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Surds and Indices
1.The cube root of .000216 is:

Explanation:

Solution:
$\begin{array}{rl}& {\left(.000216\right)}^{\frac{1}{3}}={\left(\frac{216}{{10}^{6}}\right)}^{\frac{1}{3}}\\ & ={\left(\frac{6×6×6}{{10}^{2}×{10}^{2}×{10}^{2}}\right)}^{\frac{1}{3}}\\ & =\frac{6}{{10}^{2}}\\ & =\frac{6}{100}\\ & =0.06\end{array}$

2.What should come in place of both x in the equation
$\frac{x}{\sqrt{128}}=\frac{\sqrt{162}}{x}$

Explanation:

Solution:
$\begin{array}{rl}& \text{Let}\phantom{\rule{thinmathspace}{0ex}}\frac{x}{\sqrt{128}}=\frac{\sqrt{162}}{x}\\ & \text{Then}\phantom{\rule{thinmathspace}{0ex}}{x}^{2}=\sqrt{128×162}\\ & =\sqrt{64×2×18×9}\\ & =\sqrt{{8}^{2}×{6}^{2}×{3}^{2}}\\ & =8×6×3\\ & =144\\ & \therefore x=\sqrt{144}=12\end{array}$

3.The least perfect square, which is divisible by each of 21, 36 and 66 is:

Explanation:

Solution:
L.C.M. of 21, 36, 66 = 2772.
Now, 2772 = 2 x 2 x 3 x 3 x 7 x 11
To make it a perfect square, it must be multiplied by 7 x 11.
So, required number = 22 x 32 x 72 x 112 = 213444

4.If
$3\sqrt{5}+\sqrt{125}$
= 17.88, then what will be the value of
$\sqrt{80}+6\sqrt{5}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}?$

Explanation:

Solution:
$\begin{array}{rl}& 3\sqrt{5}+\sqrt{125}=17.88\\ & ⇒3\sqrt{5}+\sqrt{25×5}=17.88\\ & ⇒3\sqrt{5}+5\sqrt{5}=17.88\\ & ⇒8\sqrt{5}=17.88\\ & ⇒\sqrt{5}=2.235\\ & \therefore \sqrt{80}+6\sqrt{5}=\sqrt{16×5}+6\sqrt{5}\\ & =4\sqrt{5}+6\sqrt{5}\\ & =10\sqrt{5}=\left(10×2.235\right)=22.35\end{array}$

5.If a = 0.1039, then the value of
$\sqrt{4{a}^{2}-4a+1}+3a$
is:

Explanation:

Solution:
$\begin{array}{rl}& \sqrt{4{a}^{2}-4a+1}+3a\\ & =\sqrt{{\left(1\right)}^{2}+{\left(2a\right)}^{2}-2×1×2a}+3a\\ & =\sqrt{{\left(1-2a\right)}^{2}}+3a\\ & =\left(1-2a\right)+3a\\ & =\left(1+a\right)\\ & =\left(1+0.1039\right)\\ & =1.1039\end{array}$

6.If
$x=\frac{\sqrt{3}+1}{\sqrt{3}-1}$
and
$y=\frac{\sqrt{3}-1}{\sqrt{3}+1},$
then the value of
$\left({x}^{2}+{y}^{2}\right)$
is?

Explanation:

Solution:
$\begin{array}{rl}& x=\frac{\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)}×\frac{\left(\sqrt{3}+1\right)}{\left(\sqrt{3}+1\right)}\\ & \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\frac{{\left(\sqrt{3}+1\right)}^{2}}{\left(3-1\right)}\\ & \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\frac{3+1+2\sqrt{3}}{2}\\ & \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=2+\sqrt{3}\\ & y=\frac{\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)}×\frac{\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)}\\ & \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\frac{{\left(\sqrt{3}-1\right)}^{2}}{\left(3-1\right)}\\ & \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\frac{3+1-2\sqrt{3}}{2}\\ & \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=2-\sqrt{3}\\ & \therefore {x}^{2}+{y}^{2}={\left(2+\sqrt{3}\right)}^{2}+{\left(2-\sqrt{3}\right)}^{2}\\ & \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=2\left(4+3\right)\\ & \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=14\end{array}$

7.A group of students decided to collect as many paise from each member of group as is the number of members. If the total collection amounts to Rs. 59.29, the number of the member is the group is:

Explanation:

Solution:
Money collected = (59.29 x 100) paise = 5929 paise.
∴ Number of members =
$\sqrt{5929}$
= 77

8.If
$\sqrt{5}=2.236,$
then the value of
$\frac{\sqrt{5}}{2}$
$-$
$\frac{10}{\sqrt{5}}$
$+$
$\sqrt{125}$
is equal to :

Explanation:

Solution:
$\begin{array}{rl}& \frac{\sqrt{5}}{2}-\frac{10}{\sqrt{5}}+\sqrt{125}\phantom{\rule{thinmathspace}{0ex}}\\ & =\frac{{\left(\sqrt{5}\right)}^{2}-20+2\sqrt{5}×5\sqrt{5}}{2\sqrt{5}}\\ & =\frac{5-20+50}{2\sqrt{5}}\\ & =\frac{35}{2\sqrt{5}}×\frac{\sqrt{5}}{\sqrt{5}}\\ & =\frac{35\sqrt{5}}{10}\\ & =\frac{7×2.236}{2}\\ & =7×1.118\\ & =7.826\end{array}$

9.
$\left(\frac{\sqrt{625}}{11}×\frac{14}{\sqrt{25}}×\frac{11}{\sqrt{196}}\right)\phantom{\rule{1pt}{0ex}}$
is equal to :

Explanation:

Solution:
$\begin{array}{rl}& \text{Give}\phantom{\rule{thinmathspace}{0ex}}\text{Expression}=\\ & \frac{25}{11}×\frac{14}{5}×\frac{11}{14}=5\end{array}$

10.
$\sqrt{0.0169×?}=1.3$

$\begin{array}{rl}& \text{Let}\sqrt{0.0169×x}=1.3\\ & \text{Then},\phantom{\rule{thinmathspace}{0ex}}0.0169x={\left(1.3\right)}^{2}=1.69\\ & ⇒x=\frac{1.69}{0.0169}=100\end{array}$