ABC26GN1382 · Square Roots and Cube Roots

Subject: General Aptitude · Chapter: Square Roots and Cube Roots · Exam: · Marks: · Difficulty:

If $\sqrt{1369}+\sqrt{.0615+x}=37.25$, then $x$ is equal to
(a)$10^{-1}$
(b)$10^{-2}$
(c)$10^{-3}$
(d)None of these
Answer
Answer (as printed): C
Explanation
$$\begin{aligned} & 37+\sqrt{.0615+x}=37.25 \Leftrightarrow \sqrt{.0615+x}=0.25 \\ & \Leftrightarrow .0615+x=(0.25)^{2}=0.0625 \\ & \Leftrightarrow x=.001=\frac{1}{10^{3}}=10^{-3} . \end{aligned}$$

Explanation as extracted from the printed page; notation may be imperfect.

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