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
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
Updated On: Aug 20, 2024
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The Correct Option is B
Solution and Explanation
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
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
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