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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

  1. $58$
  2. $50$
  3. $95$
  4. $85$
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1 Answer

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Let the two positive numbers be $x$ and $y$.

Now, $xy = 616 \quad \longrightarrow (1) $

And, $ \frac { x^{3} – y^{3} } { (x-y)^{3} } = \frac {157}{3} \quad \longrightarrow (2) $

We know that, $ \boxed{ x^{3} – y^{3} = (x – y) (x^{2} + y^{2} + xy)}$

Now, $ \frac { (x-y) (x^{2}+y^{2}+xy) } { (x-y)^{3} } = \frac{157}{3} $

$ \Rightarrow \frac { x^{2} + y^{2} + xy} { x^{2} + y^{2} -2xy} = \frac {157}{3}$

$ \Rightarrow \frac { x^{2} + y^{2} + 616} { x^{2} + y^{2} -1232} = \frac {157}{3} \quad \longrightarrow (3)$

Here, let’s say $ x^{2} + y^{2} = k $

$ \frac{k^{2} + 616} {k^{2} – 1232} = \frac {157}{3}$

$ \Rightarrow 3(k^{2} + 616) = 157 (k^{2} – 1232) $

$ \Rightarrow 3k^{2} + 1848  = 157k^{2} – 193424 $

$ \Rightarrow 154k^{2} = 195272 $

$ \Rightarrow k^{2} = 1268 $

$ \Rightarrow x^{2} + y^{2} = 1268 $

$ \Rightarrow x^{2} + y^{2} + 2xy – 2xy = 1268 $

$ \Rightarrow ( x + y)^{2} – 1232 = 1268 \quad [\because \text{From equation (1)}] $

$ \Rightarrow (x+y)^{2} = 1268+1232 $

$ \Rightarrow (x+y)^{2} = 2500 $

$ \Rightarrow (x+y)^{2} = (50)^{2} $

$ \Rightarrow \boxed{x+y = 50} $

Correct Answer $: \text{B}$
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