CAT 2019

Slot 1

QA

Question & Solution

Assume the numbers are a and b, then ab=616

We have, $\ \ \frac{\ a^3-b^3}{\left(a-b\right)^3}$ = $\ \frac{\ 157}{3}$

=> $\ 3\left(a^3-b^3\right)\ =\ 157\left(a^3-b^3+3ab\left(b-a\right)\right)$

=> $154\left(a^3-b^3\right)+3*157*ab\left(b-a\right)$ = 0

=> $154\left(a^3-b^3\right)+3*616*157\left(b-a\right)$ = 0        (ab=616)

=>$a^3-b^3+\left(3\times\ 4\times\ 157\left(b-a\right)\right)$    (154*4=616)

=> $\left(a-b\right)\left(a^2+b^2+ab\right)\ =\ 3\times\ 4\times\ 157\left(a-b\right)$

=> $a^2+b^2+ab\ =\ 3\times\ 4\times\ 157$

Adding ab=616 on both sides, we get

$a^2+b^2+ab\ +ab=\ 3\times\ 4\times\ 157+616$

=> $\left(a+b\right)^2=\ 3\times\ 4\times\ 157+616$ = 2500

=> a+b=50