CAT 2018

Slot 1

QA

Question & Solution

Given that $x^{2018}y^{2017}=\frac{1}{2}$  ... (1)

$x^{2016}y^{2019}=8$ ... (2)

Equation (2)/ Equation (1)

$\dfrac{y^2}{x^2} = \dfrac{8}{1/2}$

$\dfrac{y}{x} = 4$ or $-4$

Case 1: When $\dfrac{y}{x} = 4$

$x^{2018}(4x)^{2017}=\dfrac{1}{2}$

$x^{2018+2017}(2)^{4034}=\dfrac{1}{2}$

$x^{4035}=\dfrac{1}{(2)^{4035}}$

$x=\dfrac{1}{2}$

Since, $\dfrac{y}{x} = 4$, => y = 2

Therefore, $x^{2}+y^{3}$ = $\dfrac{1}{4}+8$ = $\dfrac{33}{4}$

Case 2: When $\dfrac{y}{x} = -4$

$x^{2018}(-4x)^{2017}=\dfrac{1}{2}$

$x^{2018+2017}(2)^{4034}=\dfrac{-1}{2}$

$x^{4035}=\dfrac{1}{(-2)^{4035}}$

$x=\dfrac{-1}{2}$

Since, $\dfrac{y}{x} = -4$, => y = 2

Therefore, $x^{2}+y^{3}$ = $\dfrac{1}{4}+8$ = $\dfrac{33}{4}$. Hence, option D is the correct answer.