ABC26GN1444 · Square Roots and Cube Roots
Subject: General Aptitude · Chapter: Square Roots and Cube Roots · Exam: 2007 · Marks: · Difficulty:
Given that $\sqrt{3}=1.732$, the value of $\frac{3+\sqrt{6}}{5 \sqrt{3}-2 \sqrt{12}-\sqrt{32}+\sqrt{50}}$ is
(a)1.414
(b)1.732
(c)2.551
(d)4.899
Answer
Explanation
Given $\exp .=\frac{3+\sqrt{6}}{5 \sqrt{3}-4 \sqrt{3}-4 \sqrt{2}+5 \sqrt{2}}=\frac{(3+\sqrt{6})}{(\sqrt{3}+\sqrt{2})}$ $$\begin{aligned} & =\frac{(3+\sqrt{6})}{(\sqrt{3}+\sqrt{2})} \times \frac{(\sqrt{3}-\sqrt{2})}{(\sqrt{3}-\sqrt{2})} \\ & =\frac{3 \sqrt{3}-3 \sqrt{2}+3 \sqrt{2}-2 \sqrt{3}}{(3-2)}=\sqrt{3}=1.732 . \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.
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