Question

The product of two positive numbers is 616. If the ratio of the difference of their cubes to the cube of their difference is 157:3, then the sum of the two numbers is

• A. 58
• B. 50
• C. 95
• D. 85

Answer 1

The Correct Option is B

Consider two numbers denoted as x and y
Given that
x × y = 616
Also , \(\frac{x³ - y³}{(x - y)³} = \frac{157}{3}\)
Let's assume that x³ - y³ = 157k and (x - y)³ = 3k
we know that 
(x - y)³ = x³ - y³ - 3xy(x - y)
⇒ (3k)³ = 157k - 3×616(3k)^1/3
⇒ 154k = 3 × 616 × (3k)^1/3 
⇒ k = 3 × 616 / 154 × (3k)1/3
⇒ k = 12 × (3k)^1/3
⇒ k³ = 12³ × 3 × k
⇒ k² = 3 × 12³
⇒ k =72

∴ x - y = (3k)^1/3 = (3×72)^1/3 = 6
Also , ( x + y )² = ( x - y )² + 4xy
⇒ ( x + y )² = 6² + 3 × 616 = 2500
⇒ ( x + y ) = 50
Therefore , Sum of the two numbers is 50