# Mathematics Questions and Answers – Finding Square Roots through Repeated Subtraction

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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

Explanation: The sum of the first n odd natural numbers is n2. 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

Explanation: The sum of the first n odd natural numbers is n2. 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

Explanation: The sum of the first n odd natural numbers is n2. 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.
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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

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

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) n2
c) √n
d) n2-1

Explanation: The sum of the first n odd natural numbers is n2. Hence the correct option would be n2 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

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

Explanation: The sum of the first n odd natural numbers is n2.
∴ 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 132 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

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

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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