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# CAT 2019 Quant Question with Solution 34

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

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

Let the two numbers be x and y

Given,

$x \times y=616$

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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