Question

The value of $\frac{\sqrt{48}+\sqrt{32}}{\sqrt{27}+\sqrt{18}}$ is

• A. $\frac{4}{3}$
• B. 4
• C. 0
• D. 2

Answer 1

Answer: (A)

$\frac{4}{3}$

$\frac{\sqrt{48}+\sqrt{32}}{\sqrt{27}+\sqrt{18}}=\frac{\sqrt{16\times 3}+\sqrt{16\times 2}}{\sqrt{9\times 3}+\sqrt{9\times 2}}$

$=\frac{4\sqrt{3}+4\sqrt{2}}{3\sqrt{3}+3\sqrt{2}}$

$=\frac{4(\sqrt{3}+\sqrt{2})}{3(\sqrt{3}+\sqrt{2})}=\frac{4}{3}$