# Power, Indices and Surds

Quantitative Aptitude Questions and Answers section on “Power, Indices and Surds” with solution and explanation for competitive examinations such as CAT, MBA, SSC, Bank PO, Bank Clerical and other examinations.

1.

Evalute :
$latex \sqrt{248+\sqrt{51+\sqrt{169}}}$
[A]21
[B]20
[C]18
[D]16

16
Since given expression is ;
$latex \sqrt{248+\sqrt{51+\sqrt{169}}}$
$latex =>\sqrt{248+\sqrt{51+13}}=\sqrt{248+\sqrt{64}}$
$latex =>\sqrt{248+8}=\sqrt{256}=16$
Hence option [D] is the right answer.

2.

If $latex a\ast b\ast c=\frac{\sqrt{\left ( a+2 \right )\left ( b+3 \right )}}{c+1}&s=2$, Then find the value of $latex 6\ast 15\ast 3$.
[A]3.28
[B]3
[C]6.25
[D]1.14

3
$latex \because a\ast b\ast c=\frac{\sqrt{\left ( a+2 \right )\left ( b+3 \right )}}{c+1}&s=2$
$latex \therefore 6\ast 15\ast 3=\frac{\sqrt{\left ( 6+2 \right )\left ( 15+3 \right )}}{3+1}&s=2$
$latex =>6\ast 15\ast 3=\frac{\sqrt{8\times 18}}{4}=\frac{\sqrt{144}}{4}=\frac{12}{4}=3&s=2$
Hence option [B] is the right answer.

3.

Find the value of $latex \sqrt{1\frac{9}{16}}&s=2$:
[A]$latex 1\frac{1}{4}&s=2$
[B]$latex 1\frac{2}{4}&s=2$
[C]$latex 1\frac{1}{3}&s=2$
[D]$latex 1\frac{5}{4}&s=2$

$latex 1\frac{1}{4}&s=2$
$latex \because \sqrt{1\frac{9}{16}}=\sqrt{\frac{25}{16}}=\frac{5}{4}=1\frac{1}{4}&s=2$
Hence option [A] is correct.

4.

What would be the squre root of 0.0009
[A]0.003
[B]0.03
[C]0.09
[D]0.0003

0.03
$latex \because \sqrt{0.0009}=\sqrt{\frac{9}{10000}}=\frac{3}{100}=0.03&s=2$
Hence option [B] is the right answer.

5.

Find the value of $latex \sqrt{\frac{0.049\times 0.016\times 0.09}{0.81\times 0.036\times 0.064}}&s=2$ :
[A]$latex \frac{0.7}{36}&s=2$
[B]$latex \frac{7}{36}&s=2$
[C]$latex \frac{0.7}{3.6}&s=2$
[D]$latex \frac{7}{0.36}&s=2$

$latex \frac{7}{36}&s=2$
Given expression is $latex \sqrt{\frac{0.049\times 0.016\times 0.09}{0.81\times 0.036\times 0.064}}&s=2$
Now since the sum of decimal places in the numerator and denominator under the redical sign is the same, so we can remove all decimals.
So now expression is ; $latex \sqrt{\frac{49\times 16\times 9}{81\times 36\times 64}}&s=2$
$latex =>\sqrt{\frac{49}{36\times 36}}=\frac{7}{36}&s=2$
Hence option [B] is the right answer.

6.

Evalute $latex \frac{\sqrt{162}}{\sqrt{288}}&s=2$:
[A]$latex \frac{2}{3}&s=2$
[B]$latex \frac{4}{5}&s=2$
[C]$latex \frac{3}{4}&s=2$
[D]$latex \frac{5}{6}&s=2$

$latex \frac{3}{4}&s=2$
$latex \because \frac{\sqrt{162}}{\sqrt{288}}=\frac{\sqrt{81}}{\sqrt{144}}=\frac{9}{12}=\frac{3}{4}&s=2$
Hence option [C] is the right answer.

7.

Find the value of $latex \sqrt{98}\times \sqrt{72}$ :
[A]86
[B]84
[C]98
[D]76

84
$latex \because \sqrt{98}\times \sqrt{72}=\sqrt{49\times 2\times 2\times 36}$
$latex =>\sqrt{7^{2}\times 2^{2}\times 6^{2}}$
$latex =>7\times 2\times 6=84$

8.

If $latex \sqrt{3}=1.732$ and $latex \sqrt{2}=1.414$, Then find the squre root of $latex \left ( \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \right )&s=2$ :
[A]3.348
[B]3.246
[C]3.146
[D]3.148

3.146
$latex \because \left ( \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \right )=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}&s=2$
$latex =>\frac{\left ( \sqrt{3}+\sqrt{2} \right )^{2}}{3-2}=\left ( \sqrt{3}+\sqrt{2} \right )^{2}&s=2$
and $latex \sqrt{\left ( \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \right )}=\left ( \sqrt{3}+\sqrt{2} \right )$
$latex \therefore \sqrt{\left ( \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \right )}=1.732+1.414=3.146$

9.

What will come instead of question mark in the following question?
$latex \frac{52}{?}=\frac{?}{13}&s=2$
[A]24
[B]27
[C]26
[D]23

26
let $latex \frac{52}{x}=\frac{x}{13}&s=2$,
Then ; $latex x^{2}=52\times 13$
$latex =>x^{2}=4\times 13\times 13$
$latex =>x^{2}=2^{2}\times 13^{2}$
$latex =>x=2\times 13=26.$

10.

The simplified form of $latex \left ( 16^{\frac{3}{2}}+16^{\frac{-3}{2}} \right )$ is :
[A]$latex \frac{16}{4097}&s=1$
[B]0
[C]$latex \frac{4097}{64}&s=1$
[D]1

$latex \frac{4097}{64}&s=1$
$latex \left ( 16^{\frac{3}{2}}+16^{\frac{-3}{2}} \right )&s=1$
$latex =>\left ( 4^{2} \right )^{\frac{3}{2}}+\frac{1}{\left ( 16 \right )^{\frac{3}{2}}}&s=1$
$latex =>4^{2\times\frac{3}{2}}+\frac{1}{4^{2\times\frac{3}{2}}}&s=1$
$latex =>4^{3}+\frac{1}{4^{3}}&s=1$
$latex =>64+\frac{1}{64}&s=1$
$latex =>\frac{4096+1}{64}=\frac{4097}{64}&s=1$
Hence option [C] is the right answer.