CAT 2019 Quant Questions with Solutions

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

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

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

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

Get one day FREE Trial of CAT online Full course FREE Registration
Also Check: 841+ CAT Quant Questions with Solutions

CAT Quant Online Course

  • 1000+ Practice Problems
  • Detailed Theory of Every Topics
  • Online Live Sessions for Doubt Clearing
  • All Problems with Video Solutions
₹ 2999

CAT 2019 Slot-1

CAT 2019 Slot-2

Try CAT 2020 online course for 1 day for FREE