If √(4096)=64, then the value of √(40.96)+√(0.4096)+√(0.004096)+√(0.00004096) up to two places of decimals is · ABC26GN1364
Subject: General Aptitude · Chapter: Square Roots and Cube Roots · Exam: S.S.C. 2005
If $\sqrt{4096}=64$, then the value of $\sqrt{40.96}+\sqrt{0.4096}+\sqrt{0.004096}+\sqrt{0.00004096}$ up to two places of decimals is
(a)7.09
(b)7.10
(c)7.11
(d)7.12
Answer
Explanation
$$\begin{aligned} & =\sqrt{\frac{4096}{10^{2}}}+\sqrt{\frac{4096}{10^{4}}}+\sqrt{\frac{4096}{10^{6}}}+\sqrt{\frac{4096}{10^{8}}} \\ & =\frac{\sqrt{4096}}{10}+\frac{\sqrt{4096}}{10^{2}}+\frac{\sqrt{4096}}{10^{3}}+\frac{\sqrt{4096}}{10^{4}} \\ & =\frac{64}{10}+\frac{64}{100}+\frac{64}{1000}+\frac{64}{10000} \\ & =6.4+0.64+0.064+0.0064 \\ & =7.1104 \approx 7.11 \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.
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