Question Bank › General Aptitude › Square Roots and Cube Roots › ABC26GN1434ABC26GN1434 · Square Roots and Cube Roots Subject: General Aptitude · Chapter: Square Roots and Cube Roots · Exam: 2007 · Marks: · Difficulty:
$\frac{1}{(\sqrt{9}-\sqrt{8})}-\frac{1}{(\sqrt{8}-\sqrt{7})}+\frac{1}{(\sqrt{7}-\sqrt{6})}-\frac{1}{(\sqrt{6}-\sqrt{5})}$ $+\frac{1}{(\sqrt{5}-\sqrt{4})}$ is equal to
(a) 0
(b) $\frac{1}{3}$
(c) 1
(d) 5
Answer Explanation Given expression $=\frac{1}{(\sqrt{9}-\sqrt{8})} \times \frac{(\sqrt{9}+\sqrt{8})}{(\sqrt{9}+\sqrt{8})}-\frac{1}{(\sqrt{8}-\sqrt{7})}$ $$\times \frac{(\sqrt{8}+\sqrt{7})}{(\sqrt{8}+\sqrt{7})}+\frac{1}{(\sqrt{7}-\sqrt{6})} \times \frac{(\sqrt{7}+\sqrt{6})}{(\sqrt{7}+\sqrt{6})}$$ $-\frac{1}{(\sqrt{6}-\sqrt{5})} \times \frac{(\sqrt{6}+\sqrt{5})}{(\sqrt{6}+\sqrt{5})}+\frac{1}{(\sqrt{5}-\sqrt{4})} \times \frac{(\sqrt{5}+\sqrt{4})}{(\sqrt{5}+\sqrt{4})}$ $$\begin{aligned} & \begin{aligned} =\frac{(\sqrt{9}+\sqrt{8})}{(9-8)}-\frac{(\sqrt{8}+\sqrt{7})}{(8-7)}+\frac{(\sqrt{7}+\sqrt{6})}{(7-6)} & \\ & \quad-\frac{(\sqrt{6}+\sqrt{5})}{(6-5)}+\frac{(\sqrt{5}+\sqrt{4})}{(5-4)} \end{aligned} \\ & \begin{array}{r} =(\sqrt{9}+\sqrt{8})-(\sqrt{8}+\sqrt{7})+(\sqrt{7}+\sqrt{6})-(\sqrt{6}+\sqrt{5}) \\ +(\sqrt{5}+\sqrt{4})=(\sqrt{9}+\sqrt{4})=3+2=5 . \end{array} \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.
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