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# Square Root and Cube Root Questions

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The square root of $$0.\bar{4}$$ is:

A$$0.\bar{6}$$

B$$0.\bar{7}$$

C$$0.\bar{8}$$

D$$0.\bar{9}$$

Explanation:

$$0.\bar{4}=\sqrt{\frac{4}{9}}=\frac{2}{3}=0.666....=0.\bar{6}$$

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The Value of $$\sqrt{\frac{0.16}{0.4}}$$ is:

A0.02

B0.2

C0.63

DNone of these

Explanation:

$$\sqrt{\frac{0.16}{0.4}}=\sqrt{\frac{0.16}{0.40}}=\sqrt{\frac{16}{40}}=\sqrt{\frac{4}{10}}=0.63$$

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The value of $$\frac{1+\sqrt{0.01}}{1-\sqrt{0.1}}$$

A0.6

B1.1

C1.6

D1.7

Explanation:

$$\frac{1+\sqrt{0.01}}{1-\sqrt{0.1}}=\frac{1+0.1}{1-0.316}=\frac{1.1}{0.684}=\frac{1100}{684}=1.6$$

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

A5.59

B7.826

C8.944

D10.062

Explanation:

$$\frac{\sqrt{5}}{2}-\frac{10}{\sqrt{5}}+\sqrt{125}=\frac{(\sqrt{5})^2-20+2\sqrt{5}\times5\sqrt{5}}{2\sqrt{5}}=\frac{5-20+50}{2\sqrt{5}}$$
$$\frac{35}{2\sqrt{5}}\times\frac{\sqrt{5}}{\sqrt{5}}=\frac{35\sqrt{5}}{10}=\frac{7}{2}\times2.236=7\times1.118=7.826$$

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

A213444

B214344

C214434

D231444

Explanation:

L.C.M of 21,36,66=2772. Now,$$2772=2\times2\times3\times3\times7\times11$$ to make it a perfect square , it must be multiplied by $$7\times11$$.So required number =$$2^2\times3^2\times7^2\times11^2=213444$$

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Find the smallest number by which 5808 should be multiplied so that the products become a perfect square

A2

B3

C7

D11

Explanation:

$$5808=2\times2\times2\times2\times3\times11\times11=2^2\times2^2\times3\times11^2$$
To make it a perfect square ,it must be multiplied by 3

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What is the least number which should be substracted from 0.000326 to make perfect square ?

A0.000002

B0.000004

C0.02

D0.04

Explanation:

$$0.000326=\frac{326}{10^6}$$,
then Required number to be substracted $$=\frac{2}{10^6}=0.000002$$

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The greatest four-digit perfect square number is:

A9000

B9801

C9900

D9981

Explanation:

Here is no explanation for this answer

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Given $$\sqrt{5}=2.2361,\sqrt{3}=1.7321,then\frac{1}{\sqrt{5}-\sqrt{3}}$$ is equal to :

A1.98

B1.984

C1.9841

D2

Explanation:

$$\frac{1}{(\sqrt{5}-\sqrt{3})}=\frac{1}{(\sqrt{5}-\sqrt{3})}\times\frac{(\sqrt{5}+\sqrt{3})}{(\sqrt{5}+\sqrt{3})}=\frac{(\sqrt{5}+\sqrt{3})}{(5-3)}=\frac{(2.2361+1.7321)}{2}$$
$$=\frac{3.9682}{2}=1.9841$$

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$$\left(2+\sqrt{2}+\frac{1}{2+\sqrt{2}}+\frac{1}{\sqrt{2}-2}\right)$$ simplifies to:

A$$2-\sqrt{2}$$

B2

C$$2+\sqrt{2}$$

D$$2\sqrt{2}$$

Explanation:

Given$$=(2+\sqrt{2})+\frac{1}{(2+\sqrt{2})}\times\frac{(2-\sqrt{2})}{(2-\sqrt{2})}-\frac{1}{(2-\sqrt{2})}\times\frac{(2+\sqrt{2})}{(2+\sqrt{2})}$$
$$=(2+\sqrt{2})+\frac{(2-\sqrt{2})}{(4-2)}-\frac{(2+\sqrt{2})}{(4-2)}$$
$$=(2+\sqrt{2})+\frac{1}{2}(2-\sqrt{2})-\frac{1}{2}(2+\sqrt{2})=2$$

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