT, TH, TT, TH

d) HH, HH, TT, TT

Answer: a

Explanation: A coin has two sides i.e. head  and tail . When a coin is tossed, we can get either head or tail. When 2 coins are tossed, the possible outcomes are HH, HT, TH, TT. Thus, there are 4 outcomes when 2 coins are tossed.

8. How many outcomes are possible when drawing a card from deck of cards?

a) 13

b) 52

c) 26

d) 1

Answer: b

Explanation: Deck of cards contains 52 cards. There are 13 cards of hearts, spades, clubs and diamonds. The 13 cards are A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K.

9. What are the possible outcomes of spinning below wheel?

mathematics-questions-answers-chance-probability-q9

a) 1

b) 5

c) 4

d) 8

Answer: c

Explanation: On spinning the wheel, one of the four sections can be selected. Hence, the number of outcomes of spinning the wheel are 4.

10. Which option represents outcome of rolling a die correctly?

a) 1, 2, 3

b) 1, 2, 3, 4, 5, 6

c) 3, 4, 5, 6

d) 1, 3, 4

Answer: b

Explanation: A die has 6 faces. The faces are numbered 1, 2, 3, 4, 5 and 6. When a die is rolled, there are 6 outcomes.

This set of Mathematics Assessment Questions for Schools focuses on “Data Handling – Getting a Result”.

1. Which statement is incorrect for outcomes when 2 coins are tossed?

a) The outcome of first coin doesn’t affect the outcome of second coin

b) The outcome of second coin doesn’t affect the outcome of first coin

c) If outcome of first coin is head then the outcome of second coin cannot be head

d) If outcome of first coin is head then the outcome of second coin can also be head

Answer: c

Explanation: A coin has two sides i.e. head and tail. When a coin is tossed, we can get either head or tail. The outcome of first toss doesn’t affect the outcome of second toss.

2. If 2 dice are rolled, which of the below outcomes is not possible when outcome is sum of 2 dice rolls?

a) 12

b) 1

c) 36

d) 72

Answer: b

Explanation: A dice has 6 faces. The faces are numbered 1, 2, 3, 4, 5 and 6. When a dice is rolled, there are 6 outcomes. When 2 dice are rolled, possible outcomes are 36 as below:

     

     

     

     

     

     .

Thus, the sum of 2 dice can be a number between 2 and 72. It can never be 1.

3. Which of the below is an invalid outcome when 3 coins are tossed?

a) TTTT

b) THT

c) HHT

d) HHH

Answer: a

Explanation: A coin has two sides i.e. head  and tail . When a coin is tossed, we can get either head or tail. When 3 coins are tossed, the possible outcomes are HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Option TTTT is possible only when 4 coins are tossed.

4. An unfair dice is rolled where the number on the faces are 1, 2, 3, 5, 5, 6. How many outcomes are possible?

a) 4

b) 3

c) 2

d) 5

Answer: d

Explanation: A die has 6 faces. The faces are numbered 1, 2, 3, 5, 5 and 6. When a die is rolled, there are 5 outcomes possible. Outcome 4 is not possible as none of the faces are marked as 4.

5. Two players toss a coin one after the other. Which option is the only correct option?

a) The chances of first player getting head is higher than chances of second player

b) The chances of second player getting head is higher than chances of first player

c) If player one gets head then player two cannot get head

d) Player one and two have equal chances of getting head

Answer: d

Explanation: A coin has two sides i.e. head  and tail . When a coin is tossed, we can get either head or tail. Player one and player two are independent of each other. They cannot influence outcome of each other.

6. A bag has 6 balls; two of each color red, blue, green. What are the possible outcomes of a ball picked from bag?

a) 6

b) 2

c) 3

d) 5

Answer: c

Explanation: A bag has 6 balls; two of each color red, blue, green. When a ball is picked from bag, it can be of red, blue or green. So possible outcome of picking ball from bag is 3.

7. Two cards are picked from well shuffled deck of 52 cards. The cards picked are of red color. Which option is the correct possibility of type of card?

a) Spade, Club

b) Heart, Club

c) Club, Diamond

d) Heart, Diamond

Answer: d

Explanation: Deck of cards contains 52 cards. There are 13 cards of hearts, spades, clubs and diamonds. The 13 cards are A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K. Heart and diamond cards are red cards and clubs and spades are black card.

8. How many outcomes are possible when drawing a queen card from deck of cards?

a) 13

b) 52

c) 4

d) 1

Answer: c

Explanation: Deck of cards contains 52 cards. There are 13 cards of hearts, spades, clubs and diamonds. The 13 cards are A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K. Queen card can belong to heart, spade, club or diamond. Hence, 4 outcomes are possible when drawing a queen card from deck of cards.

9. A bag of fruits contains apple, orange, lemons and tangerines. How many outcomes are possible to pick a fruit from bag at random?

a) 1

b) 5

c) 4

d) 8

Answer: c

Explanation: A bag contains four types of fruits apple, orange, lemons and tangerines. The outcome of picking a fruit at random from the bag is picking either of the four fruits. Hence, 4 is the correct answer.

10. What is the minimum of sum of rolling 3 dice?

a) 1

b) 2

c) 3

d) 4

Answer: c

Explanation: A die has 6 faces. The faces are numbered 1, 2, 3, 4, 5 and 6. When a die is rolled, there are 6 outcomes. The minimum of rolling a die is 1. Hence, when three dice are rolled the minimum of sum possible is 3.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Data Handling – Equally Likely Outcomes”.

1. What is the probability of getting head when a coin is tossed?

a) \

 

 \

 

 \

 

 \(\frac{3}{2}\)

Answer: a

Explanation: A coin has two sides i.e. head and tail. When a coin is tossed, we can get either head or tail. The likelihood of getting either head or tail is equal. Hence, the probability of getting head when a coin is tossed is \(\frac{1}{2}\).

2. If 2 coins are tossed, what is the probability of getting heads on both coins?

a) \

 

 \

 

 \

 

 \(\frac{3}{2}\)

Answer: c

Explanation: A coin has two sides i.e. head and tail. When a coin is tossed, we can get either head or tail. The likelihood of getting either head or tail is equal. Hence, the probability of getting head when a coin is tossed is \(\frac{1}{2}\). The probability of getting head on both the coins is \(\frac{1}{4}\).

3. If 2 coins are tossed, what is the probability of getting different outcomes on both coins?

a) \

 

 \

 

 \

 

 \(\frac{3}{2}\)

Answer: a

Explanation: A coin has two sides i.e. head and tail. When a coin is tossed, we can get either head or tail. The likelihood of getting either head or tail is equal. We want to head in one coin and tail in other coin. The four possible outcomes are HH, TH, HT, TT. Hence the probability of getting different outcomes on both coins is \(\frac{1}{2}\).

4. What is the probability of getting 6 when a die is rolled?

a) \

 

 \

 

 \

 

 \(\frac{1}{6}\)

Answer: d

Explanation: A die has 6 faces. The faces are numbered 1, 2, 3, 4, 5 and 6. When a die is rolled, there are 6 outcomes possible. Probability of getting 6 when a die is rolled is \(\frac{1}{6}\).

5. What is the probability of getting an even number when a die is rolled?

a) \

 

 \

 

 \

 

 \(\frac{1}{6}\)

Answer: b

Explanation: A die has 6 faces. The faces are numbered 1, 2, 3, 4, 5 and 6. When a die is rolled, there are 6 outcomes possible. Outcome of getting even number when a die is rolled is 3. Hence, probability of getting even number when a die is rolled is \(\frac{1}{3}\).

6. A bag has 6 balls; two of each color red, blue, green. What is the probability of picking a red ball from bag?

a) \

 

 \

 

 \

 

 \(\frac{1}{6}\)

Answer: b

Explanation: A bag has 6 balls; two of each color red, blue, green. When a ball is picked from bag, it can be of red, blue or green. So possible outcome of picking ball from bag is 3. Hence, the probability of picking a red ball from the bag is \(\frac{1}{3}\).

7. A card is picked from well shuffled deck of 52 cards. What is the probability of picking a red card?

a) \

 

 \

 

 \

 

 \(\frac{1}{26}\)

Answer: a

Explanation: Deck of cards contains 52 cards. There are 13 cards of hearts, spades, clubs and diamonds. The 13 cards are A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K. Heart and diamond cards are red cards and clubs and spades are black card. Hence, we have 26 red cards and 26 black cards in a deck. The probability of picking a red card is \(\frac{1}{2}\).

8. What is the probability of drawing a queen card from deck of cards?

a) \

 

 \

 

 \

 

 \(\frac{1}{26}\)

Answer: b

Explanation: Deck of cards contains 52 cards. There are 13 cards of hearts, spades, clubs and diamonds. The 13 cards are A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K. Queen card can belong to heart, spade, club or diamond. Hence, 4 outcomes are possible when drawing a queen card from deck of cards. Probability of picking a queen from deck of cards is \(\frac{1}{13}\).

9. A bag of fruits contains apple, orange, lemons and tangerines. What is the probability of picking up an apple?

a) \

 

 \

 

 \

 

 \(\frac{1}{3}\)

Answer: c

Explanation: A bag contains four types of fruits apple, orange, lemons and tangerines. The outcome of picking a fruit at random from the bag is picking either of the four fruits. Hence, 4 is the correct answer. Probability of picking an apple from the bag is \(\frac{1}{4}\).

10. What is the probability of picking a spade card from deck of cards?

a) \

 

 \

 

 \

 

 \(\frac{1}{3}\)

Answer: c

Explanation: Deck of cards contains 52 cards. There are 13 cards of hearts, spades, clubs and diamonds. The 13 cards are A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K. The probability of picking a spade card from deck of cards is \(\frac{1}{4}\).

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Data Handling – Outcomes as Events”.

1. What is the probability of getting 2 heads and 1 tail when 3 coins are tossed?

a) \

 

 \

 

 \

 

 \(\frac{3}{2}\)

Answer: c

Explanation: A coin has two sides i.e. head and tail. When a coin is tossed, we can get either head or tail. The likelihood of getting either head or tail is equal. The outcomes are HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Hence, the probability of getting 2 heads and 1 tail when 3 coins are tossed is \(\frac{3}{8}\).

2. What is the probability of getting a prime number when a die is rolled?

a) \

 

 \

 

 \

 

 \(\frac{3}{2}\)

Answer: a

Explanation: A die has 6 faces which are numbered 1, 2, 3, 4, 5 and 6. Out of these 2, 3 and 5 are prime numbers. \(\frac{1}{2}\).

3. A bag has 3 white and 2 black balls. What is the probability of picking a black ball from bag?

a) \

 

 \

 

 \

 

 \(\frac{2}{5}\)

Answer: d

Explanation: A bag has 3 white and 2 black balls. Picking a black ball means any of the 2 black balls can be picked. Thus, the probability of picking a black from 5 balls in the bag is \(\frac{2}{5}\).

4. What is the probability of getting 4 when a die is rolled?

a) \

 

 \

 

 \

 

 \(\frac{1}{6}\)

Answer: d

Explanation: A die has 6 faces. The faces are numbered 1, 2, 3, 4, 5 and 6. When a die is rolled, there are 6 outcomes possible. Probability of getting 4 when a die is rolled is \(\frac{1}{6}\).

5. What is the probability of getting an odd number when a die is rolled?

a) \

 

 \

 

 \

 

 \(\frac{1}{6}\)

Answer: b

Explanation: A die has 6 faces. The faces are numbered 1, 2, 3, 4, 5 and 6. When a die is rolled, there are 6 outcomes possible. Outcome of getting odd number when a die is rolled is 3. Hence, probability of getting odd number when a die is rolled is \(\frac{1}{3}\).

6. A bag has 7 balls; 2 red, 3 blue and 2 green. What is the probability of picking a red ball from bag?

a) \

 

 \

 

 \

 

 \(\frac{1}{6}\)

Answer: b

Explanation: A bag has 7 balls; 2 red, 3 blue and 2 green. When a ball is picked from bag, it can be of red, blue or green. There are 2 red balls. Hence, the probability of picking a red ball from the bag is \(\frac{2}{7}\).

7. A card is picked from well shuffled deck of 52 cards. What is the probability of picking a card of diamond?

a) \

 

 \

 

 \

 

 \(\frac{1}{26}\)

Answer: b

Explanation: Deck of cards contains 52 cards. There are 13 cards of hearts, spades, clubs and diamonds. The 13 cards are A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K. The probability of picking a card of diamond is \(\frac{1}{4}\).

8. What is the probability of drawing a king card from deck of cards?

a) \

 

 \

 

 \

 

 \(\frac{1}{26}\)

Answer: b

Explanation: Deck of cards contains 52 cards. There are 13 cards of hearts, spades, clubs and diamonds. The 13 cards are A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K. King card can belong to heart, spade, club or diamond. Hence, 4 outcomes are possible when drawing a king card from deck of cards. Probability of picking a king from deck of cards is \(\frac{1}{13}\).

9. A bag of fruits contains 2 apples, 3 orange, 4 lemons and 2 tangerines. What is the probability of picking up an apple?

a) \

 

 \

 

 \

 

 \(\frac{1}{3}\)

Answer: c

Explanation: A bag contains four types of fruits apple, orange, lemons and tangerines. The outcome of picking a fruit at random from the bag is picking either of the 11 pieces. Hence, 4 is the correct answer. Probability of picking an apple from the bag is \(\frac{2}{11}\).

10. What is the probability of picking a face card from deck of cards?

a) \

 

 \

 

 \

 

 \(\frac{13}{52}\)

Answer: a

Explanation: Deck of cards contains 52 cards. There are 13 cards of hearts, spades, clubs and diamonds. The 13 cards are A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K. There are 16 face cards. The probability of picking a face card from deck of cards is \(\frac{16}{52}\).

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Data Handling – Life Related Probability”.

1. When a die is thrown, what is the probability of getting a number greater than 5?

a) \

 

 \

 

 \

 

 \(\frac{3}{2}\)

Answer: c

Explanation: A die has 6 faces which are numbered 1, 2, 3, 4, 5 and 6. The only number greater than 5 is 6. The probability of getting a number greater than 5 when a die is rolled is \(\frac{1}{6}\).

2. What is the probability of getting product of number greater than 25 when 2 dice are rolled?

a) \

 

 \

 

 \

 

 \(\frac{3}{2}\)

Answer: c

Explanation: A die has 6 faces which are numbered 1, 2, 3, 4, 5 and 6. When 2 dice are rolled, the possible outcomes are:

     

     

     

     

     

     .

The three combinations ,  and  on multiplication have value greater than 25. Hence, the probability is \(\frac{1}{12}\).

3. A bag has 5 red, 3 white and 2 black balls. What is the probability of picking a black or red ball from bag?

a) \

 

 \

 

 \

 

 \(\frac{2}{15}\)

Answer: b

Explanation: A bag has 5 red, 3 white and 2 black balls. Picking a black or red ball means we can pick any of the 7 balls. Thus, the probability of picking a black or red ball is \(\frac{7}{13}\).

4. What is the probability of getting non-prime number when a die is rolled?

a) \

 

 \

 

 \

 

 \(\frac{1}{6}\)

Answer: b

Explanation: A die has 6 faces. The faces are numbered 1, 2, 3, 4, 5 and 6. 1,4 and 6 are non-prime numbers. Probability of getting non-prime number when a die is rolled is \(\frac{1}{2}\).

5. What is the probability of getting all heads when 4 coins are tossed?

a) \

 

 \

 

 \

 

 \(\frac{1}{16}\)

Answer: d

Explanation: A coin has two sides i.e. head and tail. When a coin is tossed, we can get either head or tail. There are 16 possibilities when 4 coins are tossed. The probability of getting all heads is \(\frac{1}{16}\).

6. A bag has 23 balls; 12 red, 5 blue and 6 green. What is the probability of picking a green ball from bag?

a) \

 

 \

 

 \

 

 \(\frac{1}{23}\)

Answer: b

Explanation: A bag has 23 balls; 12 red, 5 blue and 6 green. When a ball is picked from bag, it can be of red, blue or green. So possible outcome of picking green ball from bag is 6. Hence, the probability of picking a green ball from the bag is \(\frac{6}{23}\).

7. A card is picked from well shuffled deck of 52 cards. What is the probability of picking a club card?

a) \

 

 \

 

 \

 

 \(\frac{1}{26}\)

Answer: c

Explanation: Deck of cards contains 52 cards. There are 13 cards of hearts, spades, clubs and diamonds. The 13 cards are A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K. The probability of picking a club card is \(\frac{1}{4}\).

8. What is the probability of drawing a card of value greater than 5 and lesser than 9 from deck of cards?

a) \

 

 \

 

 \

 

 \(\frac{1}{26}\)

Answer: b

Explanation: Deck of cards contains 52 cards. There are 13 cards of hearts, spades, clubs and diamonds. The 13 cards are A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K. Card greater in value than 5 and lesser than 9 are 6, 7 and 8 and can belong to heart, spade, club or diamond. Hence, 12 outcomes are possible when drawing a card from deck of cards. Probability of picking a card greater in value than 5 and lesser than 9 from deck of cards is \(\frac{12}{52}\).

9. What is the probability of getting 2 tails and 1 head when 3 coins are tossed?

a) \

 

 \

 

 \

 

 \(\frac{1}{3}\)

Answer: c

Explanation: A coin has two sides i.e. head and tail. When a coin is tossed, we can get either head or tail.

The possible outcome when 3 coins are tossed are HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. The probability of getting 2 tails and 1 head is \(\frac{3}{8}\)

10. What is the probability of picking an ace of hearts card from deck of cards?

a) \

 

 \

 

 \

 

 \(\frac{13}{52}\)

Answer: a

Explanation: Deck of cards contains 52 cards. There are 13 cards of hearts, spades, clubs and diamonds. The 13 cards are A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K. There is one ace of heart card. The probability of picking an ace of heart card from deck of cards is \(\frac{1}{52}\).

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Properties of Squares”.

1. ________ is the square of 25.

a) 625

b) 525

c) 125

d) 655

Answer: a

Explanation: A square of any number is the product which is obtained by multiplying the number with itself. So, here 25×25=625. Hence the answer would be 625, and the others would be incorrect.

2. Which of the following is a triangular number?

a) 2

b) 4

c) 5

d) 6

Answer: d

Explanation: Triangular numbers are the numbers which when arranged in increasing order forms a shape of triangle, the triangle shown represents the number 6. Hence the only number forming the triangular number is 6. Hence the options other than 6 are incorrect.

mathematics-questions-answers-properties-squares-q2

3. Which of the following is not a triangular number?

a) 1

b) 10

c) 15

d) 20

Answer: d

Explanation: Triangular numbers are the numbers which when arranged in increasing order forms a shape of triangle. Here the options other than 20 form triangular numbers. Hence the correct option would be 20, since it does not form a triangular number.

4. If a number has 1 in it’s one’s place, what can be it’s square?

a) 91

b) 144

c) 169

d) 121

Answer: d

Explanation: We know that if a number has 1 in it’s one’s place the square of that particular number would also have 1 in it’s one’s place. Here there are two options that can be correct but the option 91 couldn’t be correct because it isn’t a perfect square. Therefore the correct option would be 121.

5. Which of the following cannot be a square of the number of with 4 at it’s one’s place?

a) 4

b) 196

c) 36

d) 144

Answer: b

Explanation: We know that the number which has 4 in it’s one’s place has it’s ending with 6 in it’s one’s place. Here 36 cannot be the correct answer since it is square of 6. Here the correct answer is 196 as it is square of 14.

6. If we add two triangular numbers what would be the result?

a) Square

b) Square Root

c) Cube

d) Cube Root

Answer: a

Explanation: If we add two triangular numbers then the sum would be a perfect square.

For example: 1 and 3 are triangular number when added to each other gives 4, 4 is a perfect square.

7. There are _____ non-square numbers between square of 5 and 6.

a) 11

b) 12

c) 13

d) 10

Answer: d

Explanation: The non-square numbers are the number which aren’t perfect squares, therefore the numbers between the square of 5 and 6  is 10 the numbers are 26, 27, 28, 29, 30, 31, 32, 33, 34 and 35.

8. If we add the first n odd numbers we get ______

a) n 2

b) 2n

c) 3n

d) n

Answer: a

Explanation: When we add first n odd numbers we get, n 2 .

For example: the first 5 odd numbers are 1, 3, 5, 7 and 9

When we add them, we get, 1+3+5+7+9+=25. We know that 25 is the square of number 5. Hence, we find that when first n numbers are added we get n 2 .

9. How can one express 144 in terms of squares?

a) 12 2 -1

b) 144-1

c) 13 2 -1

d) 142+1

Answer: a

Explanation: Here all the options other then 13 2 -1 give 143 as their answer but only 12 2 -1 gives

it in the form of squares. Hence the correct answer would be 12 2 -1.

10. If square of 11 is 121 then, what is the square of 111?

a) 121

b) 12321

c) 1234321

d) 123321

Answer: b

Explanation: The squares of the numbers containing one on all places show a very beautiful pattern.

The pattern is as follows, 11 2 =121 => 111 2 =12321. Hence students can use this pattern to calculate the squares of all the numbers having 1 on it’s all places.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Patterns in Squares”.

1. What are triangular numbers?

a) The numbers whose dot pattern can be arranged in triangles

b) The numbers which form a triangle on adding

c) The numbers which have three digits

d) The numbers which do not give perfect squares on adding

Answer: a

Explanation: Triangular numbers are the numbers which when arranged in increasing order forms a shape of triangle, the triangle shown below represents the number 6.

mathematics-questions-answers-some-interesting-patterns-q1

2. _____ is a triangular number.

a) 3

b) 2

c) 5

d) 7

Answer: a

Explanation: Triangular numbers are the numbers which when arranged in increasing order forms shape of triangle. Here, 3 is the only triangular number. Hence, 3 is the correct answer and others are incorrect.

3. If we combine two consecutive triangular numbers, we get?

a) Rational Number

b) Whole Number

c) Perfect Square

d) Prime Number

Answer: c

Explanation: If we combine two consecutive triangular numbers, we get a square number, like

1 + 3 = 4 = 2 2 Hence, we get a perfect square when we add two consecutive triangular numbers. Here options other than Perfect Square are incorrect.

4. There are ______ non-squares numbers in between 5 2 & 6 2 .

a) 10

b) 11

c) 12

d) 9

Answer: a

Explanation: The squares of the numbers 5 & 6 are 25 & 36 respectively. There are 10 numbers between 25 and 36 . Therefore, the correct answer would be 10 and the others would be incorrect.

5. ________ is the general formula to find the number of non-square numbers in between two consecutive squares.

a) n 2 +2n-1

b) n 2 -1

c) 2n+1

d) n+1

Answer: c

Explanation: If we want to find the general formula for the number of non-square numbers in between two consecutive squares, we assume to the natural numbers to be n & n+1

We square and subtract the smaller from the greater number,

 2 -n 2 =(n 2 +2n+1)-n 2 =2n+1

6. If we add first n odd numbers, we get ______

a) n

b) n-1

c) n 2

d) n 2 -1

Answer: c

Explanation: If we add first n odd numbers we get n 2 . For example

1+3+5=9=3 2 , here we have added first three odd numbers and we get 3 2 .

Hence, n 2 would be the correct answer and the other options would be incorrect.

7. What would be the square of 1111?

a) 1234321

b) 12321

c) 121

d) 1

Answer: a

Explanation: The squares of all the numbers with only 1 as its digit in all its places have a special pattern.

If we need to find the square of number 11, we can write 121. Similarly, square of number 111 is 12321. Therefore, we have the answer without calculating this huge number.

8. If we add first n numbers, we get ______

a) \

 

 n-1

c) n 2

d) n 2 -1

Answer: a

Explanation: If we add first n numbers we get \(\frac{n }{2}\). For example, if we add 1 + 2 + 3 + 4 + 5, we get 15.

Here, we added first 5 numbers. Here, \(\frac{n }{2} = \frac{5 }{2} \)= 15.

Hence, \(\frac{n }{2}\) would be the correct answer and the other options would be incorrect.

9. There are ______ non-squares numbers in between 9 2 & 10 2 .

a) 15

b) 11

c) 19

d) 9

Answer: c

Explanation: The squares of the numbers 9 & 10 are 81 & 100 respectively. There are 19 numbers between 81 and 100. Therefore, the correct answer would be 19 and the others would be incorrect.

10. _____ is not a triangular number.

a) 7

b) 6

c) 10

d) 3

Answer: a

Explanation: Triangular numbers are the numbers which when arranged in increasing order forms shape of triangle. Here, 3, 6, 10 are triangular numbers. In the given options, 7 is not a triangular number.

11. Observe the following pattern and supply the missing numbers.

112 = 1 2 1

1012 = 1 0 2 0 1

101012 = 102030201

10101012 = ___________

a) 10101

b) 102030101

c) 1020304030202

d) 1020304030201

Answer: d

Explanation: We know that 11 2 = 121. In 101, we have a 0 between the two 1. There we need to put 0 between 121 making it 10201. In 10101, we have 010 between the two 1. So, we need to put 010 between 121 making it 102030201. Similarly, for 1010101 2 we get 1020304030201.

12. Observe the following pattern and supply the missing numbers.

112 = 1 2 1

1012 = 1 0 2 0 1

10012 = 1002001

100012= ___________

a) 10101

b) 1002001

c) 100020001

d) 1002003002001

Answer: c

Explanation: We know that 11 2 = 121. In 101, we have a 0 between the two 1. There we need to put 0 between 121 making it 10201. In 1001, we have 00 between the two 1. So, we need to put 00 between 121 making it 1002001. Similarly, for 10001 2 we get 100020001.

13. What would be the square of 111?

a) 1234321

b) 12321

c) 121

d) 1

Answer: b

Explanation: The squares of all the numbers with only 1 as its digit in all its places have a special pattern.

If we need to find the square of number 11, we can write 121. Similarly, square of number 111 is 12321. Therefore, we have the answer without calculating this huge number.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Finding the Square of a Number”.

1. What can be general formula to find square of any number?

a) Misplaced & 2

b) a 2

c) a

d) b

Answer: a

Explanation: The general formula of finding square of any number is Misplaced & 2 . Option a 2 cannot be correct because if we need to find square of number such as 23 the method using the general formula Misplaced & 2 would be more easier. Hence the correct option is Misplaced & 2 .

2. How can one represent the square of 103?

a)  2

b)  2

c) 

d) 

Answer: a

Explanation: The number 103 can be represented in form 100+3, this will help us to calculate the square of the number. This method would help us eliminate the traditional method of multiplying the number with itself.

3. 89 2 = ____

a)  2

b) 90 2

c) 1 2

d) 9 2

Answer: a

Explanation: If we have to find the square of the number which is near 0 we use the second variation i.e.  2 . This would help us and make the calculations easier.

Therefore  2 = 90 2 -2×90×1+1 2

Therefore  2 = 8100-180+1

Therefore  2 =7921.

4. If a student uses  2 to calculate the square of a number, then the number is _______

a) 121

b) 112

c) 199

d) 201

Answer: c

Explanation: A student uses this variant of the formula for the numbers which are near 0. Here there is only one number which is close to 0 and hence 199 is the correct answer. The other options also can be solved using this method but that would be complicated.

5. Calculate the square of 201.

a) 40401

b) 39393

c) 40426

d) 41042

Answer: a

Explanation: In the multiple choice question student can eliminate three options by mere observation. The student should notice that option 41042 contains 2 in its unit place so it cannot be a square of number ending with 1. Similarly, the other two options 39393 & 40426 ending with 3 and 6 whereas square of numbers ending with 1 have only 1 in its unit place.

6. 34 2 =_____

a) 1156

b) 1256

c) 1356

d)1456

Answer: a

Explanation: In order to calculate the square of 34 we use the formula,

 2 , where we consider a=30 and b=4

Therefore  2 = 30 2 +2×30×4+4 2

Therefore  2 = 900+240+16

Therefore  2 = 1156.

7. How many numbers lie between the squares of 12 and 13?

a) 25

b) 26

c) 24

d) 28

Answer: a

Explanation: If we need to find the number of non-square numbers, we can use the formula .

Here when we apply this formula we get, =25. Hence the answer would be 25 and the other options would be incorrect.

8. Express the square of 11 in terms of sum of odd numbers.

a) 1+3+5+7+9+11+13+15+17+19+21

b) 1+2+3+5+7+9+11+13+15+17+19

c) 1+3+5+7+9+11+13+15+17+21+23

d) 1+2+3+4+5+6+7+8+9+10+11+12

Answer: a

Explanation: The correct option is the one with the first 11 odd numbers as the sum of first n odd numbers gives n 2 . Hence the correct answer is the one with the first 11 odd number.

9. Find the square of 25.

a) 625

b) 525

c) 635

d) 615

Answer: a

Explanation: There is a beautiful pattern followed by all the numbers ending with 5. We can get the square of any number ending with 5 without actually calculating it. All the numbers ending with 5 shows 25 at the end of their squares, and the other places can be filled by multiplying the next consecutive number. For example: 25 2 = 25 i.e.625 .

10. Find the square of 225.

a) 505625

b) 49625

c) 50625

d) 51625

Answer: c

Explanation: There is a beautiful pattern followed by all the numbers ending with 5. We can get the square of any number ending with 5 without actually calculating it. All the numbers ending with 5 shows 25 at the end of their squares, and the other places can be filled by multiplying the next consecutive number. For example: 225 2 = 25 i.e.50625 .

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Patterns in Squares of Numbers”.

1. What kind of figure does the number 15 show in triangular representation?

a) Triangular form with base 5

b) Triangular form with base 4

c) Triangular form with base 3

d) Triangular form with peak 1

Answer: a

Explanation: Triangular numbers are the numbers which form a triangle when arranged in increasing form. When the number 15 is arranged in increasing form the base of the triangle has 5 dots, which goes on decreasing and reaches the peak upto 1.

2. When we add fourth and fifth triangular number we get ______

a) 5 2

b) 6 2

c) 25 2

d) 25 3

Answer: a

Explanation: The fourth and fifth triangular numbers are 10 & 15.

10 + 15 = 25 = 5 2 . Therefore 5 2 is the correct answer and the other options are incorrect. The options should be read carefully as the options are very close.

3. There are _____ non-square numbers between 65 2 & 66 2 .

a) 130

b) 65

c) 131

d) 129

Answer: c

Explanation: In these types of questions we can use the shortcut in order to reach the answer quickly. We can use the formula . When we use this formula we consider one of the numbers as n and the hence the second number is. Here n=65.

Therefore when we substitute n=65 in the formula we get,

[2×+1]=131.

4. In some cases, there can be less than two non-square numbers between two square numbers.

a) True

b) False

Answer: a

Explanation: There are two cases where there are less than two non-square numbers between two square numbers, the cases are as follows.

Case 1:- 0 2 & 1 2 Here there is not a single number between squares.

Case 2:- 1 2 & 2 2 Here there are two numbers between the squares.

Hence the given statement is true and cannot be false.

5. We get the square of _____ number if we add first 11 odd numbers.

a) 11

b) 12

c) 5

d) 13

Answer: a

Explanation: There is an interesting pattern in squares, when we add first n odd numbers we get n 2 . So here when we add the first eleven odd numbers, we get 11 2 . Hence the correct answer would be 11 2 and the other options would be incorrect.

6. On subtracting first 25 odd numbers from first 30 odd numbers we get _____

a) 5 2

b) 375

c) 5

d) 275

Answer: c

Explanation: If we add the first 25 odd numbers, we get 625 (i.e.25 2 ) and when we add first 30 odd numbers, we get 900 (i.e.30 2 ). When we calculate the difference between the two, we get 275. Hence 275 is the correct answer and the others are incorrect.

7. A square is formed by sum of two _____________

a) Natural numbers

b) Consecutive Natural numbers

c) Alternate Natural numbers

d) Consecutive Whole numbers

Answer: b

Explanation: When we add two consecutive natural numbers, we get a square . For example: 3 2 =4+5=9.

8. Which is the lower consecutive natural number forming the square 13?

a) 85

b) 84

c) 83

d) 82

Answer: b

Explanation: The square of number 13 is formed by two consecutive natural numbers and they are 84 & 85. If we need to find the lower number  we can calculate like this, \(\frac{13^2-1}{2}=\frac{169-1}{2}=\frac{168}{2}\)=84.

9. ×=_______

a) a 2 -1

b) a 2 +1

c) a 2

d) a 2 +2

Answer: a

Explanation: We have ×. Opening the brackets, we get,

×=(a 2 +a-a-1)=(a 2 -1). Hence, we conclude this × to be the correct answer.

10. What would be the square of 11111?

a) 123454321

b) 1234321

c) 12321

d) 121

Answer: a

Explanation: The numbers which have 1 in all the digits show a beautiful pattern.

For example: 11 2 =121….111 2 =12321 and this goes on. This can be very helpful in finding squares.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Pythagorean Triplet”.

1. Which of the following is the first Pythagorean triplet?

a) 3, 4 & 5

b) 12, 13 & 15

c) 6, 8 & 10

d) 1, 2 & 3

Answer: a

Explanation: Pythagorean triples are the numbers that follow a certain pattern. The pattern is that the square of the biggest amongst three is the sum of square of other two. This condition is satisfied by only one option i.e. 3, 4 & 5, hence this is the correct option and the others are incorrect.

2. In a right angled triangle if one side forming the right angle is 6 and the hypotenuse is 10. What is the length of the other right angle forming side?

a) 8

b) 10

c) 12

d) 6

Answer: a

Explanation: In a right angle triangle we can use Pythagoras theorem. This theorem states that the square of the hypotenuse is equal to the sum of squares of the other two right angle forming sides. Let us assume the one of the sides of the right angled triangle be x.

∴ 6 2 +x 2 =10 2 ,

∴ x 2 =100-36,

∴ x=8.

3. Pythagoras theorem can only be applied on ________ triangles.

a) equilateral

b) isosceles

c) right angled

d) isosceles right angled

Answer: c

Explanation: Pythagoras theorem can only be used on right angles triangles. As a result only the option stating right angled triangles would be correct and the others would be incorrect.

4. 3, 4 & 5 are not a Pythagorean triplet.

a) True

b) False

Answer: b

Explanation: If we consider 3 2 +4 2 to be L.H.S. and 5 2 to be R.H.S. we get,

L.H.S.= 3 2 +4 2 ,

L.H.S.=25.

Now, we take R.H.S.

R.H.S.=5 2 =25.

Hence L.H.S.=R.H.S.

Hence the given triplet is a Pythagorean triplet a

5. 3, 4 & 6 is a Pythagorean triplet.

a) True

b) False

Answer: b

Explanation: If we consider 3 2 +4 2 to be L.H.S. and 6 2 to be R.H.S. we get,

L.H.S.= 3 2 +4 2 ,

L.H.S.=25.

Now, we take R.H.S.

R.H.S.= 6 2 =36.

L.H.S.≠R.H.S.

6. Which of the following is not a Pythagorean triplet?

a) 12, 16, 20

b) 21, 28, 35

c) 33, 44, 55

d) 4, 5, 6

Answer: d

Explanation: The option which has 4, 5, 6 is not a Pythagorean triplet, and the other options are the multiples of it.

7. In the below figure, ABC is a right angled triangle and the lengths AB = 12, BC = 5, What would be the length of AC?

mathematics-questions-answers-pythagorean-triplet-q7

a) 13

b) 12

c) 5

d) 169

Answer: a

Explanation: Here the given triangle is a right angled triangle, we can use Pythagoras theorem. Theorem will help us find the length of side AC.

Using Pythagoras Theorem we get,

12 2 +5 2 = AC 2 ,

AC=13.

8. Which of the following is a Pythagorean triplet?

a) 10, 6 & 8

b) 13, 14 & 15

c) 7, 8 & 10

d) 5, 2 & 3

Answer: a

Explanation: Pythagorean triples are the numbers that follow a certain pattern. The pattern is that the square of the biggest amongst three is the sum of square of other two. This condition is satisfied by only one option i.e. 10, 6 & 8, hence this is the correct option and the others are incorrect.

9. In a right angle ∆ABC, AB=12, BC=5 and Angle ABC=90°. Find AC.

a) 169

b) 13

c) 17

d) 14

Answer: b

Explanation: This triangle is a right angled triangle so we can use Pythagoras theorem. Therefore, we get

∴ AC 2 = AB 2 + BC 2

∴ AC 2 = 12 2 + 5 2

∴ AC 2 = 144 + 25

∴ AC 2 = 169

∴ AC = 13.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Square Root”.

1. If we are given the area of a square and we need to find the length of one side, what operation should we carry?

a) Square

b) Cube

c) Take Square root

d) Take Cube root

Answer: c

Explanation: The formula of area of a square,

Area= 2

So, in order to find the side where area is given we take square roots on both the sides. Therefore square root would be the right answer and the other options would be incorrect.

2. If area of a square is 144 cm 2 , then what is the side?

a) 12 cm

b) \ 12 m

d) \(\sqrt{144}\) mm

Answer: a

Explanation: The formula of area of a square,

Area= 2

So, in order to find the side where area is given we take square roots on both the sides.

Here, 144=x 2

∴x=12 cm.

3. While taking square roots we cannot include _____ numbers.

a) prime

b) natural

c) whole

d) negative

Answer: d

Explanation: While taking square roots we have to take care that we cannot have real roots for negative numbers. The roots of negative numbers are imaginary. Hence the correct option would be negative and the others are wrong.

4. What is the square root of 4?

a) -2

b) -3

c) 4

d) -4

Answer: a

Explanation: Here when we calculate the root of 4 we get two values, they are 2 & -2. We have only one of the roots as the answer. So, -2 would be the correct answer and others would be incorrect.

5. There are always two real roots of any number.

a) True

b) False

Answer: a

Explanation: Yes, there are more than two roots for every number but only two of them can be real others belong to the imaginary category. Therefore the given statement is true.

6. What are the prime factors of 16?

a) 2×4×2

b) 2×2×2×2

c) 8×2

d) 16×1

Answer: b

Explanation: Prime factors are the multiplication of all the prime numbers in order to receive the desired number at the end. Here we have the number to be factorized as 16=2 4 . Therefore the answer would 2 4 .

7. 324= _____________

a) 2×2×9×9

b) 2 2 ×3 4

c) 4×3×27

d) 12×27

Answer: b

Explanation: Prime factors are the multiplication of all the prime numbers in order to receive the desired number at the end. The prime factors of 324 are 2×2×3×3×3×3 (i.e.2 2 ×3 4 ). Therefore the answer would be 2×2×3×3×3×3 (i.e.2 2 ×3 4 ).

8. What would be the square root of 961?

a) 31

b) 32

c) 39

d) 36

Answer: a

Explanation: Since the square is ending with 1 its root has to end with either 1 or 9. So two options are ruled out and the other two are 31 & 39. When we calculate the square root we find that the square root of 961 is 31. Hence 31 is the correct answer and all others are incorrect.

9. These 2×3×7 are prime factors of some number, find the square of that particular number.

a) 1764

b) 1664

c) 1864

d) 1964

Answer: a

Explanation: 2 × 3 × 7 = 42. Now we need to find 42 2 . When we add first 42 odd numbers we get, 1764. Therefore the correct answer is 1764 and the others are incorrect.

10. \Missing open brace for subscript 21

b) 19

c) 11

d) 29

Answer: b

Explanation: If we need to find the square root of number 361 we can subtract first 19 odd numbers which would give us 0. From this we can could that 19=\(\sqrt{361}\). Hence 19 is the correct answer and the others are incorrect.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Finding Square Roots”.

1. Square root is _________ function of square.

a) converse

b) inverse

c) polynomial

d) linear

Answer: b

Explanation: Square root is inverse function of square. The statement can be derived by having a look on the graphs of both square and square root. The nature of graph of square is parabolic and that of square root is rectangular hyperbola. The other method here can be by observing the power which is 2 in case of square and \(\frac{1}{2}\) in case of square root.

2. If  2 =4 then, √4 = ________

a) -2

b) 2

c) 4

d) -4

Answer: b

Explanation: Here the correct answer is 2 and the others are incorrect. The options other then 2 cannot be correct because √, this symbol is used to indicate that the square roots are positive.

3. Find square root of 169.

a) 13

b) 14

c) -14

d) 12

Answer: a

Explanation: Here we can simply find our answer by taking square root of the number \(\sqrt{169}\)=13. There can be some smart ways to find the answer without calculating the roots. A square of number ending with 9 is a square of a number ending with 3. There is only one option ending with 3. Therefore the correct answer is 13 and the others are incorrect.

4. What is the square roots of a three digit number ending with one?

a) 1

b) 11

c) 121

d) 9

Answer: b

Explanation: If a three digit number which is ending with 1 the square root would be ending with 1 or 9. Here all the options either have 9 or 1 at the ending. The option which has 1 cannot be the answer since the square wouldn’t be a three digit number. Similarly with the options 121 and 9 their square won’t be three digit numbers either. So, the correct answer will be 11, the square of 11 is a three digit number and ends with 1.

5. Square root of 25 is 4.

a) True

b) False

Answer: b

Explanation: The given statement is mathematically wrong as square root of 25 is 5 and not 4. Hence the correct option would be false. Here is the statement would be that Square root of 16 is 4 then the statement would be true.

6. Square root of 16 is ________

a) -4

b) 4

c) 16

d) -16

Answer: b

Explanation: Since square roots cannot be negative the options stating -4 &-16 are ruled out. Now we are left with two more options from which the 16 cannot be the answer as the square root of any number cannot be that number itself . So the only answer left is 4. Hence 4 is the correct answer.

7. Find the square root of 961.

a) 31

b) 39

c) 29

d) 21

Answer: a

Explanation: Here the trick of checking the last digit of square and square roots fails because if any square has 1 in one’s place the square root can have either 1 or 9 in its ones place. So here we calculate the square root. After calculating we get the square root as 31.

8. If a perfect square ends with 1 then the square root can be ending with ___ & ___

a) 9 & 2

b) 9 & 1

c) 1 & 8

d) 1 & 2

Answer: b

Explanation: When a perfect square ends with 1 its roots can be either ending with 1 or ending with 9. Hence the options other the 9 & 1 are incorrect.

9. Match the following.

Column A Column B 

i. 4 a. 4

ii. 16 b. 3

iii. 1 c. 2

iv. 9 d. 1

Choose the correct match.

a) -; -; -; -

b) -; -; -; -

c) -; -; -; -

d) -; -; -; -

Answer: a

Explanation: Here we need be careful while choosing the correct option as the options may look similar but are not the same. We need to select squares from column A and their appropriate roots in column B. Here the correct ordered sequence of answer is -; -; -; - and the other options are incorrect.

10. Find the root of 12321.

a) 11

b) 111

c) 1111

d) 11111

Answer: b

Explanation: This question is basically based on the pattern that is shown by the squares of numbers containing 1 in all the digits. So the square root of the number 12321 would be 111 according to the pattern. Therefore the option containing 111 and all the other options would be incorrect.

This set of Mathematics Exam Questions for Schools focuses on “Finding Square Roots through Repeated Subtraction”.

1. Find the square root of 25 by repeated subtraction method.

a) 5

b) 4

c) 6

d) 2

Answer: a

Explanation: The sum of the first n odd natural numbers is n 2 . Therefore we subtract odd numbers from the square till we get 0.

Therefore  25-1=24;  24-3=21;  21-5=16;  16-7=9;  9-9=0

Since we get 0 while subtracting the fifth odd number we know that the given number is square of 5.

2. If we have to subtract 9 odd numbers from a perfect square in order to obtain 0 then the perfect square is _______

a) 9

b) 3

c) 81

d) 18

Answer: c

Explanation: The sum of the first n odd natural numbers is n 2 . Now using this property of the square we can conclude that of we have subtracted nine odd numbers from the square in order to obtain zero then the number is square of number 9. As we know that square of number 9 is 81, 81 is the correct answer and the others are incorrect.

3. How many times do we need to subtract odd numbers from the square of 31 in order to obtain 0?

a) 31

b) 0

c) 32

d) 30

Answer: a

Explanation: The sum of the first n odd natural numbers is n 2 . Keeping this property in mind we can conclude that the answer for our question is 31. As we subtract odd numbers 31 times from square of 31 we get zero.

4. What are prime factors of 120?

a) 2 × 3 × 5 × 2 × 2

b) 2 × 5 × 5 × 3 × 2

c) 12 × 10

d) 4 × 3 × 10

Answer: a

Explanation: Here there are multiple options which give 120 on multiplying the numbers but we need to find the prime factors. The option which has only prime numbers and gives the product as 120 is the correct option. The option obtaining 2 × 3 × 5 × 2 × 2 satisfies both the conditions. Hence this option is the correct answer and the others are incorrect.

5. Find the prime factors of square of 5.

a) 5 × 5

b) 6 × 5

c) 3 × 5

d) 6 × 3

Answer: a

Explanation: If the questions ask to find the prime factors of square of prime numbers then the prime factors would be the number multiplied by the number itself. Here the square of 5 is 25. Therefore the prime factors are 5×5 and rest of the options are wrong.

6. Adding the first n odd numbers we get ____

a) n

b) n 2

c) √n

d) n 2 -1

Answer: b

Explanation: The sum of the first n odd natural numbers is n 2 . Hence the correct option would be n 2 the other options cannot be correct. This is due to the property of the squares.

7. Find prime factors of 123.

a) 1 × 2 × 3

b) 1 × 62 × 2

c) 3 × 41

d) 3 × 2 × 20

Answer: c

Explanation: The prime factors of the number 123 are 3 × 41. There cannot be any other combination because the number 123 has only three factors and the factors are 1, 3 & 41.

8. If we subtract the first 13 odd numbers from 169 then the number obtained is ____

a) 0

b) 13

c) 169

d) -13

Answer: a

Explanation: The sum of the first n odd natural numbers is n 2 .

∴ 169 – 1 – 3 – 5 – 7 – 9 – 11 – 13 – 15 – 17 – 19 – 21 – 23 – 25 = 0

Therefore when we subtract first 13 odd numbers from 169 which is also known as 13 2 we obtain 0.

9. Find prime factors of 1331.

a) 10 × 12 × 13

b) 11 × 12 × 11

c) 11 × 11 × 11

d) 3 × 2 × 20

Answer: c

Explanation: The prime factors of the number 1331 are 11 × 11 × 11. There cannot be any other combination because the number 1331 has only three factors and the factors are 11 × 11 × 11.

10. Find prime factors of 69.

a) 3 × 23

b) 3 × 5 × 11

c) 3 × 9 × 11

d) 3 × 2 × 20

Answer: a

Explanation: The prime factors of the number 69 are 3 × 23. There cannot be any other combination because the number 69 has only factors and the factors are 3 × 23.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Square Roots by Prime Factorisation”.

1. What are the prime factors of 144?

a) 12×12

b) 12×2×6

c) 2×3×2×3×2×2

d) 3×2×3×2×3×2

Answer: c

Explanation: Prime factors of a number are the product of some prime numbers. Here we need to find the prime factors for 144 so we find the prime factors to be 2×3×2×3×2×2. Therefore this is the only correct option and the others are incorrect.

2. Find the prime factors of 1000.

a) 2×5×2×5×5×5

b) 2×5×2×5×2×5

c) 2×2×5×2×5×2

d) 4×5×5×2×5

Answer: b

Explanation: Prime factors of a number are the product of some prime numbers. Here we need to find the prime factors for 1000 so we find the prime factors to be 2×5×2×5×2×5. Therefore this is the only correct option and the others are incorrect.

3. Which of the prime factors would be correct for square of 9?

a) 3×3×9

b) 3 4

c) 9×9

d) 8×1

Answer: b

Explanation: To find prime factor for a square of any number we can find the prime factors of that number itself first and then write those prime factors twice. Here we need to find the prime factors of 9 2 =81, now we find the prime factors of 9, we get 3×3=3 2 . Now we square the prime factors, therefore we get 3 4 . Therefore the correct answer is 3 4 and the others are incorrect.

4. Find the square root of 225by using prime factors.

a) 3×5

b) 5×5

c) 3×3

d) 15

Answer: a

Explanation: Here we need to find the square root of 225 which is 15. Now we need to find the prime factors of 15 and they are 3×5. Therefore the correct answer is 3×5 and the others are incorrect.

5. Prime factors of 12 are _______

a) 3×4

b) 3×2

c) 3×2×2

d) 4×2×4

Answer: c

Explanation: Prime factors of a number are the product of some prime numbers. Here we need to find the prime factors for 12 so we find the prime factors to be 2×2×3. Therefore this is the only correct option and the others are incorrect.

6. Two different numbers can have same prime factors.

a) True

b) False

Answer: b

Explanation: No two different numbers can have same prime factors as we know that product of two numbers cannot be same. Therefore the above-given statement is incorrect and hence the answer would be false.

7. Square root of 324 is ______

a) 18

b) 12

c) 22

d) 24

Answer: a

Explanation: Here in order to find the square root of 324, we find the prime factors of 324.

Prime factors of 324=3×3×3×3×2×2

Now we combine the same numbers in pairs,

Prime factors of 324=3 2 ×3 2 ×2 2

Now taking roots on both the sides we get.

Prime factors of \(\sqrt{324}\)=3×3×2…..i.e.18

Therefore the correct answer is 18 and the others are incorrect.

8. Prime factors of 41 are ____

a) 1×41

b) 4×1

c) 1×4×2

d) 2×2×1

Answer: a

Explanation: Here we need to find the prime factors of 41 but from our prior knowledge we know that 41 is a prime number. So the prime factors are 41×1. Hence we get our answer 41×1. Hence we conclude that the correct answer is 41×1 and the others are incorrect.

9. What are the prime factors of prime numbers?

a) The prime number itself

b) Any number depending on the prime number

c) Prime number and the number itself

d) There cannot be any prime factors

Answer: c

Explanation: Prime numbers can be factorized by only two numbers which are 1 and the prime number itself. So, there can only be one answer which would satisfy the condition to our question. Therefore the correct answer is Prime number and the number itself and the other options are incorrect.

10. What are prime numbers?

a) A number which has one and only one factor

b) A with only 1 as factor

c) A number which has 1 and the number itself as the factors

d) All numbers are prime numbers

Answer: c

Explanation: Prime numbers are the numbers which have only two factors which are 1 and the number itself. Therefore only one option satisfies the condition of being a prime number. Hence the correct answer is a number which has 1 and the number itself as the factors and the other are incorrect.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Square Root by Division Method”.

1. If a perfect square is 144 then its root would lie in the range _____

a) 10 and 13

b) 13 and 15

c) 5 and 10

d) 12 and 16

Answer: a

Explanation: If we need to find the square root of 144 we consider the smallest three digit square number, which is 100 i.e. 10 2 , and a bigger square number than 144 is 169 i.e. 13 2 . Therefore we conclude that the range of the square root is 10 and 13.

2. Which is the smallest three digit square number?

a) 100

b) 101

c) 81

d) 121

Answer: a

Explanation: Here we have three options which are three digit number and one option is two digit. Hence the two digit option is ruled out. Now we consider the remaining three digit numbers. We observe that 100 is the smallest three digit number.

3. ______ is the greatest 2 digits perfect square.

a) 1

b) 4

c) 99

d) 81

Answer: d

Explanation: When we observe the two digit perfect squares, we find that the greatest perfect square is 81. Hence 81 is the correct answer and the other options are incorrect.

4. Biggest three digit perfect square is _____

a) 30 2

b) 999

c) 31

d) 31 2

Answer: d

Explanation: The greatest three digit perfect square is 961 which is 31 2 . Therefore the answer would be 31 2 and the other options would be incorrect.

5. Find the smallest four two digit perfect square.

a) 1

b) 4

c) 16

d) 11

Answer: c

Explanation: The smallest two digit perfect square is 16 i.e. 4 2 . Therefore the correct answer to the question is 16 and all the other options are incorrect.

6. From the following perfect square, which of the following is the greatest square?

a) 121

b) 122

c) 444

d) 400

Answer: d

Explanation: Here the options given are in the mixed form two of them are perfect squares and the other two are non-square numbers. Here the greatest square is 400 and therefore the correct answer is 400 i.e. 20 2 and the others are incorrect.

7. Which of the following is a square of a number ending with 3?

a) 139

b) 169

c) 89

d) 99

Answer: b

Explanation: There is one clue with which with we can find whether the number is a square of number ending with 3, we need to check that the square of that particular number ends with 9. Here all the options are ending with 9 but only 169=13 2 .

8. Prime factors of 444 are ______

a) 2×2×3×37

b) 37×2×3×5

c) 3×7×2×5×4

d) 41×5×7×3×2

Answer: a

Explanation: Prime factors of 444 are the product of some prime numbers. The prime factors are 2×2×3×37. Hence only 2×2×3×37 this is the correct answer the others are incorrect.

9. 91 is a perfect square.

a) True

b) False

Answer: b

Explanation: 91 is not a perfect square, therefore the given statement is incorrect. A perfect square is a number which has its roots as positive integers. Therefore the given statement is false.

10. 121 is a perfect square.

a) True

b) False

Answer: a

Explanation: 121 is a perfect square, therefore the given statement is correct. A perfect square is a number which has its roots as positive integers. This number satisfy all these requirements. Therefore the given statement is true.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Square Root on Decimals”.

1. Find the value of x=\ 4 < x > 5

b) 5 < x > 4

c) 4 > x < 5

d) 3 < x > 4

Answer: a

Explanation: We need to find the square root of 17.64 but from our prior knowledge we know that 4 2 =16 & 5 2 =25. From thus we can conclude that the square root would lie between 4 and 5. Hence the correct answer is 4 < x > 5 and the others are inncorret.

2. What would be the square of 17.66?

a) 311.8756

b) 311.8576

c) 311.5876

d) 311.5786

Answer: a

Explanation: When we calculate the square of 17.66 we get the answer as 311.8756. There are many other ways to find out the square of numbers but those ways fail to find the square on a decimal number. Therefore the correct answer is 311.8756 and the others are incorrect.

3. What would be the square root of \ 8.144

b) 8.244

c) 8.344

d) 8.544

Answer: a

Explanation: We can find the square root of any decimal numbers by division method. Here when we calculate the square root of the number 66.34 we get 8.144 as the answer. So, the correct answer would be 8.144 and the others would be incorrect.

4. Solve: \Missing open brace for subscript 11.00558

b) 11.00557

c) 11.00556

d) 11.00554

Answer: a

Explanation: Here when we calculate the square root of 121.123 we get 11.00558. Therefore the correct answer would we 11.00558 and the others are incorrect.

5. Find a perfect square near to the number 44332.

a) 210

b) 213

c) 214

d) 209

Answer: a

Explanation: Here the best way to reach our answer would be ‘option elimination’ i.e. we can try squaring all the options and by observation we can conclude that the answer would be 210.

This is because 210 2 = 44100; 213 2 = 45369; 214 2 = 45796; 209 2 = 43681

We can observe that the square of 210 is nearest to the number 44332.

6. What is the least number that should be added to 143 to make it a perfect square?

a) 0

b) 1

c) -1

d) 5

Answer: b

Explanation: The perfect square nearest to the number 143 is 12 2 =144, in order to acquire this perfect square we need to add 1 to the number. The correct answer would be 1 and others would be incorrect.

7. What would be the least number that should be subtracted from 629 to obtain a perfect square?

a) -4

b) 2

c) -2

d) 4

Answer: d

Explanation: The perfect square nearest to the number 629 is 25 2 =625, in order to acquire this perfect square we need to subtract 4 to the number. The correct answer would be 4 and others would be incorrect.

8. Which of the following is a perfect square?

a) 11.23

b) 11.00

c) 16.00

d) 16.233

Answer: c

Explanation: A perfect square is an integer obtained by multiplying an integer with itself. Here we have mixed options which contains decimals. All the options containing decimals are ruled out and the only correct option we are left with is 16. Therefore the correct answer is 16 and the others are incorrect.

9. ______ is a perfect square.

a) 91

b) 99

c) 121

d) 44

Answer: c

Explanation: A perfect square is an integer obtained by multiplying an integer with itself. Here we have mixed options that contain no square numbers. All the options containing no square numbers are ruled out and the only correct option we are left with is 121. Therefore the correct answer is 16 and the others are incorrect.

10. 122 is a perfect square.

a) True

b) False

Answer: b

Explanation: The given number 122 is not a perfect square, therefore the given statement cannot be true. Therefore the correct answer would be false.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Estimating Square Roots”.

1. What would be the square root of 901.23?

a) 30

b) 30.02

c) 30.5

d) 31

Answer: b

Explanation: We can observe that the number 901.23 is very close to the square of number 30 which is 900. Here we can estimate that the square root of 901.23 is very close to 900 and hence its square root would be closer to the square root of 900 i.e. 30. Therefore we can conclude that the answer would be 30.02.

2. Estimate the square root of 15.

a) 3.98

b) 4.58

c) 5.63

d) 6.67

Answer: a

Explanation: When we need to estimate a square root we can find the perfect square numbers. So the perfect squares near the number 15 are 9 & 16. Since the number is close to 16 the square root would be above 0.5. Therefore the correct answer would be 3.98 and the others would be incorrect.

3. _____ is the estimated square root of 123.

a) 11.03

b) 11.59

c) 11.99

d) 11.51

Answer: a

Explanation: When we need to estimate a square root we can find the perfect square numbers. So the perfect squares near the number 123 are 121 & 144. Since the number is close to 121 the square root would be below 0.5. Therefore the correct answer would be 11.03 and the others would be incorrect.

4. Which of the following is a prime square number?

a) 1

b) 2

c) 3

d) 4

Answer: a

Explanation: Here we have two options which have prime numbers. The others are ruled out, and out of two there is only one prime number which is a perfect square. Therefore the correct answer would be 1 and the others are incorrect.

5. What is the square root of 962 ?

a) 31<x>32

b) 31>x<32

c) 13<x>23

d) 13>x<23

Answer: a

Explanation: Since the number lie between the square of 31 & 32 i.e. 961 & 1024 the square root of 962 lies between 31 & 32. Hence the correct answer would be 31<x>32 and the others are incorrect.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Cubes”.

1. What can be general formula to find cube of any number?

a) Misplaced & 2

b) a 2

c) Misplaced & 3

d) b

Answer: a

Explanation: The general formula of finding square of any number is Misplaced & 3 . We need to express the given number whose cube needs to be found in the form of Misplaced &. We can then use the formula:

Misplaced & 3 = a 3 &pm; 3a 2 b+3ab 2 &pm; b 3 .

2. What form can be used to find cube of 13?

a)  3

b)  3

c) 

d) 

Answer: a

Explanation: The number 13 can be represented in form 10 + 3, this will help us to calculate the cube of the number. This method would help us eliminate the traditional method of multiplying the number with itself.

3. 89 3 = ____

a)  2

b)  3

c) 1 2

d) 9 2

Answer: b

Explanation: If we must find the cube of the 89, we use the second variation i.e. 3 . This would help us and make the calculations easier instead of multiplying the number with itself thrice.

Therefore  3 = 90 3 – 3 × 90 2 × 1 + 3 × 90 × 1 2 – 1 3

Therefore  3 = 729000 – 24300 + 270 – 1

Therefore  3 = 704969.

4. Find the cube of 32.

a) 32768

b) 37968

c) 32769

d) 35937

Answer: a

Explanation: To find cube of 32, we would use the form

Therefore  3 = 30 3 + 3 × 30 2 × 2 + 3 × 30 × 2 2 + 2 3

Therefore  3 = 27000 + 5400 + 360 + 8

Therefore  3 = 32768

5. Calculate the cube of 201.

a) 8120604

b) 8120609

c) 8120600

d) 8120601

Answer: d

Explanation: In the multiple-choice question, we can eliminate three options by mere observation. The unit place of cube of 201 should be 1. The only option which has 1 in unit place is 8120601. The other options are incorrect by mere observation.

6. 34 3 = _____

a) 39504

b) 39304

c) 35304

d) 34304

Answer: b

Explanation: In order to calculate the cube of 34 we use the formula,

 3 , where we consider a=30 and b=4

Therefore  3 = 30 3 + 3×30 2 × 4 + 3 × 30 × 4 2 + 4 3

Therefore  3 = 27000 + 10800 + 1440 + 64

Therefore  3 = 39304.

7. Which of below is not a perfect cube?

a) 125

b) 728

c) 729

d) 64

Answer: b

Explanation: Perfect cube is formed when a natural number is multiplied to itself thrice. For example, 4 is a natural number which multiplied with itself thrice gives us 64. Hence, 64 is a perfect cube. But we do not have a natural number which multiplied with itself thrice gives us 728. Hence, 728 is not a perfect cube.

8. How many perfect cubes exist between 1 and 100?

a) 4

b) 20

c) 16

d) 8

Answer: a

Explanation: Perfect cube is formed when a natural number is multiplied to itself thrice. There are 4 perfect cubes between 1 and 100. They are 1, 8, 27, 64.

9. How many perfect cubes exist between 1 and 1000?

a) 5

b) 20

c) 10

d) 15

Answer: a

Explanation: Perfect cube is formed when a natural number is multiplied to itself thrice. There are 10 perfect cubes between 1 and 1000. They are 1, 8, 27, 64, 125, 216, 343, 512, 729 and 1000.

10. 57 3 = _____

a) 1125

b) 1961

c) 3249

d) 5764

Answer: c

Explanation: In order to calculate the cube of 57 we use the formula,

 3 , where we consider a=60 and b=3

Therefore  3 = 60 3 – 3 × 60 2 × 3 + 3 × 60 × 3 2 – 3 3

Therefore  3 = 216000 – 32400 + 1620 – 27

Therefore  3 = 185193.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Patterns in Cubes”.

1. Express 2 3 as sum of consecutive odd numbers?

a) 1 + 3

b) 3 + 5

c) 5 + 7

d) 7 + 9

Answer: b

Explanation: Adding consecutive odd numbers gives us cube of a number. For example, 3 + 5 = 8 which is 2 3 . It is an interesting pattern where sum of consecutive odd numbers gives us cube of a number.

2. Express 125 as sum of consecutive numbers?

a) 21+23+25+27+29

b) 23+25+27+29+31

c) 19+21+23+25+27

d) 17+19+21+23+25

Answer: a

Explanation: Adding consecutive odd numbers gives us cube of a number. 5 3 gives us 125.

21+23+25+27+29 = 125. Sum of other options doesn’t give us 125.

3. Express 343 as sum of consecutive numbers?

a) 47+49+51+53+55+57+59

b) 41+43+45+47+49+51+53

c) 43+45+47+49+51+53+55

d) 45+47+49+51+53+55+57

Answer: c

Explanation: Adding consecutive odd numbers gives us cube of a number. 7 3 gives us 343.

43+45+47+49+51+53+55 = 343. Sum of other options doesn’t give us 343.

4. Find the cube of 4.

a) 64

b) 81

c) 111

d) 1331

Answer: a

Explanation: To find cube of 4, we need to find 4 3 . We can find cube of number by adding consecutive odd numbers.

Therefore, 4 3 = 13 + 15 + 17 + 19

Therefore, 4 3 = 64.

5. Calculate 3 3 .

a) 64

b) 81

c) 111

d) 27

Answer: d

Explanation: To find cube of 3, we need to find 3 3 . We can find cube of number by adding consecutive odd numbers.

Therefore, 3 3 = 7 + 9 + 11

Therefore, 3 3 = 27.

6. Evaluate 4 3 – 3 3 = _____

a) 39

b) 38

c) 37

d) 36

Answer: c

Explanation: 4 3 -3 3 =1+4*3*3

Therefore, 4 3 -3 3 =37. The difference between cube of consecutive numbers can be found using the above pattern.

7. Evaluate 20 3 -19 3 =_____

a) 1141

b) 1142

c) 1143

d) 1144

Answer: a

Explanation: 20 3 -19 3 =1+20*19*3

Therefore, 20 3 -19 3 =1141. The difference between cube of consecutive numbers can be found using the above pattern.

8. Evaluate 12 3 -11 3 =_____

a) 939

b) 389

c) 397

d) 396

Answer: c

Explanation: 12 3 -11 3 =1+12*11*3

Therefore, 12 3 -11 3 =397. The difference between cube of consecutive numbers can be found using the above pattern.

9. Evaluate 57 3 -56 3 =_____

a) 18890

b) 98818

c) 18900

d) 18818

Answer: d

Explanation: 57 3 -56 3 =1+57*56*3

Therefore, 57 3 -56 3 =18818. The difference between cube of consecutive numbers can be found using the above pattern.

10. Evaluate 14 3 -13 3 = _____

a) 547

b) 389

c) 559

d) 556

Answer: a

Explanation: 14 3 -13 3 =1+14*13*3

Therefore, 14 3 -13 3 =547. The difference between cube of consecutive numbers can be found using the above pattern.

This set of Mathematics Written Test Questions and Answers for Class 8 focuses on “Smallest Number which is a Perfect Cube”.

1. Which of the following numbers is a perfect cube?

a) 7

b) 9

c) 8

d) 2

Answer: c

Explanation: Perfect cube is formed when a natural number is multiplied to itself thrice. When 2 is multiplied with itself thrice, we get 8. Thus, 8 is a perfect cube. None of the other options can be obtained by multiplying a natural number with itself thrice.

2. Which of the following numbers is not a perfect cube?

a) 729

b) 216

c) 8

d) 2

Answer: d

Explanation: Perfect cube is formed when a natural number is multiplied to itself thrice. When 9 is multiplied with itself thrice, we get 729. When 6 is multiplied with itself thrice, we get 216. When 2 is multiplied with itself thrice, we get 8. Thus, 729, 216 and 8 are a perfect cube. 2 cannot be obtained by multiplying any natural number with itself thrice. Hence, 2 is not a perfect cube.

3. Which of the following numbers is a perfect cube?

a) 1331

b) 91

c) 98

d) 21

Answer: a

Explanation: Perfect cube is formed when a natural number is multiplied to itself thrice. When 11 is multiplied with itself thrice, we get 1331. Thus, 1331 is a perfect cube. None of the other options can be obtained by multiplying a natural number with itself thrice.

4. Which of the following numbers is a square and perfect cube?

a) 72

b) 125

c) 64

d) 100

Answer: c

Explanation: 8 multiplied to itself gives us 64. Thus, 64 is a square. Perfect cube is formed when a natural number is multiplied to itself thrice. When 4 is multiplied with itself thrice, we get 64. Thus, 64 is a perfect cube. Thus, 64 is square and perfect cube. None of the other options are square and perfect cube.

5. Find the smallest natural number by which 243 must be multiplied so that the product is a perfect cube.

a) 3

b) 4

c) 5

d) 6

Answer: a

Explanation: Perfect cube is formed when a natural number is multiplied to itself thrice. Factorizing 243 we get:

243 = 3×3×3×3×3

When 243 is multiplied with 3, we get 729 which is a perfect cube.

6. Find the smallest natural number by which 4394 must be multiplied so that the product is a perfect cube.

a) 3

b) 2

c) 5

d) 7

Answer: b

Explanation: Perfect cube is formed when a natural number is multiplied to itself thrice. Factorizing 4394 we get:

4394 = 2×2×13×13×13

When 4394 is multiplied with 2, we get 17576 which is 26 3 . Thus, 17576 is a perfect cube.

7. Find the smallest natural number by which 49000 must be multiplied so that the product is a perfect cube.

a) 3

b) 2

c) 5

d) 7

Answer: d

Explanation: Perfect cube is formed when a natural number is multiplied to itself thrice. Factorizing 49000 we get:

49000 = 2×2×2×5×5×5×7×7

When 49000 is multiplied with 7, we get 343000 which is 70 3 . Thus, 343000 is a perfect cube.

8. Find the smallest natural number by which 265837 must be multiplied so that the product is a perfect cube.

a) 11

b) 2

c) 5

d) 7

Answer: a

Explanation: Perfect cube is formed when a natural number is multiplied to itself thrice. Factorizing 265837 we get:

265837 = 13×13×13×11×11

When 265837 is multiplied with 11, we get 2924207 which is 143 3 . Thus, 2924207 is a perfect cube.

9. Find the smallest natural number by which 6125 must be multiplied so that the product is a perfect cube.

a) 3

b) 2

c) 5

d) 7

Answer: d

Explanation: Perfect cube is formed when a natural number is multiplied to itself thrice. Factorizing 6125 we get:

6125 = 5×5×5×7×7

When 6125 is multiplied with 7, we get 42875 which is 35 3 . Thus, 42875 is a perfect cube.

10. Find the smallest natural number by which 19652 must be multiplied so that the product is a perfect cube.

a) 3

b) 2

c) 5

d) 7

Answer: b

Explanation: Perfect cube is formed when a natural number is multiplied to itself thrice. Factorizing 19652 we get:

19652 = 2×2×19×19×19

When 19652 is multiplied with 2, we get 39304 which is 38 3 . Thus, 39304 is a perfect cube.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Cube Roots”.

1. Which symbol represents cube root of x?

a) x 2

b) x 3

c) ∛x

d) √x

Answer: c

Explanation: ∛x represents cube root of x. √x represents square root of x. x 2 represents square of x. x 3 represents cube of x.

2. Find the cube root of 226981?

a) 226981

b) 156981

c) 289981

d) 189981

Answer: a

Explanation: Factorizing 226981 we get:

226981 = 61×61×61

Cube root of 226981 is 61.

3. What is the cube root of 343?

a) 4

b) 5

c) 6

d) 7

Answer: d

Explanation: Factorizing 343 we get:

343 = 7×7×7

Cube root of 343 is 7.

4. What is the cube root of 1331?

a) 10

b) 11

c) 12

d) 13

Answer: b

Explanation: Factorizing 1331 we get:

1331 = 11×11×11

Cube root of 1331 is 11.

5. What is \ 73

b) 75

c) 77

d) 79

Answer: a

Explanation: Factorizing 389017 we get:

389017 = 73×73×73

Cube root of 389017 is 73.

6. What is the cube root of 125?

a) 4

b) 5

c) 6

d) 7

Answer: b

Explanation: Factorizing 125 we get:

125 = 5×5×5

Cube root of 125 is 5.

7. What is square and cube of 14?

a) 2744, 196

b) 196, 2744

c) 196, 2748

d) 2744, 195

Answer: b

Explanation: Square of 14 is 14×14 which gives us 196.

Cube of 14 is 14×14×14 which gives us 2744.

8. Evaluate \ 45

b) 55

c) 65

d) 85

Answer: d

Explanation: Factorizing 614125 we get:

614125 = 5×5×5×17×17×17

Cube root of 614125 is 85.

9. What is the cube root of 729000?

a) 40

b) 50

c) 90

d) 70

Answer: c

Explanation: Factorizing 729000 we get:

729000 = 9×9×9×10×10×10

Cube root of 729000 is 90.

10. What is the cube root of 50653?

a) 34

b) 35

c) 36

d) 37

Answer: d

Explanation: Factorizing 50653 we get:

50653 = 37×37×37

Cube root of 50653 is 37.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Cube Root through Prime Factorisation”.

1. Find the prime factors of 6.

a) 2, 3

b) 2, 5

c) 3, 5

d) 5, 7

Answer: a

Explanation: A prime number can only be divided by 1 or itself, so it cannot be factored any further. On multiplying 2 and 3, we get 6. 2 and 3 cannot be factored further. Hence, 2 and 3 are prime factors of 6.

2. Find prime factors of 72.

a) 2×2×7×7

b) 2×2×2×3×3

c) 2×3×5×7×7

d) 2×2×2×3×7

Answer: b

Explanation: When factorizing 72 we get,

72 = 2×36

72 = 2×2×18

72 = 2×2×2×9

72 = 2×2×2×3×3.

3. Which number should be multiplied to 392 to obtain perfect cube?

a) 7

b) 9

c) 8

d) 2

Answer: a

Explanation: Perfect cube is formed when a natural number is multiplied to itself thrice. Factorizing 392 we get:

392 = 2×2×2×7×7

When 392 is multiplied with 7, we get a perfect cube.

4. Find the cube root of 729.

a) 3

b) 7

c) 5

d) 9

Answer: d

Explanation: Factorizing 729 we get:

729 = 3×243

729 = 3×3×81

729 = 3×3×9×9

729 = 9×9×9

Thus, cube root of 729 is 9.

5. What multiplied to 68600 gives a perfect cube?

a) 7

b) 2

c) 5

d) 9

Answer: c

Explanation: Perfect cube is formed when a natural number is multiplied to itself thrice. Factorizing 68600 we get:

68600 = 2×2×2×5×5×7×7×7

When 68600 is multiplied with 5, we get a perfect cube.

6. What is the cube root of 216000?

a) 60

b) 70

c) 80

d) 90

Answer: a

Explanation: Factorizing 216000 we get:

21600 = 2×2×2×3×3×3×10×10×10

The cube root of 216000 is 60.

7. Which of below is not a perfect cube?

a) 1331

b) 2197

c) 4913

d) 5508

Answer: d

Explanation: Perfect cube is formed when a natural number is multiplied to itself thrice. For example, 11 is a natural number which multiplied with itself thrice gives us 1331. Hence, 1331 is a perfect cube. But we do not have a natural number which multiplied with itself thrice gives us 5508. Hence, 5508 is not a perfect cube.

8. Evaluate \ 35

b) 25

c) 45

d) 15

Answer: b

Explanation: ∛x represents cube root of number x. Factorizing 15625 we get:

15625 = 25×25×25

The cube root of 15625 is 25.

9. Evaluate \ 51

b) 21

c) 29

d) 39

Answer: b

Explanation: ∛x represents cube root of number x. Factorizing 9261 we get:

9261 = 21×21×21

The cube root of 9261 is 21.

10. 67 3 = _____

a) 300733

b) 300753

c) 300763

d) 300793

Answer: c

Explanation: In order to calculate the cube of 67 we use the formula,

 3 , where we consider a=70 and b=3

Therefore  3 = 70 3 – 3 × 70 2 × 3 + 3 × 70 × 3 2 – 3 3

Therefore  3 = 343000 – 44100 + 1890 – 27

Therefore  3 = 300763.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Cube Root of a Cube Number”.

1. Find cube root of 91125.

a) 35

b) 45

c) 55

d) 57

Answer: b

Explanation: We can find cube root of number by estimation if we know for certain that cube root of number is natural number. Start making group of 3 digits from right. Thus, we get 91 and 125. Cube root of 125 is 5 which forms the unit place. 91 lies between 64 and 125. Hence, we would consider it to be 4 which forms tens place. Hence, cube root of 91125 is 45.

2. Find cube root of 6859.

a) 17

b) 18

c) 19

d) 20

Answer: c

Explanation: We can find cube root of number by estimation if we know for certain that cube root of number is natural number. Start making group of 3 digits from right. Thus, we get 6 and 859. Cube root of 859 would have unit place 9 as unit place of 859 is 9. 6 lies between 1 and 8. Cube root of 6 would like between 1 and 2. Hence, we would consider it to be 1 which forms tens place. Hence, cube root of 6859 is 19.

3. Find cube root of 19683.

a) 27

b) 37

c) 47

d) 57

Answer: a

Explanation: We can find cube root of number by estimation if we know for certain that cube root of number is natural number. Start making group of 3 digits from right. Thus, we get 19 and 683. Cube root of 683 would have unit place as 7 because 683 has unit place as 3. 19 lies between 8 and 27. Hence, we would consider it to be 2 which forms tens place. Hence, cube root of 19683 is 27.

4. Find cube root of 42875.

a) 25

b) 57

c) 55

d) 35

Answer: d

Explanation: We can find cube root of number by estimation if we know for certain that cube root of number is natural number. Start making group of 3 digits from right. Thus, we get 42 and 875. Cube root of 875 would have unit digit as 5. 42 lies between 27 and 64. Hence, we would consider it to be 3 which forms tens place. Hence, cube root of 42875 is 35.

5. Find cube root of 857375.

a) 85

b) 97

c) 95

d) 65

Answer: c

Explanation: We can find cube root of number by estimation if we know for certain that cube root of number is natural number. Start making group of 3 digits from right. Thus, we get 857 and 375. Cube root of 375 would have unit digit as 5. 857 lies between 729 and 1000. Hence, we would consider it to be 9 which forms tens place. Hence, cube root of 857375 is 95.

6. What is the cube root of 216000?

a) 60

b) 70

c) 80

d) 90

Answer: a

Explanation: We can find cube root of number by estimation if we know for certain that cube root of number is natural number. Start making group of 3 digits from right. Thus, we get 216 and 000. Cube root of 0 would have unit digit as 0. Cube root of 216 is 6 which would form tens place. Hence, the cube root of 216000 is 60.

7. Find cube root of 300763.

a) 27

b) 37

c) 67

d) 57

Answer: c

Explanation: We can find cube root of number by estimation if we know for certain that cube root of number is natural number. Start making group of 3 digits from right. Thus, we get 300 and 763. Cube root of 763 has unit place as 7 as unit place of 763 is 3. 300 lies between 216 and 343. Hence, we would consider it to be 6 which forms tens place. Hence, cube root of 300763 is 67.

8. Evaluate \ 35

b) 25

c) 45

d) 15

Answer: b

Explanation: ∛x represents cube root of number x. We can find cube root of number by estimation if we know for certain that cube root of number is natural number. Start making group of 3 digits from right. Thus, we get 15 and 625. Cube root of 625 has unit place as 5 as unit place of 625 is 5. 15 lies between 8 and 27. Hence, we would consider it to be 2 which forms tens place. Hence, cube root of 15625 is 25.

9. Evaluate \ 51

b) 21

c) 29

d) 39

Answer: b

Explanation: ∛x represents cube root of number x. We can find cube root of number by estimation if we know for certain that cube root of number is natural number. Start making group of 3 digits from right. Thus, we get 9 and 261. Cube root of 261 has unit place as 1 as unit place of 261 is 1. 9 lies between 8 and 27. Hence, we would consider it to be 2 which forms tens place. Hence, the cube root of 9261 is 21.

10. Find cube root of 79507.

a) 45

b) 47

c) 95

d) 43

Answer: d

Explanation: We can find cube root of number by estimation if we know for certain that cube root of number is natural number. Start making group of 3 digits from right. Thus, we get 79 and 507. Cube root of 507 would have unit digit as 3 as unit digit of 507 is 7. 79 lies between 64 and 125. Hence, we would consider it to be 4 which forms tens place. Hence, cube root of 79507 is 43.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Ratios and Percentages”.

1. A class has 30 girls and 40 boys. What is ratio of number of girls to total students in class?

a) \

 

 \

 

 \

 

 \(\frac{4}{3}\)

Answer: a

Explanation: A class has 30 girls and 40 boys. So total number of students in the class is 70. The ration of number of girls to total students in the class is

\(\frac{Number \,of \,girls}{Total \,number \,of \,students}=\frac{30}{70}=\frac{3}{7}\).

2. 40% of group likes tea as a beverage. What is the ratio of people liking tea as a beverage in the group?

a) \

 

 \

 

 \

 

 \(\frac{4}{3}\)

Answer: b

Explanation: 40% of group likes tea as a beverage. Ratio of people liking tea as a beverage in the group is

\(\frac{Tea \,loving \,people}{Total \,people \,in \,the \,group}=\frac{40}{100}=\frac{2}{5}\).

3. Train travels at 45 kmph and bus travels at 30 kmph. What is the ratio of speed of train with respect to bus?

a) \

 

 \

 

 \

 

 \(\frac{5}{2}\)

Answer: c

Explanation: Train travels with 45 kmph and bus travels at 30 kmph. The ratio of speed of train with respect to speed of bus is

\(\frac{Speed \,of \,train}{Speed \,of \,bus}=\frac{45}{30}=\frac{3}{2}\).

4. Train travels at 45 kmph and bus travels at 30 kmph. What is the ratio of speed of bus with respect to train?

a) \

 

 \

 

 \

 

 \(\frac{5}{2}\)

Answer: c

Explanation: Train travels with 45 kmph and bus travels at 30 kmph. The ratio of speed of bus with respect to speed of train is

\(\frac{Speed \,of \,bus}{Speed \,of \,train}=\frac{30}{45}=\frac{2}{3}\).

5. Rakesh has Rs. 50 and Karan has Rs. 1.5. Find the ratio of money with Rakesh and money with Karan.

a) \

 

 \

 

 \

 

 \(\frac{500}{2}\)

Answer: b

Explanation: Rakesh has Rs. 50 and Karan has Rs. 1.5.

Ratio of money with Rakesh and money with Karan is

\(\frac{money \,with \,Rakesh}{money \,with \,Karan}=\frac{50}{1.5}=\frac{500}{15}=\frac{100}{3}\).

6. Rakesh has Rs. 10 and Karan has 10 paise. Find the ratio of money with Rakesh and money with Karan.

a) \

 

 \

 

 \

 

 \(\frac{5}{2}\)

Answer: b

Explanation: Rakesh has Rs. 10 and Karan has 10 paisa. To derive ratio, quantities should be in converted to same unit. Here, one quantity is in rupees and other in paisa. We will convert both the quantities to paisa to derive ratio.

Ratio of money with Rakesh and money with Karan is

\(\frac{money \,with \,Rakesh}{money \,with \,Karan}=\frac{1000}{10}=\frac{100}{1}\).

7. Kiran has Rs. 130 left after using 65% of total money she had on shopping. Find total money with Kiran before shopping.

a) 100

b) 10

c) 65

d) 200

Answer: d

Explanation: Kiran has Rs. 130 left after using 65% of total money she had on shopping. Let the total money with Kiran be Rs. X. Total money with Kiran can be found as below

130=\(\frac{65}{100}\) × X

Therefore, X=\(\frac{130×100}{65}\)

Therefore, X=\(\frac{130×20}{13}\)

Therefore, X=\(\frac{130×20}{13}\)

Therefore, X=200

Thus, total money with Kiran before shopping is Rs. 200.

8. Ticket to Rameshwar by train is Rs. 1200 and ticket to Delhi is Rs. 900 from Mumbai. What is the ratio of cost to reach Delhi with respect to Rameshwar from Mumbai?

a) \

 

 \

 

 \

 

 \(\frac{5}{2}\)

Answer: a

Explanation: Ticket to Rameshwar by train is Rs. 1200 and ticket to Delhi is Rs. 900 from Mumbai. The ratio of ticket to Delhi with respect to ticket to Rameshwar is

\(\frac{cost \,to \,Delhi}{cost \,to \,Rameshwar}=\frac{900}{1200}=\frac{3}{4}\).

9. Sam is 78 kg and Sameer is 91 kg. Find the ratio of weight of Sam and Sameer.

a) \

 

 \

 

 \

 

 \(\frac{6}{7}\)

Answer: d

Explanation: Sam is 78 kg and Sameer is 91 kg. The ratio of weight of Sam and Sameer is

\(\frac{weight \,of \,Sam}{weight \,of \,Sameer}=\frac{78}{91}=\frac{6}{7}\).

10. Satish has 15 kg sugar and Ray has 250 gm. Find the ratio of sugar with of Satish and Ray.

a) \

 

 6

c) \

 

 \(\frac{15}{25}\)

Answer: b

Explanation: Satish has 15 kg of sugar and Ray has 250 gm. We must have both quantities in the same unit. 15 kg converted to grams is 1500gm. The ratio of sugar with Satish and Ray is

\(\frac{Sugar \,with \,Satish}{Sugar \,with \,Ray}=\frac{1500}{250}=\frac{30}{5}\)=6.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Finding Increase and Decrease of Percentage”.

1. The price of bike is Rs. 75000 last year. The price increased by 20% this year. What is the price of bike this year?

a) Rs.50000

b) Rs.60000

c) Rs.80000

d) Rs.90000

Answer: d

Explanation: Price of bike last year was Rs. 75000. The price increased by 20%.

Increase in price = \(\frac{20}{100}\)×75000=15000

New price = Old price + Increase in price = 75000 + 15000 = 90000.

2. The price of bike is Rs. 75000 last year. The price decreased by 20% this year. What is the price of bike this year?

a) Rs.50000

b) Rs.60000

c) Rs.80000

d) Rs.90000

Answer: b

Explanation: Price of bike last year was Rs. 75000. The price decreased by 20%.

Decrease in price = \(\frac{20}{100}\)×75000=15000

New price = Old price + Decrease in price = 75000 – 15000 = 60000.

3. An item is marked as Rs. 6000. There is a 25% discount on the item. What is the amount of discount received on the item?

a) Rs.6000

b) Rs.1000

c) Rs.1500

d) Rs.2500

Answer: c

Explanation: Marked price of item Rs. 6000. Discount of item is 25%.

Discount = \(\frac{25}{100}\)×6000=1500.

4. An item is marked as Rs. 6000. There is a 25% discount on the item. What is the discounted price of item?

a) Rs.6000

b) Rs.4500

c) Rs.1500

d) Rs.2500

Answer: b

Explanation: Marked price of item is Rs. 6000. Discount on item is 25%.

Discount = \(\frac{25}{100}\)×6000=1500

Discounted price = Marked price – Discount = 6000 – 1500 = 4500.

5. An item was marked for Rs. 5000 but sold for Rs. 4500. What is the % discount offered on the product?

a) 5

b) 10

c) 15

d) 20

Answer: b

Explanation: Marked price of item is Rs. 5000. Discounted price of item is Rs. 4500.

Discount = Marked price – Selling price = 5000 – 4500 = 500

% discount=\(\frac{discount}{marked price}=\frac{500}{5000}\)=10.

6. The list price of shirt is Rs. 2500. There is 30% discount on the item. What is the sell price?

a) Rs.1750

b) Rs.4500

c) Rs.1500

d) Rs.2500

Answer: a

Explanation: List price of shirt is Rs. 2500. Discount on item is 30%.

Discount = \(\frac{30}{100}\)×2500=750

Discounted price = Marked price – Discount = 2500 – 750 = 1750.

7. The list price of shirt is Rs. 2500. There is 30% discount on the item. What is the discount offered on the shirt?

a) Rs.1750

b) Rs.4500

c) Rs.1500

d) Rs.750

Answer: d

Explanation: List price of shirt is Rs. 2500. Discount on item is 30%.

Discount = \(\frac{30}{100}\)×2500=750.

8. Zayn bought tickets to concert for Rs. 1000. He wants to sell them at a premium of 20%. What is the selling price of tickets?

a) Rs.1750

b) Rs.1200

c) Rs.1500

d) Rs.800

Answer: b

Explanation: Buying price of tickets is Rs. 1000. Premium on tickets is 20%

Premium = \(\frac{20}{100}\)×1000=200

Selling price = Buying price + Premium = 1000 + 200 = 1200.

9. Zayn bought tickets to concert for Rs. 2700. He wants to sell them at a discount of 15%. What is the selling price of tickets?

a) Rs.2295

b) Rs.1200

c) Rs.2700

d) Rs.3105

Answer: a

Explanation: Buying price of tickets is Rs. 2700. Discount on tickets is 15%

Discount = \(\frac{15}{100}\)×2700=405

Selling price = Buying price – Discount = 2700 – 405 = 2295.

10. Zayn bought tickets to concert for Rs. 3500. He wants to sell them at a discount of 15%. What is the discount in Rs.?

a) Rs.1525

b) Rs.350

c) Rs.525

d) Rs.1050

Answer: c

Explanation: Buying price of tickets is Rs. 3500. Discount on tickets is 15%

Discount = \(\frac{15}{100}\)×3500=525.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Finding Discounts”.

1. A bag marked at a price of Rs. 1600 is available at a discount of 45%. What is the discount given?

a) Rs 740

b) Rs 720

c) Rs 800

d) Rs 760

Answer: b

Explanation: Discount offered = 45%

Marked price of bag = Rs. 1600

Discount = 45%

= 45% of 1600

= \(\frac{45}{100}\) × 1600 = 720

The discount given is Rs. 720

2. At a clothing store, Ameena buys a pair of jeans and some t-shirts marked at a price of Rs. 2500 at 50% discount. How much does she need to pay to the shopkeeper?

a) Rs 1600

b) Rs 1250

c) Rs 1750

d) Rs 1800

Answer: b

Explanation: Discount offered = 50%

Marked price of all the items = Rs. 2500

Discount = 50%

= 50% of 2500

= \(\frac{50}{100}\) × 2500 = 1250

The discount given is Rs. 1250

Sale Price = Marked Price – Discount

= 2500 – 1250 = 1250

Ameena has to pay Rs. 1250 to the shopkeeper.

3. At a shopping bonanza a 55% discount is offered on all its items. What would be the sale price of a shoe marked at Rs. 5000?

a) Rs 2250

b) Rs 2500

c) Rs 5000

d) Rs 2750

Answer: a

Explanation: Discount offered = 55 %

Marked price of shoe = Rs. 5000

Discount = 55%

= 55% of 5000

= \(\frac{55}{100}\) × 5000 = 2750

The discount given is Rs. 2750.

Sales price = Marked Price – discount

= 5000 – 2750

= 2250

The sale price of the shoe will be Rs. 2250

4. An almirah is marked is marked at Rs. 17,000 is sold at Rs. 10,000. What is the discount given?

a) 48.9%

b) 45.7%

c) 42.6%

d) 41.17 %

Answer: d

Explanation: Marked price of almirah = Rs. 17000

Sale price of the almirah = Rs. 10000

Discount = Marked price – Sale price

= 17000 – 10000

= 7000

Discount % = \(\frac{7000}{17000}\) × 100 = 41.17%

5. 10 books each of Rs. 300 are sold at Rs. 2500. What is the percentage of the discount given?

a) 11.2 %

b) 17.9 %

c) 16.67%

d) 16.85%

Answer: c

Explanation: Marked price of 1 book = Rs. 300

Marked price of 10 books = Rs. 3000

Sale price of the almirah = Rs. 2500

Discount = Marked price – Sale price

= 3000 – 2500

= 500

Discount % = \(\frac{500}{3000}\) × 100 = 16.67 %

6. A dinner set is sold at Rs. 4500 after giving a discount of 15%. What is the marked price of the dinner set?

a) Rs 4600

b) Rs 4700

c) Rs 4000

d) Rs 4800

Answer: b

Explanation: Sale price of the dinner set = Rs. 4500

Discount given = 15%

Let the marked price be Rs. x

Discount = 15% of x

= \(\frac{15}{100}\) × x = \(\frac{5x}{100}\)

Sale price = Marked price – discount

4500 = x – \(\frac{5x}{100}\)

4500 = \(\frac{95x}{100}\)

x = \(\frac{4500 × 100}{95}\) = 4736.84 ∼ Rs. 4700

The marked price of the dinner set is Rs. 4700

7. An article was sold after offering two successive discounts of 250 and 340. If the mark price of is 1000 then what is the selling price of the article?

a) Rs 495

b) Rs 500

c) Rs 300

d) Rs 465

Answer: a

Explanation: Discount given = 250 and 340.

Marked price = Rs.1000

First discount % = \(\frac{250}{1000}\) × 100 = 25%

Second discount % = \(\frac{340}{1000}\) × 100 = 34%

We know that, if when an article is sold after two successive discounts of x% and y%, then the final selling price = \(\frac{Marked \, Price ×  × }{100 × 100}\)

Here two successive discounts of 25% and 34% are given,

Therefore the sales price = \(\frac{1000××}{100×100}\) = 495

The selling price of the article is Rs. 495

8. A pen is sold for Rs. 1500 after offering two successive discounts of 5% and 7%. What is the marked price of the shirt?

a) Rs 1500

b) Rs 1600

c) Rs 1700

d) Rs 1200

Answer: c

Explanation: We know that, if when an article is sold after two successive discounts of x% and y%, then the final selling price = \(\frac{Marked \, Price ×  × }{100 × 100}\)

Here two successive discounts of 5% and 7% are given,

Let the marked price be a

Therefore, the sales price = \(\frac{a××}{100×100} = \frac{1767a}{2000}\)

Selling price = 1500

1500 = \(\frac{1767a}{2000}\)

a = 1697.79 ∼ 1700

The marked price of the article is Rs. 1700

9. The marked price of a mobile is Rs 40000. The shopkeeper allowed a discount of 30% but the customer bargained for 45%. So the shopkeeper allowed 30% and 15% successive discounts. Which of the following statements is true?

a) The shopkeeper benefitted the deal by 1200

b) The customer gained by 1800

c) The shopkeeper had a loss of 1000

d) Both parties benefitted equally

Answer: b

Explanation: Case I: The shopkeeper offers a 45% discount

Marked price = Rs 40000

Discount = 45%

= 45% of 40000

= \(\frac{45}{100}\) × 40000 = 18000

Sale price = Marked price – discount = 40000 – 18000 = 22000

Case II: The shopkeeper offers two successive discounts of 30% and 15%

Here two successive discounts of 30% and 15% are given,

The marked price is 40000

Therefore, the sales price = \(\frac{40000××}{100×100}\) = 23800

Clearly, the customer benefitted from the second deal by 1800 rupees.

10. The cost price of a vintage clock is Rs 12000 and its marked price is Rs 15000. If the shopkeeper sells it at a gain of 9%, then hat is the rate of discount offered by him?

a) 9.8%

b) 11.3%

c) 10%

d) 12.8 %

Answer: d

Explanation: Let the discount given by him be x %.

To find out the discount given we first have to calculate the selling price of the clock.

Selling price when cost price and profit % are given  = cost price 

 

Selling price = 12000

 

 = 12000 = 13080

Now, discount = Marked price – selling price = 15000 – 13080 = 1920

Discount % = \(\frac{discount}{marked \,price}\) × 100 = \(\frac{1920}{15000}\) × 100 = 12.8%

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Estimations in Percentage”.

1. Calculate 10% of 557.65.

a) 557.65

b) 55.765

c) 5.5765

d) 0.55765

Answer: b

Explanation: 10% of any number is shifting decimal places one place to the left. 10% of 557.65 means shifting decimal one place to left making it 55.765.

2. Calculate 15% of 882.

a) 88.2

b) 44.1

c) 132.3

d) 882

Answer: c

Explanation: 10% of any number is shifting decimal places one place to the left. 10% of 882 means shifting decimal one place to left making it 88.2. 5% any number is \(\frac{1}{2}\) of 10% of any number. Thus, 5% of 882 means \(\frac{1}{2}\) × 88.2 = 44.1. Thus, 15% of 882 = 88.2 + 44.1 = 132.3.

3. Calculate 25% of 776.

a) 776

b) 388

c) 97

d) 194

Answer: d

Explanation: 25% of any number is dividing the number by 4.

\(\frac{1}{4}\)×776=194.

4. Calculate 20% of 1266.

a) 253.2

b) 126.6

c) 63.3

d) 31.1

Answer: a

Explanation: 10% of 1266 is moving one decimal to the left which gives us 126.6. 20% is twice the number obtained by 10%. Thus, 20% of 1266 is 253.2.

5. A bill in a shop is Rs. 533.6. Shopkeeper offered 10% discount. What is the amount to be paid to shopkeeper?

a) Rs.500

b) Rs.480

c) Rs. 150

d) Rs.200

Answer: b

Explanation: 10% of 533.6 is moving one decimal to the left which gives us 53.36.

Discounted amount = 533.36 – 53.36

By approximation, discounted amount = 533 – 53 = 480.

6. The list price of dress is Rs. 3000. There is 30% discount on the item. What is the sell price?

a) Rs.2200

b) Rs.900

c) Rs.2100

d) Rs.2500

Answer: c

Explanation: List price of dress is Rs. 3000. Discount on item is 30%.

10% of 3000 is moving one decimal point to the left which gives us 300.

30% is thrice 10% giving us 900.

Hence, sell price = list price – discount = 2100.

7. The list price of dress is Rs. 1500. There is 20% discount on the item. What is the discount offered on the dress?

a) Rs.300

b) Rs.450

c) Rs.150

d) Rs.750

Answer: a

Explanation: List price of shirt is Rs. 1500. Discount on item is 20%.

10% of 1500 is 150.

20% is twice 10% which gives us 300.

8. Zayn bought tickets to concert for Rs. 5000. He wants to sell them at a premium of 20%. What is the selling price of tickets?

a) Rs.5000

b) Rs.1200

c) Rs.4500

d) Rs.6000

Answer: d

Explanation: Buying price of tickets is Rs. 5000. Premium on tickets is 20%

10% of 5000 is 500. 20% of any number is twice 10% of number which gives us 1000.

Selling price = Buying price + Premium = 5000 + 1000 = 6000.

9. Zayn bought tickets to concert for Rs. 2700. He wants to sell them at a discount of 15%. What is the selling price of tickets?

a) Rs.2295

b) Rs.1200

c) Rs.2700

d) Rs.3105

Answer: a

Explanation: Buying price of tickets is Rs. 2700. Discount on tickets is 15%

10% of 2700 is 270.

5% of any number is \(\frac{1}{2}\) of 10% of number. 5% of 2700 is 135.

15% of number is sum of 10% of number and 5% of number = 270 + 135 = 405.

Selling price = Buying price – Discount = 2700 – 405 = 2295.

10. Zayn bought tickets to concert for Rs. 3500. He wants to sell them at a discount of 15%. What is the discount in Rs.?

a) Rs.1525

b) Rs.350

c) Rs.525

d) Rs.1050

Answer: c

Explanation: Buying price of tickets is Rs. 3500. Discount on tickets is 15%.

10% of 3500 is 350.

5% of any number is \(\frac{1}{2}\) of 10% of number. 5% of 3500 is 175.

15% of number is sum of 10% of number and 5% of number = 350 + 175 = 525.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Price Related to Buying and Selling ”.

1. Rohan bought bike for Rs. 75000. He wants to sell it by making 20% profit. What is the selling price of bike?

a) Rs.50000

b) Rs.60000

c) Rs.80000

d) Rs.90000

Answer: d

Explanation: Price of bike is Rs. 75000. The price increased by 20%.

Profit = \(\frac{20}{100}\)×75000=15000

Selling price = Buying price + profit = 75000 + 15000 = 90000.

2. The price of bike is Rs. 75000. The seller offered discount of 20% this year. What is the discounted price of bike?

a) Rs.50000

b) Rs.60000

c) Rs.80000

d) Rs.90000

Answer: b

Explanation: Price of bike is Rs. 75000. The discount offered is 20%.

Discount in price = \(\frac{20}{100}\)×75000=15000

Selling price = Buying price – discount = 75000 – 15000 = 60000.

3. An item is marked as Rs. 8000. There is a 25% discount on the item. What is the amount of discount received on the item?

a) Rs.6000

b) Rs.1000

c) Rs.1500

d) Rs.2000

Answer: d

Explanation: Marked price of item Rs. 8000. Discount of item is 25%.

Discount = \(\frac{25}{100}\)×8000=2000.

4. An item is marked as Rs. 6000. There is a 40% discount on the item. What is the discounted price of an item?

a) Rs.6000

b) Rs.3600

c) Rs.1500

d) Rs.2500

Answer: b

Explanation: Marked price of item is Rs. 6000. Discount on item is 25%.

Discount = \(\frac{40}{100}\)×6000=2400

Discounted price = Marked price – Discount = 6000 – 2400 = 3600.

5. An item was marked for Rs. 5000 but sold for Rs. 3500. What is the % loss on the product?

a) 5

b) 10

c) 30

d) 20

Answer: c

Explanation: Marked price of item is Rs. 5000. Selling price of item is Rs. 3500.

Discount = Marked price – Selling price = 5000 – 3500 = 1500

% discount=\(\frac{discount}{marked \,price}=\frac{1500}{5000}\)×100=30.

6. The cost price of shirt is Rs. 2500. The shopkeeper sold 10 shirts at 15% profit. What is the total profit made?

a) Rs.1750

b) Rs.4500

c) Rs.1500

d) Rs.2500

Answer: a

Explanation: Cost price of shirt is Rs. 2500. The shopkeeper sold 10 shirts. Thus, total cost price is Rs. 25000.

Profit = \(\frac{15}{100}\)×25000=3750

Total profit made by shopkeeper is R. 1750.

7. The cost price of house is Rs. 2500000. The house repair costed Raj Rs. 150000. The house was sole for 20% profit. What is the selling price of house?

a) Rs.318000

b) Rs.2580000

c) Rs.2500000

d) Rs.3180000

Answer: d

Explanation: Cost price of house is Rs. 2500000. Cost of repair is Rs. 150000. Thus, total cost price of house is Rs. 2650000.

Profit = \(\frac{20}{100}\)×2650000=530000.

Selling price = Cost price + profit = 3180000.

8. Zayn bought tickets to concert for Rs. 1700. He wants to sell them at a profit of 22%. What is the selling price of tickets?

a) Rs.1750

b) Rs.2074

c) Rs.1700

d) Rs.1800

Answer: b

Explanation: Buying price of tickets is Rs. 1700. Premium on tickets is 22%

Profit = \(\frac{22}{100}\)×1700=374

Selling price = Buying price + Profit = 1700 + 374 = 2074.

9. Rehan bought bulbs for Rs. 25. He wants to sell them at a discount of 15%. What is the selling price of bulbs?

a) Rs.22.95

b) Rs.21.25

c) Rs.27

d) Rs.31.46

Answer: b

Explanation: Buying price of bulbs is Rs. 25. Discount on tickets is 15%

Discount = \(\frac{15}{100}\)×25=3.75

Selling price = Buying price – Discount = 25 – 3.75 = 21.25.

10. Riya bought apples for Rs. 350. She sold them for Rs. 400. Find the percentage of profit made by Riya.

a) 14.29%

b) 15.23%

c) 5.25%

d) 10.50%

Answer: a

Explanation: Cost price of apples is Rs. 350. Selling price of apples is Rs. 400.

Profit = Selling price – Cost price = 400 – 350 = 50

%Profit = \(\frac{50}{350}\)×100=14.29.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Sales Tax/Value Added Tax”.

1. What will be the buying price of a soap at Rs. 100 when 5% sales tax is added?

a) Rs 105

b) Rs 110

c) Rs 112

d) Rs 134

Answer: a

Explanation: Cost price of soap = Rs. 100

Sales tax = 5%

= 5% of 100

= \(\frac{5}{100}\) × 100

= Rs. 5

Buying price = Cost price + sales tax = 100 + 5 = Rs.105

2. If 5% VAT is included in the price of a shampoo bottles bought for Rs. 550 then what is the original price of the shampoo?

a) Rs 420

b) Rs 520

c) Rs 650

d) Rs 700

Answer: b

Explanation: Let the original price of the shampoo be Rs. x

Sales tax = 5%

= 5% of x

= \(\frac{5}{100}\) × x

= Rs. \(\frac{5x}{100}\)

Buying price = Cost price + Sales tax

550 = x + \(\frac{5x}{100}\)

550 = \(\frac{105x}{100}\)

x = \(\frac{550×100}{105}\) = 523.80 ∼ 520

The original price of the shampoo is Rs. 520.

3. What will be the buying price of 12 kg of sugar at Rs. 20/kg when a 9% sales tax is added?

a) Rs 378.9

b) Rs 350.5

c) Rs 348.8

d) Rs 398.0

Answer: c

Explanation: Cost of one kg sugar = Rs. 20

Cost of 12 kg sugar = 20 × 12 = 320

The cost price of sugar = Rs. 320

Sales tax = 9%

= 9% of 320

= \(\frac{9}{100}\) × 320

= 28.8

Buying price = Cost price + Sales tax = 320 + 28.8 = 348.8.

4. If you buy a hair dryer for Rs. 10000 including 10% VAT what will be the price of the hair dryer before VAT was added?

a) Rs 9393

b) Rs 9292

c) Rs 9191

d) Rs 9090

Answer: d

Explanation: Let the original price of the hair dryer be Rs. x

VAT added = 10%

= 10% of x

= \(\frac{10}{100}\) × x

= \(\frac{x}{10}\)

Buying price = cost price + sales tax

10000 = x + \(\frac{x}{10}\)

10000 = \(\frac{11x}{10}\)

x = \(\frac{10000×10}{11}\) = 9090.90 ∼ 9090

5. What will be amount Shridhar has to pay for a watch costing Rs. 18000 when 8% sales tax is added?

a) Rs 19440

b) Rs 18850

c) Rs 17560

d) Rs 16540

Answer: a

Explanation: Cost price of watch = Rs. 18000

Sales tax = 8%

= 8% of 18000

= \(\frac{8}{100}\) × 18000

= Rs. 1440

Buying price = Cost price + sales tax = 18000 + 1440 = Rs. 19440

6. The cost of a hover board was 15000 at the shop. What will be the bill amount after a sales tax of 10% is added?

a) Rs 16160

b) Rs 15150

c) Rs 14540

d) Rs 17170

Answer: b

Explanation: Cost price of hover board = Rs. 15000

Sales tax = 10%

= 10% of 15000

= \(\frac{10}{100}\) × 15000

= Rs. 1500

Buying price = Cost price + sales tax = 15000 + 1500 = Rs. 15150

7. The bill amount of a cutlery set is Rs. 5000 including 9% sales tax. What is its original price?

a) Rs 4800

b) Rs 4700

c) Rs 5000

d) Rs 4500

Answer: d

Explanation: Let the original price of the cutlery be Rs. x

VAT added = 9%

= 9% of x

= \(\frac{9}{100}\) × x

= \(\frac{9x}{100}\)

Buying price = cost price + sales tax

5000 = x + \(\frac{9x}{100}\)

5000 = \(\frac{109x}{100}\)

x = \(\frac{5000×100}{109}\) = 4587.15 ∼ 4500

8. The cost of 1 kg of flour is Rs. 30 and 1 kg of sugar Rs. 35. What will be the bill amount if 10 kg of sugar and 15 kg of flour is bought with 5% of sales tax added?

a) Rs 866.25

b) Rs 890.98

c) Rs 865.56

d) Rs 834.56

Answer: a

Explanation: Cost of 1 kg flour = Rs. 30

Cost of 10 kg sugar = 30 × 10 = 300

Cost of 1 kg sugar = Rs. 35

Cost of 15 kg sugar = 35 × 15 = 525

Total cost price = 525 + 300 = 825

Sales tax = 5%

= 5% of 825

= \(\frac{5}{100}\) × 825 = 41.25

Buying price = cost price + sales price = 825 + 41.25 = Rs. 866.25

9. What will be the bill amount of two sets of TV at Rs. 50000 each with a 7% of sales tax included?

a) Rs. 117000

b) Rs. 107000

c) Rs. 128000

d) Rs. 135000

Answer: c

Explanation: Cost of 1 TV set = Rs. 50000

Cost of 2 TV sets = 50000 × 2 = 100000

Sales tax = 7%

= 7% of 100000

= \(\frac{7}{100}\) × 100000 = 7000

Buying price = cost price + sales price = 100000 + 7000 = Rs. 107000

10. If 6% VAT is added in the price of a lawn mower at Rs. 15000 then what is the original price of the lawn mower?

a) Rs 19000

b) Rs 16000

c) Rs 15000

d) Rs 14000

Answer: d

Explanation: Let the original price of lawn mower be Rs x.

Sales tax = 6%

= 6% of x

= \(\frac{6x}{100}\)

Buying price = cost price + sales price

15000 = x + \(\frac{6x}{100}\)

15000 = \(\frac{106x}{100}\)

x = \(\frac{15000 × 100}{106}\) = 14150.94 ∼ 14000

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Compound Interest”.

1. Calculate the simple interest if the principal amount is 50000 and the rate is 2% for 4 years.

a) 4000

b) 400

c) 40000

d) 40

Answer: a

Explanation: S.I. = PNR/100

⇒ S.I. = 50000 × 2 × 4/100 = 4000.

2. Find the Compound Interest on Rs. 1000 for two years at 2% per annum.

a) 20

b) 20.5

c) 20.4

d) 20.6

Answer: c

Explanation: Principal for the first year = Rs. 1000

Interest for the first year = 1000 × 2 × 1/100 = Rs. 20

Amount at the end of one year = 1000 + 20 = Rs. 1020

Interest for second year = 1020 × 2 × 1/100 = Rs. 20.4

Principal for the second year = Rs. 1020

Amount at the end of one year = 1000 + 20. 4 = Rs. 1040.4

C.I. = Amount – Principal = 1040.4 – 1020 = Rs. 20.4.

3. Evaluate the compound interest on Rs. 10101 for 3 years at the rate of 9% per annum compounded annually.

a) 2980

b) 30000

c) 10101

d) 33333

Answer: a

Explanation: A = P 

 

\) n

⇒ A = 10101 

 

 3 = 10101 

 

 3 = Rs. 13081.08

C.I. = A – P = 13081.08 – 10101 = Rs. 2980.08.

4. Vidhya lent Rs. 5000 to Kavya for 3 years at the rate of 5% per annum compound interest. Calculate the amount that Vidhya will get after 3 years.

a) 5789

b) 5788.12

c) 5788.13

d) 5788

Answer: c

Explanation: A = P 

 

\) n

Amount for 3 years = 5000 

 

 3 = 5000 

 

 3 = 5788.13.

5. A farmer gets a loan of Rs. 100000 against his fixed deposits. If the rate of interest is 1.5 paise per rupee per annum, calculate the compound interest payable after 2 years.

a) 22250

b) 42250

c) 52250

d) 32250

Answer: d

Explanation: R = 1.5 paise per rupee per annum = 1.5 × 100 paise per hundred rupee per annum

= 1.5 × \

 

 

 2 = 100000 

 

 2 = 132250

C.I. = 132250 – 100000 = Rs. 32250.

6. Calculate the compound interest at the rate of 6% per annum for 2 years on the principle which in 2 years at the rate of 2% per annum gives Rs. 8000 as simple interest.

a) 50000

b) 49440

c) 59440

d) 49000

Answer: b

Explanation: P = 8000 × \

 

 

 2 = 400000 

 

 2 = Rs. 449440

C.I. = 449440 – 400000 = Rs. 49440.

7. Compute the compound interest on Rs. 16000 for 2 years 10% per annum when compounded half yearly.

a) 18600

b) 17640

c) 18640

d) 17600

Answer: b

Explanation: A = P 

 

 2n

⇒ A = 16000 

 

 2 = 16000 

 

 2 = Rs. 17640.

8. Find the amount on Rs. 5000 at the rate of 20% per annum for 18 months when interest is compounded half yearly.

a) 6644

b) 6666

c) 6000

d) 6655

Answer: d

Explanation: n = 18/12 = 3/2

A = P 

 

 2n = 5000 

 

 2n = Rs. 6655.

9. If the amount is Rs. 400 and Principal is Rs. 100 which is compounded half yearly for 1 year, calculate the rate of interest.

a) 10

b) 200

c) 2

d) 20

Answer: b

Explanation: A = P 

 

 2n

⇒ 400 = 100 

 

 2

⇒ 2 = 1 + \(\frac{R}{200}\)

R = 200.

10. Calculate the compound interest on Rs. 4000 for 2 years at 20% per annum when compounded annually.

a) 1856.4

b) 1756.4

c) 1846.4

d) 1746.4

Answer: a

Explanation: A = P 

 

 2n = 4000

 

 4 = Rs. 5856.4

C.I. = 5856.4 – 4000 = Rs. 1856.4.

This set of Mathematics Aptitude Test for Schools focuses on “Rate Compounded Annually or Half Yearly”.

1. What will be the amount to be paid at the end of 7 years on Rs. 3500 at 2% per annum compounded annually?

a) 4428.5

b) 3456.89

c) 4229.56

d) 4020.39

Answer: d

Explanation: Principal = Rs. 3500

Rate = 2%

Time = 7 years

Amount = P 

 

 t = 3500 

 

 7 = 3500 

 

 7 = 3500 × 1.148 = 4020.39

2. What will be the amount to be paid at the end of 3 year on Rs. 500 at 10% per annum compounded half-yearly?

a) 657.12

b) 638.14

c) 638.15

d) 645.78

Answer: b

Explanation: Interest is compounded half-yearly

Principal = Rs 500

Rate = 10%

Time = 3 year

Since, interest is charged half-yearly,

Rate = 9 %

= \

 

 

 t

= 500 

 

 6

= 500 

 

 5

= 638.14

3. A man took loan of Rs. 65,250 from HSBC Bank. If the rate of interest is 9% p.a., then what will be the difference in the amounts he would be paying after 1 year if the interest is compounded quarterly and compounded annually?

a) 72.4

b) 70

c) 69.05

d) 65.5

Answer: c

Explanation: Case 1: Interest is compounded half-yearly

Principal = Rs 65250

Rate = 9%

Time = 1 year

Since, interest is charged half-yearly,

Rate = 9 %

= \

 

 

 t

= 65250 

 

 2

= 65250 

 

 2

= 71254.63

Case 2: Interest is compounded quarterly

Principal = Rs 65250

Rate = 9%

Time = 1 year

Since, interest is charged quarterly,

Rate = 9 %

= \

 

 

 t

= 65250 

 

 4

= 65250 

 

 4

= 71323.68

Difference in the amounts in both cases = 71323.68 – 71254.63 = 69.05

4. What will be the time period for a sum taken for 3 years at 12% p.a. compounded half-yearly?

a) 6

b) 3

c) 3.5

d) 4

Answer: a

Explanation: The interest is charged half-yearly

So, the time will be multiplied by 2

Time = 3

= 2 × 3 = 6 half – yearly

The interest will be charged 6 times in 3 years.

5. What will be the rate for a sum taken for 5 years at 8% p.a. compounded half-yearly?

a) 8%

b) 5%

c) 4%

d) 7%

Answer: c

Explanation: The interest is charged half-yearly

So, the rate will be divided by 2

Rate = 8%

= \(\frac{8}{2}\) = 4%

6. What will be the time period for a sum taken for 4 years 4 months at 10% p.a. compounded quarterly?

a) \

 

 18

c) 4 years 4 months

d) 1 year

Answer: b

Explanation: The interest is charged quarterly,

So, time is multiplied by 4

Time = 4 years 4 months = 4 \(\frac{1}{3}\) years

= 4 × 4 \(\frac{1}{3}\) = 4 × \(\frac{13}{3}\) = \(\frac{52}{3}\) or 18 quaters approx

7. What will be the rate for a sum taken for 1 ½ years at 10% p.a. compounded quarterly?

a) 2

b) \

 

 1 ½ %

d) 2 ¼ %

Answer: b

Explanation: Since, the interest is compounded quarterly

Rate = 10%

= \(\frac{10}{4}\) = \(\frac{5}{2}\)%

8. A sum is taken for 2 years at 12% per annum. If the interest is compounded every 6 months, how many times will the interest be charged in 1 year?

a) 1

b) 3

c) 2

d) 4

Answer: d

Explanation: The interest is charged every six months,

Rate = 12 %

= \(\frac{12}{2}\) = 6%

Time = 2 years

= 2 × 2 = 4 half – periods

The interest will be charged four times in two years.

9. A sum is taken for 1 ½ years at 10% per annum. If the interest is compounded every 3 months, how many times will the interest be charged in 1 year?

a) 1 ½ years

b) 6

c) 5

d) 4

Answer: b

Explanation: The interest is compounded after every 3 months,

Rate = 10 %

= \(\frac{10}{4}\) = \(\frac{5}{2}\)%

Time = 1 ½ years or \(\frac{3}{2}\) years

= 4 × \(\frac{3}{2}\) = 6 quaters

The interest will be charged six times in one and a half year.

10. What will be the amount to be paid at the end of 18 months on Rs. 12000 at 4% per annum compounded quarterly?

a) 12000

b) 12738.24

c) 13567

d) 13789

Answer: b

Explanation: Principal: Rs. 12000

Rate: 4%

Time: 18 months or 1 ½ year

As the rate is compounded quarterly,

Rate = \

 

 

 

 t

= 12000 

 

 6

= 12000 

 

 6

= 12738.24

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Applications of Compound Interest Formula – 1”.

1. The population of a city is increasing at the rate of 10% per annum. Calculate the population of the city on this basis after 3 years if the current population is 10000.

a) 12100

b) 13100

c) 13310

d) 1210

Answer: c

Explanation: Here, P = Initial population = 10000

R = Rate of growth of population = 10%

n = number of years

Population after 3 years = P 

 

 n = 10000 

 

 3 = 10000 

 

 3 = 13310.

2. The population of the town is 30000. The annual birth rate is 5% and the annual death rate is 3%. Calculate the population after 2 years.

a) 61212

b) 612

c) 312

d) 31212

Answer: d

Explanation: P = Initial population = 30000

R = 5 – 3 = 2%

n = 2

Population after 2 years = P 

 

 n = 30000 

 

 2 = 30000 

 

 2 = 31212.

3. The population of a village was 10000 four years ago. If it had increased by 2%, 2.5%, 3%, 2% in last four years. Find the present population of the village.

a) 267.903

b) 260

c) 266.903

d) 266

Answer: a

Explanation: Present Population = 10000 

 

 

 

 

= 10000 

 

 

 

 

 = 267.903.

4. The population of a sate decreases every year at the rate 4% per annum. The population of the state 3 years ago was 150000. Find present population.

a) 576000

b) 138240

c) 57600

d) 13824

Answer: b

Explanation: Population three years ago = 150000

Rate of decrease of population = 4% per annum

Present population = 150000 

 

 3 = 150000 

 

 2 = 138240.

5. If the population of a city has been increasing at the rate of 10%. The present population of the city is 9680000. Find its population 2 years ago.

a) 800000

b) 8000

c) 80000

d) 8000000

Answer: d

Explanation: Let the population 2 years ago be P.

Present Population = P 

 

 2

9680000 = P 

 

 2

P = 8000000.

6. The production of spare parts rose to 25000 from 16000 in 2 years. Find the rate of growth per annum.

a) 25

b) 20

c) 250

d) 200

Answer: a

Explanation: Previous production = 16000

Present production = 25000

25000 = 16000 

 

 2

\(\frac{5}{4}\) = 1 + \(\frac{R}{100}\)

R = 25.

7. The bacteria in a culture grows by 5% in the first hour, decreases by 10% in the second hour and increases by 2% in the third hour. If the original count of the bacteria is 50000, Calculate the count of bacteria at the end of 3 hours.

a) 48000

b) 48195

c) 24097

d) 24095

Answer: b

Explanation: P = Original count of bacteria = 50000

Count of bacteria after 3 hours = 50000 

 

 

 

 = 48195.

8. Manish opened a cafe with an initial investment of Rs. 64000. In the first year he incurred the loss of 10%. During second year he gained the profit of 4% and in the third year he gained the profit of 15%. Calculate his net profit gained in the entire three years.

a) 68889.5

b) 4889.6

c) 68889.61

d) 68889.60

Answer: b

Explanation: Initial investment = Rs. 64000

Loss in first year = 10%

Profit in second year = 4%

Profit in third year = 15%

profit = 64000 

 

 

 

 = 64000 

 

 

 

\) = 68889.60.

Net profit = 68889.60 – 64000 = 4889.6.

9. 900 workers were employed to construct a dam in four years. At the end of first year, 5% workers were retrenched. At the end of the second year 2% of the workers were retrenched. To complete the construction in time, the number of workers were increased to 10% at the end of third year. How many workers were working at the end of fourth year.

a) 945

b) 1000

c) 921

d) 989

Answer: c

Explanation: Initial number of workers = 900

Reduction of workers at the end of first year = 5%

Reduction of workers at the end of second year = 2%

Increase of workers at the end third year = 10%

Number of workers working during fourth year = 900 

 

 

 

= 900 

 

 

 

\) = 921.69 ≅ 921 workers.

10. An engine has initially 24000 ml oil. The amount of oil has to be increased at the rate of 5% every six month. Find the time-period at the end of which the amount of oil becomes 27783ml.

a) n = \

 

 n = \

 

 n = \

 

 n = \(\frac{1}{2}\)

Answer: a

Explanation: P = Initial amount of oil = 24000 ml

R = rate of increase = 5% every 6 months = 10% per annum.

Let the total time be n years. Then,

A = P 

 

^{2n}\)

27783 = 24000 

 

^{2n}\)

\

 

 

^{2n}\)

 

 3 = 

 

^{2n}\)

2n = 3

n = \(\frac{3}{2}\)

Hence, required time-period is \(\frac{3}{2}\) years.

This set of Mathematics Objective Questions and Answers for Class 8 focuses on “Applications of Compound Interest Formula – 2”.

1. A company increased the production of cars from 15625 in 2002 to 27000 in 2005. Find the annual rate of growth of production of the cars.

a) 15%

b) 2%

c) 10%

d) 20%

Answer: d

Explanation: Let the annual rate of growth be R% per annum. Then,

⇒ 27000 = 15625 

 

^3\)

⇒ \

 

 

^3\)

⇒ 

 

^3\) = 

 

^3\)

⇒ \(\frac{30}{25}\) = 1 + \(\frac{R}{100}\)

⇒ \(\frac{1}{5}\) = \(\frac{R}{100}\)

⇒ R = 20%.

2. Fe-57 decays at the constant rate in such a way that it reduces to 50% in 5568 years. Find the age of an old rod which the iron  is only 12.5% of the original.

a) 16700

b) 16705

c) 16704

d) 16000

Answer: c

Explanation: Let the rate of decay be R% per annum and the age of rod be n years. Let the original amount of iron in the rod be P. Then in 5568 years the amount left is \

 

 

 

^{5568}\)

⇒ \

 

 

^{5568}\)

After n years, the iron left in the rod is 12.5% of P

⇒ \

 

 

^n\)

⇒ \

 

 

^n\)

⇒ 

 

^3\) = 

 

^n\)

⇒ 

 

^{16704}\) = 

 

^n\)

⇒ n = 5568 × 3 = 16704 years.

Hence the age of the rod is 16707 years.

3. The present population of a village is 10000. If it increases at the rate of 2% per annum, find the population after 2 years.

a) 10404

b) 10004

c) 10402

d) 10400

Answer: a

Explanation: Initial population = 10000

⇒ Let after 2 years be P.

⇒ P = 10000 

 

^2\) = 10000 

 

^2\) = 10404.

4. The initial population of a country is 1000000. If the birth rate and the death rate is 10% and 4% respectively, then find the population of the country after 2 years.

a) 1126000

b) 1023600

c) 1123000

d) 1123600

Answer: d

Explanation: Initial population = 1000000

Increment in population = 10%

Decrement in population = 4%

Net growth in population = 10-4 = 6%

Population after 2 years = 1000000 

 

^2\) = 1000000 

 

^2\) = 1123600.

5. Two years ago the population of a village was 4000. If the annual increase during the two successive years be at the rate of 2% & 4%, find the present population.

a) 7074

b) 7080

c) 7072

d) 7000

Answer: c

Explanation: Present population = 4000 

 

 

 = 4000 

 

 

 = 7072.

6. The annual rate of growth in population of a certain city is 10%. If its present population is 532400, find the population 3 years ago.

a) 404000

b) 300000

c) 400400

d) 400000

Answer: d

Explanation: Population 3 years ago = P 

 

^3\)

⇒ 532400 = P 

 

^3\).

7. Vidhya started a business with initial investment of 500000. She incurred a loss of 2% in the first year and in the second year she gained the profit of 10%. Calculate her net profit earned.

a) 39000

b) 38000

c) 41000

d) 40000

Answer: a

Explanation: Initial investment = 500000

Profit = 500000 

 

 

 = 500000 

 

 

 = 539000

⇒ Net profit earned = 500000 – 539000 = 39000.

8. An apartment of two floors is constructed at the cost of Rs. 5000000. It is depreciating at the rate of 20% per annum. Find its value 3 years after construction.

a) 2400000

b) 2460000

c) 2500000

d) 2560000

Answer: d

Explanation: V 0 = Rs. 5000000

Value after 3 years = V 0 

 

^3\) = 5000000 

 

^3\) = 5000000 

 

^3\) = Rs. 2560000.

9. A new two-wheeler costs Rs. 90000. Its price depreciates at the rate of 10% a year during first two years and the rate of 25% a year thereafter. Find the cost of the two-wheeler after 3 years.

a) 55000

b) 546750

c) 50460

d) 540000

Answer: b

Explanation: Cost of the vehicle = Rs 90000

Rate of depreciation in the first two years = 10%

Rate of depreciation in the second year = 25%

⇒ Price of the vehicle after 3 years = 90000 

 

^2\) 

 

 = 90000 

 

^2\)

 

 = Rs. 546750.

10. The present price of a cycle is Rs 8000. If its value is decreasing every year by 10% then find the price before 3 years.

a) 10977

b) 10976

c) 10975

d) 10974

Answer: d

Explanation: Let the price of the cycle be Rs. P before 3 years. Then its present value is Rs P 

 

^3\)

But the present price is Rs. 8000

⇒ 8000 = P 

 

^3\)

⇒ 8000 = P 

 

^3\)

⇒ P = 10974.

11. The value of a gadget worth Rs.10000 is depreciating at the rate of 10% per annum. In how many years will its value be reduced to Rs 6561?

a) 5

b) 3

c) 4

d) 2

Answer: c

Explanation: Present value = Rs. 10000

Depreciated value = Rs. 6561

Rate of depreciation = 10%

⇒ 6562 = 10000 

 

^n\)

⇒ \

 

 

^n\)

⇒ 

 

^n\) = 

 

^n\)

⇒ n = 4.

12. The value of a land increases every year at the rate of 8%. If its value at the end of 3 years be Rs 6000000, find its original value at the beginning of these years.

a) 4762994

b) 3762994

c) 3000000

d) 4000000

Answer: a

Explanation: R = 8% and n = 3

⇒ 6000000 = P 

 

^3\)

⇒ 6000000 = P 

 

^3\)

⇒ P = 4762994.

13. Max bought a gadget for Rs. 21000. If the cost of the gadget after 2 years depreciates to Rs. 18162, find the rate of depreciation.

a) 20%

b) 5%

c) 15%

d) 7%

Answer: d

Explanation: Let R be the rate of depreciation

⇒ 18162 = 21000 

 

^2\)

⇒ \

 

 

^2\)

⇒  2 = 

 

^2\)

⇒  = 

 

⇒ R = 7%.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Algebraic Expressions – Monomials, Binomials and Polynomials”.

1. Identify the given expression: 3x + 5x.

a) Binomial

b) Monomial

c) Trinomial

d) Four-term polynomial

Answer: b

Explanation: An algebraic expression is an expression which consists variables, constants along with algebraic operations like addition, subtraction, multiplication. An algebraic expression containing a single term is a monomial. Example: 9y, 5xy, etc. Now, from the given expression, 3x + 5x = 8x, which is a single term.

2. Identify the given expression: 4xy.

a) Monomial

b) Binomial

c) Trinomial

d) Four-term polynomial

Answer: a

Explanation: Algebraic expression is an expression which consists variables, constants along with algebraic operations like addition, subtraction, multiplication. An algebraic expression containing a single term is a monomial. Example: 9y, 4xy, etc.

3. Identify the expression: 9xy + 4x.

a) Monomial

b) Trinomial

c) Four-term polynomial

d) Binomial

Answer: d

Explanation: Algebraic expression is an expression which consists variables, constants along with algebraic operations like addition, subtraction, multiplication. An algebraic expression containing two terms is a binomial. Example: 9y + 7xy, 10x – 5y, etc.

4. Identify the expression: 9x – 6y + 2xy.

a) Monomial

b) Binomial

c) Trinomial

d) Four-term polynomial

Answer: c

Explanation: Algebraic expression is an expression which consists variables, constants along with algebraic operations like addition, subtraction, multiplication. An algebraic expression containing three terms is a trinomial. Example: 9y + 7xy – 2x, 8x + 7y + 20z, etc.

5. Identify the term of the expression: 3x 2 + 4xy.

a) 3x 2 , 4xy

b) 3x 2

c) 4xy

d) 3, 4

Answer: a

Explanation: A term is the part of the algebraic expression. The given expression has two terms.

6. Identify the all coefficients of the expression: x 2 – 12xy + 2.

a) 1, -12, 2

b) -1, 12, -2

c) 1

d) -12

Answer: d

Explanation: In a term of an algebraic expression, constant value before any variable term is called coefficient. The given expression can be written as 1x 2 – 12xy + 2. So, 1 and -12 are coefficients of x 2 and xy respectively.

7. Identify all the terms and coefficients of the expression: y 3 – 10x 2 y + 14x.

a) Terms: y 3 , -10x 2 y, 14x

Coefficients: 1, -10, 14

b) Terms: y 3 , -10x 2 y, 14x

Coefficients: 1, 10, 14

c) Coefficients: y 3 , -10x 2 y, 14x

Terms: 1, -10, 14

d) Coefficients: y 3 , -10x 2 y, 14x

Terms: 1, 10, 14

Answer: a

Explanation: A term is the part of the algebraic expression. In a term of an algebraic expression, constant value before any variable term is called coefficient.

8. Which of the following expressions have -1, 2 as the coefficients?

a) 12 – x

b) -x + 2y

c) x – 2y

d) xy

Answer: b

Explanation: A term is the part of the algebraic expression. In a term of an algebraic expression, constant value before any variable term is called coefficient. Among the given expressions, -x + 2y has -1 and 2 as coefficients.

9. Which of the following expressions have -x, x 2 as the terms?

a) x

b) x 3 – x

c) x 2

d) x 2 – x

Answer: d

Explanation: Algebraic expression is a combination of terms along with algebraic operations like addition, subtraction, multiplication. Example: 9y + 7xy – 2x is an expression with three terms, 8x + 7y is an expression with two term, etc.

10. Identify the coefficients of the following expression: x 2-a + 2y.

a) 1, 2

b) x 2-a

c) -2

d) -1, -2

Answer: a

Explanation: Algebraic expression is a combination of terms along with algebraic operations like addition, subtraction, multiplication. In a term of an algebraic expression, constant value before any variable term is called coefficient. The given expression can be written as 1x 2-a + 2y with 1 and 2 as coefficients of x 2-a and y respectively.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Like and Unlike Terms, Addition and Subtraction of Algebraic Expressions”.

1. Identify the like terms: 3ab – 4b 2 – 6ab.

a) 3ab, -6ab

b) -4b 2 , 3ab

c) 3ab, 6ab

d) -4b 2 , 6ab

Answer: a

Explanation: An algebraic expression is combination of terms with algebraic operations like addition, subtraction, multiplication. The terms having same literal factors are called like terms. Like terms are formed from same variables and the powers of the variables are similar. But, coefficients of the like terms can be different. Eg: 2x, 3x form a pair of like terms; 5ax, -ax form a pair of like terms.

2. Identify the like terms: 4xy – 3y + 44y.

a) -3y, 44y

b) 4xy, 44y

c) 44y, -3y

d) 3y, 44y

Answer: c

Explanation: An algebraic expression is combination of terms with algebraic operations like addition, subtraction, multiplication. The terms having same literal factors are called like terms. Like terms are formed from same variables and the powers of the variables are similar. But, coefficients of the like terms can be different. Eg: 2x, 3x form a pair of like terms; 5ax, -ax form a pair of like terms.

3. Identify all the unlike terms: 2a – 3b – 4c.

a) 2a, 4c

b) 2a, 3b, 4c

c) 2a, -3b, 4c

d) 2a, -3b, -4c

Answer: d

Explanation: An algebraic expression is combination of terms with algebraic operations like addition, subtraction, multiplication. The terms which are not having same literal factors are called unlike terms. Unlike terms have different variables and the powers of the variables can be different. But, coefficients of the like terms can be different. Eg: 3x, 4yx form a pair of like terms; 5ax, -ax form a pair of like terms.

4. Identify the unlike terms: 3a 2 b + 4b 2 a – ba 2 .

a) 3a 2 b, 4b 2 a

b) 3a 2 b, -ba 2

c) 3a 2 b, ba 2

d) 4b 2 a, -ba 2

Answer: b

Explanation: An algebraic expression is combination of terms with algebraic operations like addition, subtraction, multiplication. The terms which are not having same literal factors are called unlike terms. Unlike terms have different variables and the powers of the variables can be different. But, coefficients of the like terms can be different. Eg: 3x, 4yx form a pair of like terms; 5ax, -ax form a pair of like terms.

5. What will be the result if the expressions 2x – 4x 2 y and 16y 2 x + 2x 2 y are added?

a) 2x – 4x 2 y + 18x 2 y

b) 2x – 2x 2 y + 16y 2 x

c) 2x – 2x 2 y – 16y 2 x

d) 2x + 4x 2 y + 18x 2 y

Answer: b

Explanation: The terms of an algebraic expression having same literal factors are called like terms and those terms with different literal factors are called unlike terms. While addition or subtraction of polynomials, like terms are added/subtracted first and then unlike terms are handled. In the given expressions, -4x 2 y and 2x 2 y are like terms.

⇒ (2x – 4x 2 y) + (16y 2 x + 2x 2 y) = 2x – 4x 2 y + 2x 2 y + 16y 2 x 

⇒ (2x – 4x 2 y) + (16y 2 x + 2x 2 y) = 2x – 2 x 2 y + 16y 2 x .

6. Find the value the following expression: 1/6x + 2/5xy – 13y – 7/5xy + 1/6x.

a) 2/3x – 9/5xy – 13y

b) 2/3x + 1xy – 13y

c) 2/3x – 1xy – 13y

d) -1xy – 13y

Answer: c

Explanation: 1/6x + 2/5xy – 13y – 7/5xy + 1/6x = 1/6x + 1/6x + 2/5xy – 7/5xy – 13y 

= 2/3x – 1xy – 13y .

7. Subtract the following expression: 22ab – 30a 2 b + 2b from 50a 2 b – 6b 2 a – 5b.

a) 22ab – 80a 2 b – 6b 2 – 7b

b) 80a 2 b – 6b 2 + 7b – 22ab

c) 80a 2 b – 6b 2 – 7b – 22ab

d) 80a 2 b + 6b 2 – 7b

Answer: c

Explanation: (50a 2 b – 6b 2 a – 5b) – (22ab – 30a 2 b + 2b) = 50a 2 b – 6b 2 a – 5b – 22ab + 30a 2 b + 2b

= 50a 2 b + 30a 2 b – 6b 2 – 5b – 2b 

= 80a 2 b – 6b 2 – 7b22ab .

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Multiplication of Algebraic Expressions”.

1. Find the value when the given polynomials 3p and (24x + 14y 2 ) are multiplied.

a) 72px + 42y 2

b) 72px + 42y 2 p

c) 72px + 42yp

d) 114py 2 x

Answer: b

Explanation: In the given question, a monomial is multiplied with a binomial.

3p(24x + 14y 2 ) = 3p + 3p(42y 2 )

= 72px + 42y 2 p.

2. What is the resultant polynomial obtained when the given polynomials (a 3 ) and (51cb 2 + 14abc) are multiplied?

a) 51cb 2 a 3 + 14a 4 bc

b) 51a 3 cb + 14a 4 bc

c) 51cba 2 + 14a 3 bc

d) 51abc + 14a 3 bc

Answer: a

Explanation: In the given question, a monomial is multiplied with a binomial.

a 3 (51cb 2 + 14abc) = 51cb 2 (a 3 ) + 14abc(a 3 )

= 51cb 2 a 3 + 14a 4 bc.

3. Find the product of the polynomials 82x and (2y + 2x 2 yz + 5x).

a) 164(xy + x 3 yz) + 410x

b) 164(xy 2 + x 3 yz) + 410x 2

c) 164(xy + x 2 yz) + 410x 2

d) 164(xy + x 3 yz) + 410x 2

Answer: d

Explanation: In the given question, a monomial is multiplied with a trinomial.

82x(2y + 2x 2 yz + 5x) = 82x + 82x(2x 2 yz) + 82x

= 164(xy + x 3 yz) + 410x 2 .

4. Find the product of the polynomials 2pq and .

a) 18pqx + 4pqxy + 4px

b) 22pqx + 4pqxy

c) 18pqx + 4pqxy + 4pq

d) 22pqx + 4pqx

Answer: b

Explanation: In the given question, a monomial is multiplied with a trinomial.

2pq = 2pq + 2pq +2pq

= 18pqx + 4pqxy + 4pqx

= 22pqx + 4pqxy.

5. What is the value obtained when [a] is subtracted from 2b?

a) 28ab

b) 48ab + 28bc

c) 28bc

d) -48ab + 28bc

Answer: c

Explanation: In the given question, the terms are operated according to BODMAS Rule.

[2b – a] = [24ab + 16bc – 24ab + 12ac] = 28bc.

6. What is the value obtained when 48x 2 – 22z + 13xz is subtracted from 12x?

a) 24xy – 13xz + 144x 2

b) 24xy + 13xz – 144x 2

c) 24xy + 13xz + 144x 2

d) 24xy – 13xz – 144x 2

Answer: d

Explanation: In the given question, the terms are operated according to BODMAS Rule.

[12x – (48x 2 – 22z + 13xz)] = 24xy – 96x 2 – 48x 2 + 22z – 13xz = 24xy – 13xz – 144x 2 .

7. Find the polynomial obtained by adding 2q and 14s.

a) 488qp + 216pqs – 196s 2

b) 488qp – 216pqs – 196s 2

c) 488qs – 216pqs – 196s 2

d) 488qs + 216pqs – 196s 2

Answer: c

Explanation: In the given question, the terms are operated according to BODMAS Rule.

2q + 14s = 180sq – 216pqs + 308qs – 196s 2

= 488qs – 216pqs – 196s 2 .

8. What is the resultant polynomial if 12x 2 y + 22 and (24xy – 2x 2 )4y are added?

a) 4x 2 y – 96x 2 + 22

b) 108x 2 y + 4x 2 + 22

c) 108x 2 y + 4x 2 – 22

d) 4x 2 y + 96x 2 + 22

Answer: d

Explanation: In the given question, the terms are operated according to BODMAS Rule.

12x 2 y + 22 + (24xy – 2x 2 )4y = 12x 2 y – 8 x 2 y + 96x 2 + 22 = 4x 2 y + 96x 2 + 22.

9. Find the result of the addition of the polynomials 12x(4y – 16y 2 ) and  (12y + x 2 ).

a) 48xy – 192xy 2 + 408y 2 – 34x 2 y – 264y – 22x 2

b) 48xy – 158xy 2 + 408y 2 – 264y – 22x 2

c) 48xy – 192xy 2 + 408y 2 + 34x 2 y – 264y – 22x 2

d) 48xy + 192xy 2 + 408y 2 + 34x 2 y – 264y – 22x 2

Answer: c

Explanation: In the given question, the terms are operated according to BODMAS Rule.

12x(4y – 16y 2 ) +  (12y + x 2 ) = 48xy – 192xy 2 + 408y 2 + 34x 2 y – 264y – 22x 2 .

10. What is the value attained when (6w 2 + 4)  is subtracted from  (4w 2 – z)?

a) 124w 2 z + 85z + 2z 2 – 3z

b) 140w 2 z – 24w 2 + 91z – 2z 2 – 8

c) 124w 2 z – 85z + 2z 2 – 3z

d) 140w 2 z – 24w 2 – 91z – 2z 2 – 8

Answer: a

Explanation: In the given question, the terms are operated according to BODMAS Rule.

(6w 2 + 4)  –  (4w 2 – z) = 132w 2 z – 12w 2 + 88z – 8 – [8w 2 z – 2z 2 – 12w 2 + 3z]

= 132w 2 z – 8w 2 z – 12w 2 + 12w 2 + 88z – 3z – 8 + 2z 2 = 124w 2 z + 85z + 2z 2 – 3z.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Identity”.

1. Select the term which is not an identity.

a)  2 = a 2 + b 2

b)  2 = a 2 + b 2 + 2ab

c)  2 = a 2 + b 2 – 2ab

d)  = a 2 + 3a + 2

Answer: a

Explanation: Identities are the special type of equations which are true for every value of the variables  2 = a 2 + b 2 is only true for a = 1, b = 0; a = 0, b = 1.

2. Find the value of  2 using standard identity.

a) 4a 2 + 2 + 2a

b) 4a 2 + 2 – 4a

c) 4a 2 – 2 + 4a

d) 4a 2 + 2 + 4a

Answer: d

Explanation: Using the identity  2 = a 2 + b 2 + 2ab

⇒  2 =  2 +  2 + 2

⇒  2 = 4a 2 + 2 + 4a.

3. Find the value of  2 using the standard identity.

a) m 2 + 4n 2 – 8mn

b) m 2 + 16n 2 + 4mn

c) m 2 + 16n 2 + 8mn

d) m 2 + 4n 2 + 8mn

Answer: c

Explanation: We know that  2 = a 2 + b 2 + 2ab.

⇒  2 =  2 +  2 + 2

⇒  2 = m 2 + 16n 2 + 8mn.

4. Calculate the value of  2 .

a) 121x 2 + 4y 2 – 22xy

b) 121x 2 + 4y 2 – 44xy

c) 121x 2 + 4y 2 + 44xy

d) 121x 2 + 4y 2 + 22xy

Answer: b

Explanation: The expression is of the form  2 where, a = 11x and b = 2y and  2 = a 2 + b 2 – 2ab.

⇒  =  2 +  2 – 2

⇒  = 121x 2 + 4y 2 – 44xy.

5. Evaluate the value of  2 .

a) 4q 2 + 9p 2 + 12pq

b) 4q 2 + 9p 2 – 12pq

c) 4q 2 + 9p 2 + 6pq

d) 4q 2 + 9p 2 – 6pq

Answer: b

Explanation: The expression is of the form  2 where, a = 2q and b = 3p and  2 = a 2 + b 2 – 2ab.

⇒  2 =  2 +  2 – 2

⇒  2 = 4q 2 + 9p 2 – 12pq.

6. Using the standard identity, find the value of  2 .

a) 15.11

b) 16.11

c) 15.21

d) 16.21

Answer: c

Explanation: 3.9 can be written as  and  2 = a 2 + b 2 – 2ab.

⇒  2 =  2 +  2 – 2

⇒  2 = 16 + 0.01 – 0.8 = 15.21.

7. Find the value of  2 .

a) 152.77

b) 153.66

c) 153.76

d)152.76

Answer: c

Explanation: We can use the identity  2 = a 2 + b 2 + 2ab.

⇒  2 =  2 +  2 + 2

⇒  2 = 144 + 0.16 + 9.6 = 153.76.

8. Using the standard identity, find the value of .

a) 4m 2 + 10m + 5

b) 4m 2 + 10m + 6

c) 4m 2 + 10m – 6

d) 4m 2 + 10m – 5

Answer: b

Explanation: The expression is of the form . Now,  = x 2 + x + ab.

⇒  =  2 + 2m + 

⇒  = 4m 2 + 10m + 6.

9. Evaluate the value .

a) x 2 + 13x – 13

b) x 2 + 13x + 13

c) x 2 + 13x – 12

d) x 2 + 13x + 12

Answer: d

Explanation: The expression is of the form . Now,  = x 2 + x + ab.

⇒  = x 2 + x + 12

⇒  = x 2 + 13x + 12.

10. Using the standard identity, find the value of 201 × 204.

a) 4104

b) 401004

c) 40104

d) 41004

Answer: d

Explanation: 201 and 204 can be written as  and  respectively.  = x 2 + x + ab

⇒  =  2 + 200 + 

⇒  = 40000 + 1000 + 4 = 41004.

11. Calculate the value of 904 × 902.

a) 815408

b) 8105408

c) 810548

d) 81548

d) 815400

Answer: a

Explanation: 904 and 902 can be written as  and  respectively. Also,  = x 2 + x + ab.

⇒  =  2 + 900 + 4

⇒  = 810000 + 5400 + 8 = 815408.

12. Find the value of .

a) 1000190857

b) 1000019857

c) 10019857

d) 100019857

Answer: d

Explanation: We know that  = a 2 – b 2 .

⇒  2 –  2 = 100020001 – 144

⇒  2 –  2 = 100019857.

13. Find the value of  2 –  2 .

a) 9600

b) 960

c) 96000

d) 900

Answer: a

Explanation: a 2 – b 2 = 

⇒  2 –  2 = 

⇒  2 –  2 = 100 = 9600.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Visualising Solid Shapes – Faces Edges and Vertices”.

1. Name all the adjacent angles of the given quadrilateral.

a) ∠A, ∠C; ∠B, ∠D

b) ∠A, ∠C

c) ∠A, ∠B; ∠A, ∠D; ∠B, ∠C; ∠C, ∠D

d) ∠A, ∠B; ∠A, ∠D

Answer: c

Explanation: Two angles of a quadrilateral which have a common side as an arm are called adjacent angles. ∠A, ∠B; ∠A, ∠D; ∠B, ∠C; ∠C, ∠D are the adjacent angles of the given quadrilateral.

2. Which of the following are the opposite angles of the given quadrilateral?

a) ∠A, ∠B; ∠A, ∠D; ∠B, ∠C; ∠C, ∠D

b) ∠A, ∠C; ∠B, ∠D

c) ∠A, ∠B

d) ∠A, ∠D; ∠B, ∠C

Answer: b

Explanation: Two angles of a quadrilateral which are not adjacent angles are opposite angles.

∠A, ∠C; ∠B, ∠D are the opposite angles of the quadrilateral.

3. Match the following:

     A            B

   1. Adjacent angles      a. ∠X, ∠Y

   2. Opposite angle       b. ∠Y, ∠Z

              c. ∠A, ∠Y     

              d. ∠A, ∠X  

a) 1-a, b, d; 2-c

b) 1-a, b; 2-d

c) 1-b, d; 2-c

d) 1-a, b, c; 2-d

Answer: a

Explanation: When two angles of a quadrilateral have a common side as an arm, they are called adjacent angles and the non-adjacent angles of a quadrilateral are opposite angles.

4. Match the following:

     A            B

    1. Adjacent sides       a. KL, LM

    2. Opposite sides       b. KL, MN

              c. KN, LM

              d. MN, NK 

a) 1-b, c; 2-a, d

b) 1-b, d; 2-a, c

c) 1-a, b; 2-c, d

d) 1-a, d; 2-b, c

Answer: d

Explanation: Two sides of a quadrilateral having a common end point are called adjacent sides of the quadrilateral. Two sides of a quadrilateral which are not having a common end point are called opposite sides of the quadrilateral.

5. Match the pairs:

     A            B

   1. Adjacent sides        a. ∠D, ∠B

   2. Opposite Angle       b. AB, DA

   3. Opposite Side        c. ∠D, ∠C

   4. Adjacent Angle      d. AB, DC 

a) 1-c, 2-a, 3-d, 4-b

b) 1-b, 2-c, 3-d, 4-a

c) 1-b, 2-a, 3-d, 4-c

d) 1-c, 2-a, 3-d, 4-b

Answer: c

Explanation: Two sides of a quadrilateral having a common end point are called adjacent sides of the quadrilateral. Two sides of a quadrilateral which are not having a common end point are called opposite sides of the quadrilateral. Two angles of a quadrilateral which have a common side as an arm are called adjacent angles. Two angles of a quadrilateral which are not adjacent angles are opposite angles.

6. Which are the interior and exterior points of the quadrilateral?

a) Interior point: L, Exterior point: M

b) Interior Point: P, Exterior point: O

c) Interior point: O, Exterior point: K

d) Interior point: L, Exterior point: O

Answer: a

Explanation: Points inside the quadrilateral are called interior points of the quadrilateral. Points outside the quadrilateral are called the exterior points of the quadrilateral.

7. Find ∠B using the information given in the figure.

a) 10°

b) 100°

c) 110°

d) 50°

Answer: b

Explanation: By Angle sum property of Quadrilateral,

∠A + ∠B + ∠C + ∠D = 360°

⇒ 110° + ∠B + 120° + 30° = 360°

⇒ ∠B = 100°.

8. If ∠P + ∠Q = 240 and ∠R = 90, find the remaining angle of the quadrilateral PQRS.

a) 30°

b) 40°

c) 50°

d) 60°

Answer: a

Explanation: By Angle sum property of Quadrilateral,

∠P + ∠Q + ∠R + ∠S = 360°

240° + 90° + ∠S = 360°

⇒ ∠S = 30°.

9. Find the value of ∠N from the given figure.

a) 110°

b) 30°

c) 90°

d) 100°

Answer: d

Explanation: Sum of interior angles of a pentagon is 540°

⇒ ∠K + ∠L + ∠M + ∠N + ∠O = 540°

⇒ ∠O = 540° – 440°

⇒ ∠O = 100°.

10. Find the remaining angle of the given figure.

a) 260°

b) 270°

c) 300°

d) 290°

Answer: d

Explanation: Sum of interior angles of a hexagon is 720°

⇒ ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 720°

⇒ ∠F = 720° – 43°

⇒ ∠F = 290°.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Mensuration – Area of Trapezium – 1”.

1. Find the area of the given rectangle.

a) 16 cm 2

b) 100 cm 2

c) 625 cm 2

d) 50 cm 2

Answer: b

Explanation: Area of rectangle = l × b = 25 × 4 = 100 cm 2 .

2. Calculate the area of the given figure.

a) 150cm 2

b) 125cm 2

c) 225cm 2

d) 100cm 2

Answer: c

Explanation: Area of square =  2 =  2 = 225cm 2 .

3. Find the area of the given figure.

a) 20cm 2

b) 50cm 2

c) 15cm 2

d) 25cm 2

Answer: d

Explanation: Area of triangle = \(\frac{1}{2}\) × b × h = \(\frac{1}{2}\) × 10 × 5 = 25cm 2 .

4. What is the area of the given circle? 

a) 78.5cm 2

b) 78cm 2

c) 76.5cm 2

d) 77cm 2

Answer: a

Explanation: Area of circle = π 2 = 3.14 2 = 78.5cm 2 .

5. Find the area of the given parallelogram.

a) 142cm 2

b) 284cm 2

c) 144cm 2

d) 288cm 2

Answer: d

Explanation: Area of parallelogram = b × h = 12 × 24 = 288cm 2 .

6. Match the following: 

 

A B 

i)

a) 616cm 2

ii)

b) 143cm 2

iii)

c) 121cm 2

iv)

d) 5cm 2

a) i – c; ii – d; iii – b; iv – d

b) i – c; ii – d; iii – b; iv – b

c) i – c; ii – d; iii – b; iv – c

d) i – c; ii – d; iii – b; iv – a

Answer: d

Explanation: i) Area of parallelogram = b × h = 11 × 11 = 121cm 2

ii) Area of triangle = \(\frac{1}{2}\) × b × h = \(\frac{1}{2}\) × 2 × 5 = 5cm 2

iii) Area of rectangle = l × b = 11 × 13 = 143cm 2

iv) Area of circle = π 2 = \

 

 2 = 616cm 2 .

7. Which of the following option is correct if a,b and c are the areas of the given figures ,  and  respectively? 

a) a < b < c

b) a > b > c

c) a < b > c

d) b < c > a

Answer: a

Explanation: Area of triangle = \(\frac{1}{2}\) × b × h = \(\frac{1}{2}\) × 2 × 5 = 5cm 2

Area of rectangle = l × b = 5 × 2 = 10cm 2

Area of circle = π 2 = \

 

 2 = 154cm 2 .

8. Find the area of the given figure.

a) 3.14cm 2

b) 31.4cm 2

c) 3140cm 2

d) 314cm 2

Answer: d

Explanation: Area of semicircle = π \

 

 2 = 314cm 2 .

9. Find the area of the Lawn.

a) 500m 2

b) 101m 2

c) 505m 2

d) 504m 2

Answer: c

Explanation: Area of lawn can be found by subtracting the area of inner rectangle from square.

Hence, Area of lawn = Area of square – Area of rectangle =  2 –  = 505m 2 .

10. Calculate the area of the given figure.

a) 14.28cm 2

b) 12.28cm 2

c) 7.14cm 2

d) 15cm 2

Answer: a

Explanation: Area of the given figure = Area of semicircle – Area of triangle = π \(\frac{^2}{2} + \frac{1}{2}\) × b × h

= π\(\frac{^2}{2} + \frac

{1}{2}\) × 4 × 4 = 14.28cm 2 .

This set of Mathematics Aptitude Test for Class 8 focuses on “Mensuration – Area of Trapezium – 2”.

1. Find the area of the given figure. 

a) 26m 2

b) 13.28m 2

c) 26.28m 2

d) 27.28m 2

Answer: c

Explanation: Area of the given figure = Area of square + 2 × semicircle + 2 × triangle =  2 + π \(\frac{^2}{2} + \frac{1}{2}\) × 4 × 2 = 26.28m 2 .

2. What is the area of the given figure? 

 

a) 70m 2

b) 35m 2

c) 74m 2

d) 60m 2

Answer: a

Explanation: Area = Area of rectangle – Area of circle = l × b – π 2 = 16 × 14 – \

 

 2 = 70m 2 .

3. Find the area and perimeter of the given trapezium, respectively.

a) 30m 2 , 15m

b) 15m 2 , 30m

c) 15m 2 , 15m

d) 30m 2 , 30m

Answer: d

Explanation: Area of trapezium = \(\frac{1}{2}\) × (p 1 + p 2 ) × h  [(p 1 + p 2 ) = Σ of ∥sidesoftrapezium]

= \

 

 × 3 = 30m 2

Perimeter = 4 + 8 + 6 + 12 = 30m.

4. Calculate the area of the given figure.

a) 7cm 2

b) 7.5cm 2

c) 6.5cm 2

d) 8cm 2

Answer: b

Explanation: Area of general quadrilateral = \(\frac{1}{2}\) × (h 1 + h 2 ) × d  [d is the length of diagonal]

= \

 

 × 6 = 7.5cm 2 .

5. The Area of trapezium shaped garden is 225m 2 . The distance between the two parallel sides is 25m. The length of one parallel side is 11m, what is the length of the other parallel side?

a) 7m

b) 8m

c) 6m

d) 5m

Answer: a

Explanation: Area of trapezium = \(\frac{1}{2}\) × (p 1 + p 2 ) × h  [(p 1 + p 2 ) = Σ of ∥ sidesoftrapezium]

⇒ 225m 2 = \(\frac{1}{2}\) × (11 + p 2 ) × 25

⇒ p 2 = 7m.

6. The Area of trapezium is 800m 2 . The lenghts of the parallel sides is 28mand 12m. Find the height of the trapezium.

a) 22m

b) 42m

c) 40m

d) 20m

Answer: c

Explanation: Area of trapezium = \(\frac{1}{2}\) × (p 1 + p 2 ) × h  [(p 1 + p 2 ) = Σ of ∥ sidesoftrapezium]

⇒ 800m 2 = \

 

 × h

⇒ h = 40m.

7. In a trapezium shaped ground, there is a square shaped lawn. The side of the lawn is 12m. The length of the parallel sides is 24m and 40m.The height of the ground is 30m. Find the area of the shaded portion.

a) 806m 2

b) 336m 2

c) 1776m 2

d) 816m 2

Answer: d

Explanation: Area of the shaded portion = Area of trapezium – Area of square

= \(\frac{1}{2}\) × (p 1 + p 2 ) × h –  2 = \

 

 × 30 –  2 = 816m 2 .

8. Find the area of the given Figure.

a) 42m 2

b) 24m 2

c) 14m 2

d) 7m 2

Answer: c

Explanation: Area = Area of trapezium + Area of triangle = \(\frac{1}{2}\) × (p 1 + p 2 ) × h + \

 

 

 × 2 + \(\frac{1}{2}\) × 4 × 3 = 14cm 2 .

9. What is the area of the given figure?

a) 10cm 2

b) 13cm 2

c) 12cm 2

d) 11cm 2

Answer: b

Explanation: Area = Area of trapezium 1 + Area of trapezium 2

= \(\frac{1}{2}\) × (p 1 + p 2 ) × h 1 + \(\frac{1}{2}\) × (p 3 + p 4 ) × h 2 = \

 

 × 2 + \

 

 × 2 = 13cm 2 .

10. Find length of AE if area of the given figure is 15cm 2 .

a) 9 cm

b) 8 cm

c) 7 cm

d) 6 cm

Answer: a

Explanation: Area of given figure = Area of trapezium + Area of triangle

⇒ 15cm 2 = \(\frac{1}{2}\) × (p 1 + p 2 ) × h 1 + \(\frac{1}{2}\) × b × h

⇒ 15cm 2 = \

 

 × 2 + \

 

 = 9 cm.

11. Compare the area of given trapezium.

a) B > A

b) B < A

c) B = A = 7cm 2

d) B = A = 9cm 2

Answer: a

Explanation: Area of A = \(\frac{1}{2}\) × (p 1 + p 2 ) × h 1 = \

 

 × 2 = 7cm 2

Area of B = \(\frac{1}{2}\) × (p 1 + p 2 ) × h 1 = \

 

 × 3 = 9cm 2

⇒ Area of B > Area of A.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Mensuration – Area of General Quadrilaterals”.

1. Find the area of the given quadrilateral.

a) 26cm 2

b) 27cm 2

c) 29cm 2

d) 25cm 2

Answer: b

Explanation: Area of general quadrilateral = \(\frac{1}{2}\) × (h 1 + h 2 ) × d  [d – diagonal of quadrilateral]

= \

 

 × 9

= 27cm 2 .

2. Find the height of the quadrilateral if the area is 144cm 2 .

a) 24 cm

b) 12 cm

c) 22 cm

d) 11 cm

Answer: a

Explanation: Area of general quadrilateral = \(\frac{1}{2}\) × (h 1 + h 2 ) × d  

⇒ 144 = \

 

 × d

⇒ d = 24 cm.

3. The area of quadrilateral PQRS is 228cm 2 . Find length of QA.

a) 16 cm

b) 9 cm

c) 8 cm

d) 18 cm

Answer: d

Explanation: Area of general quadrilateral = \(\frac{1}{2}\) × (h 1 + h 2 ) × d  

⇒ 228 = \(\frac{1}{2}\) × (h 1 + 20) × 12

⇒ l = 18 cm.

4. Find the area of the shaded region. The length of the diagonal DB is 8 cm. The perpendicular distances from the vertices to the diagonal DB are 4 cm and 3 cm. The length of segment BC is 6 cm. 

a) 21.87cm 2

b) 9.87cm 2

c) 13.87cm 2

d) 17.87cm 2

Answer: c

Explanation: Area of shaded region = Area of general quadrilateral – Area of semicircle

= \(\frac{1}{2}\) × (h 1 + h 2 ) × d – \

 

 2 = \

 

 × 8 – \

 

 2 = 13.87cm 2 .

5. Find the area of pentagon PQRSA. The perpendicular distances from the vertices to the diagonal PR are 5 cm and 2 cm.

Given: l = 10 cm, l = 8 cm, l = 7cm.

a) 7cm 2

b) 6cm 2

c) 8cm 2

d) 5cm 2

Answer: a

Explanation: Area of pentagon PQRSA = Area of Quadrilateral PQRS – Area of Triangle PAS

= \(\frac{1}{2}\) × (h 1 + h 2 ) × d – \

 

 

 × 10 – \(\frac{1}{2}\) × 8 × 7 = 7cm 2 .

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Mensuration – Surface Area of a Cube, Cuboid and Cylinder”.

1. Find the surface area of the given cuboid.

a) 44 cm 2

b) 22 cm 2

c) 40 cm 2

d) 88 cm 2

Answer: d

Explanation: Surface area cuboid = 2 = 2 = 88 cm 2 .

2. What is the surface area of the given cube?

a) 46 cm 2

b) 96 cm 2

c) 48 cm 2

d) 92 cm 2

Answer: b

Explanation: Surface area cube = 6 a 2 = 6 (4 2 ) = 96 cm 2 .

3. Find the surface area of the given cylinder.

a) 113.14 cm 2

b) 113.16 cm 2

c) 112.16 cm 2

d) 112.14 cm 2

Answer: a

Explanation: Curved Surface area of cylinder = 2πrh + 2πr 2 = 2πr = 2 × \

 

= 113.14 cm 2 .

4. A gift in the form of cube has to be wrapped with gift paper whose external measures are 15 cm. Find the area of gift paper needed.

a) 1250 cm 2

b) 1354 cm 2

c) 1350 cm 2

d) 1150 cm 2

Answer: c

Explanation: Surface area cube = 6 a 2 = 6 (15 2 ) = 1350 cm 2 .

5. The internal measurement of a cuboidal room 15m × 6m × 4m.The room has to be painted along with the ceiling. If cost of painting is Rs 17 per m 2 , Find the total cost to be paid.

a) Rs 4916

b) Rs 3916

c) Rs 6916

d) Rs 5916

Answer: d

Explanation: Surface area cuboid = 2 = 2 = 348 m 2

Total cost to be paid = 348 × 17 = Rs 5916.

6. A building has 6 cylindrical pillars whose radius is 2m and height is 12m. Each pillar has to be painted and the cost of painting is Rs 22 per meter sq. Find the cost of painting the curved surface area of the 6 pillars.

a) Rs. 3872

b) Rs. 4872

c) Rs. 5872

d) Rs. 2872

Answer: a

Explained: Curved Surface area of cylinder = 2πrh + 2πr 2 = 2πr = 2 × \

 

= 176 m 2

The total cost of painting = 176 × 22 = Rs. 3872.

7. Find the side of the cube whose surface area is 384m 2 .

a) 5 m

b) 6 m

c) 8 m

d) 7 m

Answer: c

Explanation: Surface area cube = 6 a 2

⇒ 384 = 6 (a 2 )

⇒ a = 8 m.

8. If the surface area of a cuboid is 108 cm 2 , find the height of the cuboid. The length and breadth the cuboid is 6 and 3 cm respectively.

a) 8 cm

b) 4 cm

c) 2cm

d) 6 cm

Answer: b

Explanation: Surface area cuboid = 2

⇒ 108 = 2

⇒ h = 4 cm.

9. The curved surface of the cylinder is 88 cm 2 . The height of the cylinder is 5 m, what is the radius of the cylinder?

a) 6 cm

b) 4 cm

c) 1 cm

d) 2 cm

Answer: b

Explanation: Curved Surface area of cylinder = 2πrh + 2πr 2 = 2πr

⇒ 88 = 2 × \

 

⇒ h = 4cm.

10. Match the pairs.

A B

1)

a) 96 cm 2

2)

b) 132 cm 2

3)

c) 28 cm 2

a) 1-c, 2-a, 3-b

b) 1-c, 2-a, 3-a

c) 1-c, 2-b, 3-b

d) 1-c, 2-b, 3-a

Answer: a

Explanation: Surface area cuboid = 2 = 2 = 28 cm 2

Surface area cube = 6 a 2 = 6 (4 2 ) = 96 cm 2

Curved Surface area of cylinder = 2πrh + 2πr 2 = 2πr = 2 × \

 

= 132 cm 2 .

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Mensuration – Volume of Cube, Cuboid and Cylinder”.

1. Find the volume of the given figure.

a) 2400 cm 3

b) 1600 cm 3

c) 2200 cm 3

d) 1400 cm 3

Answer: a

Explanation: Volume of a cuboid = l × b × h = 20 × 15 × 8 = 2400 cm 3 .

2. What is the volume of the given figure?

a) 2642 cm 3

b) 2644 cm 3

c) 2744 cm 3

d) 2742 cm 3

Answer: c

Explanation: Volume of cube = a 3 = 14 3 = 2744 cm 3

3. Find the capacity of the cylinder. 

 

a) 1366 m 3

b) 1276 m 3

c) 1376 m 3

d) 1386 m 3

Answer: d

Explanation: Volume of cylinder = πr 2 h = \(\frac{22}{7}\) × 7 2 × 9 = 1386 m 3 .

4. A cuboid box is having measurements 5m × 8m × 12m. How many cube boxes having side 2m can be kept inside the cuboid box?

a) 80

b) 120

c) 240

d) 60

Answer: d

Explanation: No. of cube boxes = \(\frac{Volume \,of\, cuboid\, box}{Volume\, of\, cube\, box} = \frac{l × b × h}{a^3} = \frac{5×8×12}{2^3}\) = 60

5. A cylindrical tank is having height 6m and radius 4 m. How much litres of water can be filled in the tank? 

a) 301.43

b) 301.46 m 3

c) 301.44 m 3

d) 301.45 m 3

Answer: c

Explanation: Capacity of the tank = πr 2 h = 3.14 × 4 2 × 6 = 301.44 m 3 .

6. A cuboid is having volume 3674.16 cm 3 . Find the unknown data.

a) 12.5 cm

b) 12.6 cm

c) 11.6 cm

d) 11.5 cm

Answer: b

Explanation: Volume of a cuboid = l × b × h

3674.16 = 16.2 × 18 × x

x = 12.6

7. Find the height of the cylinder whose volume is 14.08 m 3 and radius is 4 m.

a) 0.54 m

b) 0.56 m

c) 0.14 m

d) 0.28 m

Answer: d

Explanation: Volume of cylinder = πr 2 h

14.08 = \(\frac{22}{7}\) × 4 2 × x

X = 0.28 m.

8. If the edge of a cube is doubled then how many times the volume will increase?

a) 1

b) 2

c) 4

d) 8

Answer: d

Explanation: Volume of cube = a 3

If side is doubled,

Volume of cube =  3 = 8a 3 .

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Powers with Negative Exponents”.

1. What is the multiplicative inverse of 10 -3 ?

a) 100

b) -100

c) \

 

 1000

Answer: d

Explanation: Multiplicative inverse of a n = \(\frac{1}{a^n}\)

⇒ Multiplicative inverse of 10 -3 = 10 3 = 1000.

2. What is the value of 10 -6 ?

a) -0.000010

b) -0.0000010

c) 0.000010

d) 0.0000010

Answer: c

Explanation: a -m = \(\frac{1}{a^m}\)

⇒ 10 -6 = 0.000010.

3. Find the value of 3 -2 .

a) \

 

 –\

 

 8

d) \(\frac{1}{8}\)

Answer: a

Explanation: a -m = \(\frac{1}{a^m}\)

⇒ 3 -2 = \(\frac{1}{3^2} = \frac{1}{9}\).

4. Find the unknown value: \

 

 -4

b) -3

c) 3

d) 4

Answer: b

Explanation: a -m = \(\frac{1}{a^m}\)

⇒ 9 -3 = \(\frac{1}{9^3} = \frac{1}{729}\).

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Laws of Exponents”.

1. Find the value of \

 

 \

 

 \

 

 16

d) 64

Answer: d

Explanation: \(\frac{1}{a^{-m}}\) = a m

⇒ \(\frac{1}{8^{-2}}\) = 8 2 = 64.

2. What is 2 -3 ?

a) \

 

 \

 

 4

d) 8

Answer: b

Explanation: a -m = \(\frac{1}{a^m}\)

⇒ 2 -3 = \(\frac{1}{2^3}\) = \(\frac{1}{8}\).

3. Find the value of 16 -2 × 16 14 .

a) 16 12

b) 16 -12

c) 16 14

d) 16 -14

Answer: a

Explanation: a m × a n = a m + n

⇒ 16 -2 × 16 14 = 16 -2 + 14 = 16 12 .

4. Evaluate \

 

\).

a) 21 25

b) 21 -25

c) 21 24

d) 21 -24

Answer: c

Explanation: \(\frac{a^m}{a^n}\) = a m – n

⇒ \

 

\) = 21 29 – 5 = 21 24 .

5. Match the pairs.

A B

1) 2 3 × 8 3 a. 1

2) \(\frac{8^3}{2^3}\) b. 16 3

3) (2 3 ) 0 c. 4 3

d. 8

a) 1-b, 2-c, 3-d

b)1-b, 2-c, 3-a

c) 1-c, 2-b, 3-a

d) 1-c, 2-b, 3-d

Answer: b

Explanation: a m × b m =  m

⇒ 2 3 × 8 3 =  3 = 16 3

Now, \

 

 

^m\)

⇒ 

 

^3\) = 4 3

Also, (a m ) n = a mn , a 0 = 1

⇒ (2 3 ) 0 = 2 3 × 0 = 2 0 = 1.

6. Find the value of (2 3 ) 2 .

a) 128

b) 16

c) 32

d) 64

Answer: d

Explanation: (a m ) n = a mn

⇒ (2 3 ) 2 = 2 6 = 64.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Use of Exponents to Express Small Numbers in Standard Form”.

1. Express the following in the standard form: 0.000000021.

a) 2.1 × 10 -7

b) 2.1 × 10 -9

c) 2.1 × 10 -10

d) 2.1 × 10 -8

Answer: d

Explanation: 0.000000021 = 2.1 × 10 -8 .

2. What is the standard form of 80460000?

a) 8.046 × 10 7

b) 8.046 × 10 6

c) 8.046 × 10 9

d) 8.046 × 10 8

Answer: a

Explanation: 80460000 = 8.046 × 10 7 .

3. Find the usual form of 2.33 × 10 3 .

a) 0.002033

b) 0.0000233

c) 0.00233

d) 0.000233

Answer: c

Explanation: 2.33 × 10 -3 = 0.00233.

4. What is the Usual form of 16.8 × 10 5 ?

a) 16800000

b) 1680000

c) 168000

d) 16800

Answer: b

Explanation: 16.8 × 10 5 = 1680000.

This set of Mathematics Multiple Choice Questions & Answers  focuses on “Comparing Very Small and Large Number”.

1. Mass of object A is 120000 and that of B is 0.240. Compare their masses.

a) Mass of Object B is approximately is 10 5 times larger than A

b) Mass of Object A is approximately is 10 5 times larger than B

c) Mass of Object A is approximately is 10 6 times larger than B

d) Mass of Object B is approximately is 10 6 times larger than A

Answer: c

Explanation: \(\frac{mass\, of\, object\, A}{mass\, of\, object\, B} = \frac{1.2 × 10^5}{2.4 × 10^{-1}} = \frac{1.2 × 10^6}{2.4}\)

⇒ Mass of Object A is approximately is 10 6 times larger than B.

2. Compare the radius of circle P (r 1 = 10.2 × 10 -6 ) with the radius of circle B (r 1 = 0.2 × 10 -5 ).

a) r 1 = 6.1r 2

b) r 2 = 5r 1

c) r 1 = 5r 2

d) r 1 = 5.1r 2

Answer: d

Explanation: \(\frac{r_1}{r_2} = \frac{10.2 × 10^{-6}}{0.2 × 10^{-5}} = \frac{10.2 × 10^{-1}}{0.2}\) = \(\frac{1.02}{0.2}\) = 5.1.

3. Mass of a neutron is 1.675 × 10 -27 and mass of proton is 1.672 × 10 -27 . Find the total mass.

a) 3.347 × 10 -27

b) 3.337 × 10 -27

c) 3.447 × 10 -27

d) 3.357 × 10 -27

Answer: a

Explanation: Total mass = mass of neutron + mass of proton = 1.675 × 10 -27 + 1.672 × 10 -27 =  × 10 -27 = 3.347 × 10 -27 .

4. Distance between A to C is 2.1 × 10 11 and that of A to B is 3.12 × 10 9 . Find the distance between B and C.

a) 2.1312 × 10 11

b) 2.0412 × 10 11

c) 2.1312 × 10 11

d) 2.1312 × 10 10

Answer: c

Explanation: Distance between B and C = AC – AB = 2.1 × 10 11 – 3.12 × 10 9 =  × 10 11 = 2.1312 × 10 11 .

5. Compare volumes of cube A having side 2 × 10 2 m and cube B having side 1 × 10 1 m.

a) V 1 = 800V 2

b) V 1 = 8000V 2

c) V 2 = 8000V 1

d) V 2 = 800V 1

Answer: b

Explanation: volume of cube A = (2 × 10 2 ) 3 = 8 × 10 6 m 3 .

⇒ Volume of cube B = (10 1 ) 3 = 10 3

⇒ \(\frac{V_1}{V_2} = \frac{8 × 10^6}{10^3}\) = 8000.