Question

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

• A. 35 and 46
• B. 35 and 50
• C. 40 and 55
• D. 45 and 60

Hint:

When both sum and difference of squares are given, add the equations first to eliminate one variable quickly.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The sum of the squares of two numbers is 3341 and the difference of their squares is 891. We need to find the two numbers.

Step 2: Key Formula or Approach:
Let the numbers be \(x\) and \(y\). Given: \[ x^2+y^2=3341 \] \[ x^2-y^2=891 \] Using addition of equations: \[ (x^2+y^2)+(x^2-y^2)=3341+891 \]

Step 3: Detailed Explanation:
Add the two equations: \[ 2x^2=4232 \] \[ x^2=2116 \] \[ x=\sqrt{2116}=46 \] Now substitute into: \[ x^2+y^2=3341 \] \[ 2116+y^2=3341 \] \[ y^2=3341-2116 \] \[ y^2=1225 \] \[ y=\sqrt{1225}=35 \] Therefore, the two numbers are: \[ 35 \text{ and } 46 \]

Step 4: Final Answer:
Hence, the correct option is: \[ \boxed{\text{(A) 35 and 46}} \]

Answer 2

Since both the sum and the difference of the squares are given, each candidate pair of numbers can be tested directly by squaring both numbers and checking whether they satisfy both conditions at once.

1. 35 and 46: \( 46^2 = 2116 \) and \( 35^2 = 1225 \). Sum: \( 2116 + 1225 = 3341 \), matching the given sum. Difference: \( 2116 - 1225 = 891 \), matching the given difference as well.
2. 35 and 50: \( 50^2 = 2500 \) and \( 35^2 = 1225 \). Sum: \( 2500+1225 = 3725 \), which does not equal the required 3341.
3. 40 and 55: \( 55^2 = 3025 \) and \( 40^2 = 1600 \). Sum: \( 3025+1600=4625 \), well above the required 3341.
4. 45 and 60: \( 60^2 = 3600 \) and \( 45^2 = 2025 \). Sum: \( 3600+2025=5625 \), far higher than the required 3341.

Only the pair 35 and 46 satisfies both the sum-of-squares condition and the difference-of-squares condition simultaneously.

Therefore, the correct answer is 35 and 46.