Question 32

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waecmaths question: 

If $\frac{\sqrt{2}+\sqrt{3}}{\sqrt{3}}$is simplified as $m+n\sqrt{6}$, find the value of (m + n)

Option A: 

$\frac{1}{3}$

Option B: 

$\frac{2}{3}$

Option C: 

$1\tfrac{1}{2}$

Option D: 

$1\tfrac{1}{3}$

waecmaths solution: 

$\begin{align}  & \frac{\sqrt{2}+\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{2}+\sqrt{3}}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{6}+3}{3} \\ & \frac{\sqrt{2}+\sqrt{3}}{\sqrt{3}}==\frac{3+\sqrt{6}}{3}=1+\frac{\sqrt{6}}{3}=1+\frac{1}{3}\sqrt{6} \\ & m=1,\text{ }n=\frac{1}{3} \\ & m+n=1+\frac{1}{3}=\frac{4}{3}=1\frac{1}{3} \\\end{align}$

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