ABC26GN1413 · Square Roots and Cube Roots

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

The square root of $\frac{(0.75)^{3}}{1-0.75}+\left[0.75+(0.75)^{2}+1\right]$ is
(a)1
(b)2
(c)3
(d)4
Answer
Answer (as printed): B
Explanation
$$\begin{aligned} \sqrt{\frac{(0.75)^{3}}{(1-0.75)}} & +\left[0.75+(0.75)^{2}+1\right] \\ & =\sqrt{\frac{(0.75)^{3}+(1-0.75)\left[(1)^{2}+(0.75)^{2}+1 \times 0.75\right]}{1-0.75}} \\ & =\sqrt{\frac{(0.75)^{3}+\left[(1)^{3}-(0.75)^{3}\right]}{1-0.75}} \\ & =\sqrt{\frac{1}{0.25}}=\sqrt{\frac{100}{25}}=\sqrt{4}=2 . \end{aligned}$$

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

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