SSC CHSL 2021 PYQ — Question 72

SSC CHSL - 15 April 2021 - Shift 1 • Quantitative Aptitude • Q72

If x + y = 5 and 1x+1y=209\frac{1}{x} + \frac{1}{y}= \frac{20}{9}, then the value of (x3 + y3) will be:

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Correct answer: A

Explanation

The correct answer is option 1 i.e.  3654\frac{365}{4}.

(x3+y3)=(x+y)(x2−xy+y2)(x^3 + y^3) = (x + y)(x^2 - xy + y^2)

(x+y)2=x2+2xy+y2(x + y)^2 = x^2 + 2xy + y^2

1x+1y=209\rm \frac{1}{x}+\frac{1}{y}=\frac{20}{9}

⇒(y+x)xy\frac{(y+x)}{xy} = 20/9 

⇒ 5xy\frac{5}{xy} = 20/9 

⇒ xy = 9/4.....(1) 

Also, (x + y)2 = x2 + 2xy + y2

⇒ 52 - 2xy =  x2 + y2 

⇒ 25 - 2 × 94\frac{9}{4} =  x2 + y2     (From 1) 

⇒ 25 - 92\frac{9}{2} =  x2 + y2 

⇒ (50−9)2\frac{(50-9)}{2} =  x2 + y2 

⇒ x2 + y2 = 412\frac{41}{2} .....(2) 

Now, (x3 + y3) = (x + y)(x2 - xy + y2) = 5 × (412\frac{41}{2} - 9/4) = 5 × ​​​​734​​​​\frac{73}{4} = ​​​​3654​​​​\frac{365}{4}