When the square of the sum of two numbers are added to the square of their difference, we get 416. The difference between the square of the sum and square of the difference is 384. What are the numbers ?
18 and 24
12 and 16
8 and 12
10 and 12
Answer is Right!
Answer is Wrong!
This question was previously asked in
UPSC CISF-AC-EXE – 2022
Condition 1: The square of the sum added to the square of the difference is 416.
(a + b)² + (a – b)² = 416.
Using the identities (a+b)² = a² + 2ab + b² and (a-b)² = a² – 2ab + b², we get:
(a² + 2ab + b²) + (a² – 2ab + b²) = 416.
2a² + 2b² = 416.
Dividing by 2, we get: a² + b² = 208. (Equation 1)
(a + b)² – (a – b)² = 384.
Using the identities again:
(a² + 2ab + b²) – (a² – 2ab + b²) = 384.
a² + 2ab + b² – a² + 2ab – b² = 384.
4ab = 384.
Dividing by 4, we get: ab = 96. (Equation 2)
1) a² + b² = 208
2) ab = 96
We can check the options given to see which pair of numbers satisfies both equations.
Option C gives the numbers 8 and 12.
Let’s test if a=8 and b=12 satisfy the equations:
Equation 1: 8² + 12² = 64 + 144 = 208. (Satisfied)
Equation 2: 8 * 12 = 96. (Satisfied)
Since the numbers 8 and 12 satisfy both conditions derived from the problem statement, they are the correct numbers. Alternatively, we can solve the system using the identities (a+b)² = a²+b²+2ab and (a-b)² = a²+b²-2ab.
(a+b)² = 208 + 2(96) = 208 + 192 = 400 => a+b = √400 = 20 (assuming positive numbers).
(a-b)² = 208 – 2(96) = 208 – 192 = 16 => a-b = √16 = 4 (assuming a>b).
Solving a+b=20 and a-b=4: Adding gives 2a=24 => a=12. Substituting gives 12+b=20 => b=8. The numbers are 12 and 8.