# Aptitude Question ID: 107033

By how much does $\sqrt{12}+\sqrt{18}$ exceed $\sqrt{3}+\sqrt{2}$ ?
[A]$2\left ( \sqrt{3}-\sqrt{2} \right )$
[B]$\sqrt{3}+2\sqrt{2}$
[C]$2\left ( \sqrt{3}+\sqrt{2} \right )$
[D]$\sqrt{2}-4\sqrt{3}$

$\mathbf{\sqrt{3}+2\sqrt{2}}$
$\because \left ( \sqrt{12}+\sqrt{18} \right )-\left ( \sqrt{3}+\sqrt{2} \right )$
$= (2\sqrt{3}-\sqrt{3})+(3\sqrt{2}-\sqrt{2})$
$= \sqrt{3}+2\sqrt{2}$
Hence option [B] is correct answer.