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CAT 2019 Quant Question with Solution 34

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
Show Answer

Correct Answer: Option: 2

Let the two numbers be x and y


$x \times y=616$

Also, $\frac{x^{3}-y^{3}}{(x-y)^{3}}=\frac{157}{3}$

Let $x^{3}-y^{3}=157 k$ and $(x-y)^{3}=3 k$

we know that

$(x-y)^{3}=x^{3}-y^{3}-3 x y(x-y)$

$\Rightarrow (3 k)^{3}=157 k-3 \times 616(3 k)^{1 / 3}$

$\Rightarrow 154 k=3 \times 616 \times(3 k)^{1 / 3}$

$\Rightarrow k=\frac{3 \times 616}{154} \times(3 k)^{1/3}$

$\Rightarrow k= 12 \times(3 k)^{1 / 3}$

$\Rightarrow k^{3}= 12^{3} \times 3 \times k$

$\Rightarrow k^{2}= 3 \times 12^{3}$

$\Rightarrow k= 72$

Therefore, $x-y={{(3k)}^{1/3}}={{(3\times 72)}^{1/3}}=6$

Also, ${{\left( x+y \right)}^{2}}={{\left( x-y \right)}^{2}}+4xy$

$\Rightarrow {{\left( x+y \right)}^{2}}={{6}^{2}}+3\times 616=2500$

$\Rightarrow \left( x+y \right)=50$

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