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Given that $x^{2018}y^{2017}=1/2$ and $x^{2016}y^{2019}=8$, the value of $x^2+y^3$ is

  1. $35/4$
  2. $37/4$
  3. $31/4$
  4. $33/4$
in Quantitative Aptitude retagged by
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Given that,

  • $x^{2018} y^{2017} = \frac {1}{2} \quad \longrightarrow (1)$
  • $x^{2016} y^{2019} =8 \quad \longrightarrow (2)$

Divide the equation $(1)$ by equation $(2),$ we get

$ \frac {x^{2018} y^{2017} }{x^{2016} y^{2019}} = \frac{ \frac{1}{2}}{8}$

$ \Rightarrow \frac {x^{2}}{y^{2}} = \frac {1}{16}$

$ \Rightarrow \boxed {\frac {x}{y} = \pm \frac{1}{4}}$

$ \Rightarrow \boxed {x = \pm \frac{y}{4}} \quad \longrightarrow (3)$

Put the value of $x$ in equation $(2),$ we get

$x^{2016} y^{2019} =8$

$ \Rightarrow \left( \pm \frac{y}{4} \right)^{2016} y^{2019} = 8$

$ \Rightarrow y^{2016} \ast y^{2019} = 4^{2016} \ast 8$

$ \Rightarrow y^{4035} = \left( 2^{2} \right)^{2016} \ast 8$

$ \Rightarrow y^{4035} = 2^{4032} \ast 2^{3}$

$ \Rightarrow y^{4035} = 2^{4035}$

$ \Rightarrow \boxed {y=2}$

Put the value of $y$ in equation $(3),$ we get

$ x = \pm \frac {y}{4} = \pm \frac {2}{4}$

$ \boxed {x= \pm \frac {1}{2}}$

Now, $ x^{2} + y^{3} = \left( \pm \frac {1}{2} \right)^{2} + (2)^{3}$

$ = \dfrac {1}{4} + 8$

$ = \dfrac {1+32}{4}$

$ = \dfrac {33}{4} $

Correct Answer $: \text{D}$

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