MCQ
The value of $\frac{\sqrt{48}+\sqrt{32}}{\sqrt{27}+\sqrt{18}}$, is
- ✓$\frac{4}{3}$
- B$4$
- C$3$
- D$\frac{3}{4}$
✓
Answer
Correct option: A.
$\frac{4}{3}$
$=\frac{\sqrt{48}+\sqrt{32}}{\sqrt{27}+\sqrt{18}}$
$=\frac{\sqrt{4 \times 4 \times 3}+\sqrt{4 \times 4 \times 2}}{\sqrt{3 \times 3 \times 3}+\sqrt{3 \times 3 \times 2}}$
$=\frac{4 \sqrt{3}+4 \sqrt{2}}{3 \sqrt{3}+3 \sqrt{2}}$
$=\frac{4(\sqrt{3}+\sqrt{2})}{3(\sqrt{3}+\sqrt{2})}$
$=\frac{4}{3}$
$=\frac{\sqrt{4 \times 4 \times 3}+\sqrt{4 \times 4 \times 2}}{\sqrt{3 \times 3 \times 3}+\sqrt{3 \times 3 \times 2}}$
$=\frac{4 \sqrt{3}+4 \sqrt{2}}{3 \sqrt{3}+3 \sqrt{2}}$
$=\frac{4(\sqrt{3}+\sqrt{2})}{3(\sqrt{3}+\sqrt{2})}$
$=\frac{4}{3}$
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