SarkariPYQ

SSC CHSL 2021 PYQ — Question 69

SSC CHSL - 16 April 2021 - Shift 1 • Quantitative Aptitude • Q69

Back to paper

If (xy+1)=4,({{x} \over y}+1)=4, then what is the value of (x2+y2y2)({{x^2+y^2} \over y^2})  ?

Show answer & explanation

Correct answer: C

Explanation

The correct answer is Option 3 i.e. 10.

Given:

(xy+1)=4({{x} \over y}+1)=4

Explanation:

According to the given question,

(xy+1)=4({{x} \over y}+1)=4

⇒ xy=3{{x} \over y}= 3

Now, squarying both sides of equation (xy+1)=4({{x} \over y}+1)=4, we get

⇒ (x2+y2y2)({{x^2+y^2} \over y^2}) + 2(xy)2({x \over y}) = 16

⇒ (x2+y2y2)({{x^2+y^2} \over y^2}) + 6 = 16

 ⇒ (x2+y2y2)({{x^2+y^2} \over y^2}) = 10

Hence the required answer is 10.