Question 5

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Jambmaths question: 

if $\tfrac{y}{2}=x$, evaluate $\left( \tfrac{{{x}^{3}}}{{{y}^{3}}}+\tfrac{1}{2} \right)\times \left( \tfrac{1}{2}-\tfrac{{{x}^{2}}}{{{y}^{2}}} \right)$

Option A: 

$\frac{5}{8}$

Option B: 

$\frac{5}{2}$

Option C: 

$\frac{5}{4}$

Option D: 

$\frac{5}{16}$

Jamb Maths Solution: 

$\begin{align}  & \left( \frac{{{x}^{3}}}{{{y}^{3}}}+\frac{1}{2} \right)\div \left( \frac{1}{2}-\frac{{{x}^{2}}}{{{y}^{2}}} \right)=\left( \frac{{{(\tfrac{y}{2})}^{3}}}{{{y}^{3}}}+\frac{1}{2} \right)\div \left( \frac{1}{2}-\frac{{{(\tfrac{y}{2})}^{2}}}{{{y}^{2}}} \right) \\ & =\left( \frac{{{y}^{3}}}{8{{y}^{3}}}+\frac{1}{2} \right)\div \left( \frac{1}{2}-\frac{{{y}^{2}}}{4{{y}^{2}}} \right) \\ & =\left( \frac{1}{8}+\frac{1}{2} \right)\div \left( \frac{1}{2}-\frac{1}{4} \right) \\ & =\frac{5}{8}\div \frac{1}{4}=\frac{5}{8}\times \frac{4}{1}=\frac{5}{2} \end{align}$

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