Question:medium

Sum of squares of two numbers is 3341 and difference is 891. Find numbers.

Show Hint

When both sum and difference of squares are given, add the equations first to eliminate one variable quickly.
Updated On: Jul 14, 2026
  • 35 and 46
  • 35 and 50
  • 40 and 55
  • 45 and 60
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: Let s = x + y and d = x - y, and write both given facts in terms of s and d.
Since \( x^2 - y^2 = (x+y)(x-y) = sd \), the given difference of squares gives \[ sd = 891 \] Also, \( s^2 + d^2 = (x+y)^2 + (x-y)^2 = 2(x^2+y^2) \), so the given sum of squares gives \[ s^2 + d^2 = 2 \times 3341 = 6682 \]

Step 2: Combine these to find s + d and s - d.
\[ (s+d)^2 = s^2 + 2sd + d^2 = 6682 + 2(891) = 8464 \] \[ s + d = \sqrt{8464} = 92 \] \[ (s-d)^2 = s^2 - 2sd + d^2 = 6682 - 1782 = 4900 \] \[ s - d = \sqrt{4900} = 70 \]

Step 3: Solve for s and d, then for x and y.
Adding the two results: \( 2s = 92 + 70 = 162 \), so \( s = 81 \). Subtracting: \( 2d = 92 - 70 = 22 \), so \( d = 11 \). Since \( s = x + y = 81 \) and \( d = x - y = 11 \): \[ x = \frac{81+11}{2} = 46, \quad y = \frac{81-11}{2} = 35 \] So the two numbers are \[ \boxed{35 \text{ and } 46} \]
Was this answer helpful?
0