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

View Answer

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 (i) 25-1=24; (ii) 24-3=21; (iii) 21-5=16; (iv) 16-7=9; (v) 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

View Answer

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

View Answer

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

View Answer

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

View Answer

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

View Answer

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

View Answer

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

View Answer

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

View Answer

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

View Answer

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.

**Sanfoundry Global Education & Learning Series – Mathematics – Class 8**.

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