Class 8 RD Sharma Solutions – Chapter 3 Squares and Square Roots – Exercise 3.1 | Set 2
Chapter 3 Squares and Square Roots – Exercise 3.1 | Set 1
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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