CAT 2019 Shift 1 Paper

Question : 83 of 100

Marks: +1, -0

The product of two positive numbers is 616. If the ratio of the difference of their cubesto the cube of their difference is 157:3, then the sum of the two numbers is

Validate

Solution:

Let the two numbers be

{x}

and

{y}

Given,

x × y=616

Also,

\;{x^{3}-y^{3}}/{(x-y)^{3}}=\;{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)

⇒(3 k)^{3}=157 k-3 × 616(3 k)^{1 {/} 3}

⇒ 154 k=3 × 616 ×(3 k)^{1{/}3}

⇒ k=\;{3 × 616}/{154} ×(3 k)^{1{/}3}

⇒ k=12 ×(3 k)^{1 {/} 3}

⇒ k^{3}=12^{3} × 3 × k

⇒ k^{2}=3 × 12^{3}

⇒ k=72

Therefore,

x-y=(3 k)^{1 {/} 3}=(3 × 72)^{1{/} 3}=6

Also,

(x+y)^{2}=(x-y)^{2}+4 x y

⇒(x+y)^{2}=6^{2}+3 × 616=2500

⇒(x+y)=50

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