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Class 8 RD Sharma Solutions – Chapter 3 Squares and Square Roots – Exercise 3.1 | Set 2
• Last Updated : 06 Apr, 2021

### Question 9. Find the greatest number of two digits which is a perfect square.

Solution:

Greatest two-digit number is 99

99 = 81+18

= 9×9 + 18

18 is the remainder

Perfect square number is 99 – 18 = 81

Therefore, the greatest number of two digits which is perfect square is 81

### Question 10. Find the least number of three digits which is a perfect square.

Solution:

Least three-digit number is 100

100 = 10 × 10

100 itself is the square of 10

Therefore, the least number of three digits which is perfect square is 100

### Question 11. Find the smallest number by which 4851 must be multiplied so that the product becomes a perfect square.

Solution:

Prime factorization of 4851

4851 = 3×3×7×7×11

By grouping the prime factors

= (3×3) × (7×7) × 11

11 is left out

Therefore, the smallest number by which 4851 must be multiplied so that the product becomes a perfect square is 11.

### Question 12. Find the smallest number by which 28812 must be divided so that the quotient becomes a perfect square.

Solution:

Prime factorization of 28812

28812 = 2×2×3×7×7×7×7

By grouping the prime factors

= (2×2) × 3 × (7×7) × (7×7)

3 is left out

Therefore, the smallest number by which 28812 must be divided so that the quotient becomes a perfect square is 3.

### Question 13. Find the smallest number by which 1152 must be divided so that it becomes a perfect square. Also find the number whose square is the resulting number.

Solution:

Prime factorization of 1152

1152 = 2×2×2×2×2×2×2×3×3

By grouping the prime factors

= (2×2) × (2×2) × (2×2) × (3×3) × 2

Therefore, the smallest number by which 1152 must be divided so that the quotient becomes a perfect square is 2.

The number after division, 1152/2 = 576

Prime factors for 576 = 2×2×2×2×2×2×3×3

By grouping the prime factors

= (2×2) × (2×2) × (2×2) × (3×3)

= (2×2×2×3) × (2×2×2×3)

= 242

Therefore, the resulting number is square of 24.

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