Find the value of $\rm (\sqrt{98}-\sqrt{72}+\sqrt{50})\div \sqrt{18}?$

Question

Question

This question was previously asked in

NPCIL Assistant Grade-1 (Advance) Official Paper-II (Held On: 08 Sept, 2018)

Answer 1

Given:

$\dfrac{(\sqrt{98}-\sqrt{72}+\sqrt{50})}{\sqrt{18}}$

Formula used:

$\sqrt{a} = a^{1/2}$

Calculation:

$\sqrt{98} = \sqrt{49 \times 2} = 7\sqrt{2}$

$\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}$

$\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$

$\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$

⇒ $\dfrac{(7\sqrt{2}-6\sqrt{2}+5\sqrt{2})}{3\sqrt{2}} = \dfrac{6\sqrt{2}}{3\sqrt{2}}$ = 2

∴ The correct answer is option (3).