Surds and Indices

1.The cube root of .000216 is:

1. .6
2. .06
3. 77
4. 87

Answer:Option 1

Explanation:

Solution:

\begin{aligned} {(.000216)}^{\frac{1}{3}} = {(\frac{216}{10^{6}})}^{\frac{1}{3}} \\ = {(\frac{6 \times 6 \times 6}{10^{2} \times 10^{2} \times 10^{2}})}^{\frac{1}{3}} \\ = \frac{6}{10^{2}} \\ = \frac{6}{100} \\ = 0.06 \end{aligned}

$\eqalign{ & {\left( {.000216} \right)^{\frac{1}{3}}} = {\left( {\frac{{216}}{{{{10}^6}}}} \right)^{\frac{1}{3}}} \cr & = {\left( {\frac{{6 \times 6 \times 6}}{{{{10}^2} \times {{10}^2} \times {{10}^2}}}} \right)^{\frac{1}{3}}} \cr & = \frac{6}{{{{10}^2}}} \cr & = \frac{6}{{100}} \cr & = 0.06 \cr}$

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2.What should come in place of both x in the equation

\frac{x}{\sqrt{128}} = \frac{\sqrt{162}}{x}

$\frac{x}{{\sqrt {128} }} = \frac{{\sqrt {162} }}{x}$

1. 12
2. 14
3. 144
4. 196

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Let \frac{x}{\sqrt{128}} = \frac{\sqrt{162}}{x} \\ Then x^{2} = \sqrt{128 \times 162} \\ = \sqrt{64 \times 2 \times 18 \times 9} \\ = \sqrt{8^{2} \times 6^{2} \times 3^{2}} \\ = 8 \times 6 \times 3 \\ = 144 \\ ∴ x = \sqrt{144} = 12 \end{aligned}

$\eqalign{ & {\text{Let}}\,\frac{x}{{\sqrt {128} }} = \frac{{\sqrt {162} }}{x} \cr & {\text{Then}}\,{x^2} = \sqrt {128 \times 162} \cr & = \sqrt {64 \times 2 \times 18 \times 9} \cr & = \sqrt {{8^2} \times {6^2} \times {3^2}} \cr & = 8 \times 6 \times 3 \cr & = 144 \cr & \therefore x = \sqrt {144} = 12 \cr}$

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

1. 213444
2. 214344
3. 214434
4. 231444

Answer:Option 1

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

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4.If

3 \sqrt{5} + \sqrt{125}

$3\sqrt 5 + \sqrt {125}$ = 17.88, then what will be the value of

\sqrt{80} + 6 \sqrt{5} ?

$\sqrt {80} + 6\sqrt 5 \,\,?$

1. 13.41
2. 20.46
3. 21.66
4. 22.35

Answer:Option 1

Explanation:

Solution:

\begin{aligned} 3 \sqrt{5} + \sqrt{125} = 17.88 \\ \Rightarrow 3 \sqrt{5} + \sqrt{25 \times 5} = 17.88 \\ \Rightarrow 3 \sqrt{5} + 5 \sqrt{5} = 17.88 \\ \Rightarrow 8 \sqrt{5} = 17.88 \\ \Rightarrow \sqrt{5} = 2.235 \\ ∴ \sqrt{80} + 6 \sqrt{5} = \sqrt{16 \times 5} + 6 \sqrt{5} \\ = 4 \sqrt{5} + 6 \sqrt{5} \\ = 10 \sqrt{5} = (10 \times 2.235) = 22.35 \end{aligned}

$\eqalign{ & 3\sqrt 5 + \sqrt {125} = 17.88 \cr & \Rightarrow 3\sqrt 5 + \sqrt {25 \times 5} = 17.88 \cr & \Rightarrow 3\sqrt 5 + 5\sqrt 5 = 17.88 \cr & \Rightarrow 8\sqrt 5 = 17.88 \cr & \Rightarrow \sqrt 5 = 2.235 \cr & \therefore \sqrt {80} + 6\sqrt 5 = \sqrt {16 \times 5} + 6\sqrt 5 \cr & = 4\sqrt 5 + 6\sqrt 5 \cr & = 10\sqrt 5 = \left( {10 \times 2.235} \right) = 22.35 \cr}$

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5.If a = 0.1039, then the value of

\sqrt{4 a^{2} - 4 a + 1} + 3 a

$\sqrt {4{a^2} - 4a + 1} + 3a$ is:

1. 0.1039
2. 0.2078
3. 1.1039
4. 2.1039

Answer:Option 1

Explanation:

Solution:

\begin{aligned} \sqrt{4 a^{2} - 4 a + 1} + 3 a \\ = \sqrt{{(1)}^{2} + {(2 a)}^{2} - 2 \times 1 \times 2 a} + 3 a \\ = \sqrt{{(1 - 2 a)}^{2}} + 3 a \\ = (1 - 2 a) + 3 a \\ = (1 + a) \\ = (1 + 0.1039) \\ = 1.1039 \end{aligned}

$\eqalign{ & \sqrt {4{a^2} - 4a + 1} + 3a \cr & = \sqrt {{{\left( 1 \right)}^2} + {{\left( {2a} \right)}^2} - 2 \times 1 \times 2a} + 3a \cr & = \sqrt {{{\left( {1 - 2a} \right)}^2}} + 3a \cr & = \left( {1 - 2a} \right) + 3a \cr & = \left( {1 + a} \right) \cr & = \left( {1 + 0.1039} \right) \cr & = 1.1039 \cr}$

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6.If

x = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}

$x = \frac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}$ and

y = \frac{\sqrt{3} - 1}{\sqrt{3} + 1},

$y = \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}},$ then the value of

(x^{2} + y^{2})

$\left( {{x^2} + {y^2}} \right)$ is?

1. 10
2. 13
3. 14
4. 15

Answer:Option 1

Explanation:

Solution:

\begin{aligned} x = \frac{(\sqrt{3} + 1)}{(\sqrt{3} - 1)} \times \frac{(\sqrt{3} + 1)}{(\sqrt{3} + 1)} \\ = \frac{{(\sqrt{3} + 1)}^{2}}{(3 - 1)} \\ = \frac{3 + 1 + 2 \sqrt{3}}{2} \\ = 2 + \sqrt{3} \\ y = \frac{(\sqrt{3} - 1)}{(\sqrt{3} + 1)} \times \frac{(\sqrt{3} - 1)}{(\sqrt{3} - 1)} \\ = \frac{{(\sqrt{3} - 1)}^{2}}{(3 - 1)} \\ = \frac{3 + 1 - 2 \sqrt{3}}{2} \\ = 2 - \sqrt{3} \\ ∴ x^{2} + y^{2} = {(2 + \sqrt{3})}^{2} + {(2 - \sqrt{3})}^{2} \\ = 2 (4 + 3) \\ = 14 \end{aligned}

$\eqalign{ & x = \frac{{\left( {\sqrt 3 + 1} \right)}}{{\left( {\sqrt 3 - 1} \right)}} \times \frac{{\left( {\sqrt 3 + 1} \right)}}{{\left( {\sqrt 3 + 1} \right)}} \cr & \,\,\,\,\,\, = \frac{{{{\left( {\sqrt 3 + 1} \right)}^2}}}{{\left( {3 - 1} \right)}} \cr & \,\,\,\,\,\, = \frac{{3 + 1 + 2\sqrt 3 }}{2} \cr & \,\,\,\,\,\, = 2 + \sqrt 3 \cr & y = \frac{{\left( {\sqrt 3 - 1} \right)}}{{\left( {\sqrt 3 + 1} \right)}} \times \frac{{\left( {\sqrt 3 - 1} \right)}}{{\left( {\sqrt 3 - 1} \right)}} \cr & \,\,\,\,\,\, = \frac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{{\left( {3 - 1} \right)}} \cr & \,\,\,\,\,\, = \frac{{3 + 1 - 2\sqrt 3 }}{2} \cr & \,\,\,\,\,\, = 2 - \sqrt 3 \cr & \therefore {x^2} + {y^2} = {\left( {2 + \sqrt 3 } \right)^2} + {\left( {2 - \sqrt 3 } \right)^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2\left( {4 + 3} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 14 \cr}$

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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:

1. 57
2. 67
3. 77
4. 87

Answer:Option 1

Explanation:

Solution:

Money collected = (59.29 x 100) paise = 5929 paise.
∴ Number of members =

\sqrt{5929}

$\sqrt {5929}$ = 77

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8.If

\sqrt{5} = 2.236,

$\sqrt 5 = 2.236,$ then the value of

\frac{\sqrt{5}}{2}

$\frac{{\sqrt 5 }}{2}$

-

$-$

\frac{10}{\sqrt{5}}

$\frac{{10}}{{\sqrt 5 }}$

+

$+$

\sqrt{125}

$\sqrt {125}$ is equal to :

1. 5.59
2. 7.826
3. 8.944
4. 10.062

Answer:Option 1

Explanation:

Solution:

\begin{aligned} \frac{\sqrt{5}}{2} - \frac{10}{\sqrt{5}} + \sqrt{125} \\ = \frac{{(\sqrt{5})}^{2} - 20 + 2 \sqrt{5} \times 5 \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 \times 2.236}{2} \\ = 7 \times 1.118 \\ = 7.826 \end{aligned}

$\eqalign{ & \frac{{\sqrt 5 }}{2} - \frac{{10}}{{\sqrt 5 }} + \sqrt {125} \, \cr & = \frac{{{{\left( {\sqrt 5 } \right)}^2} - 20 + 2\sqrt 5 \times 5\sqrt 5 }}{{2\sqrt 5 }} \cr & = \frac{{5 - 20 + 50}}{{2\sqrt 5 }} \cr & = \frac{{35}}{{2\sqrt 5 }} \times \frac{{\sqrt 5 }}{{\sqrt 5 }} \cr & = \frac{{35\sqrt 5 }}{{10}} \cr & = \frac{{7 \times 2.236}}{2} \cr & = 7 \times 1.118 \cr & = 7.826 \cr}$

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9.

(\frac{\sqrt{625}}{11} \times \frac{14}{\sqrt{25}} \times \frac{11}{\sqrt{196}})

$\left( {\frac{{\sqrt {625} }}{{11}} \times \frac{{14}}{{\sqrt {25} }} \times \frac{{11}}{{\sqrt {196} }}} \right){\kern 1pt}$ is equal to :

1. 5
2. 6
3. 8
4. 11

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Give Expression = \\ \frac{25}{11} \times \frac{14}{5} \times \frac{11}{14} = 5 \end{aligned}

$\eqalign{ & {\text{Give}}\,{\text{Expression}} = \cr & \frac{{25}}{{11}} \times \frac{{14}}{5} \times \frac{{11}}{{14}} = 5 \cr}$

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10.

\sqrt{0.0169 \times ?} = 1.3

$\sqrt {0.0169 \times ?} = 1.3$

1. 10
2. 100
3. 1000
4. None of these

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Let \sqrt{0.0169 \times x} = 1.3 \\ Then, 0.0169 x = {(1.3)}^{2} = 1.69 \\ \Rightarrow x = \frac{1.69}{0.0169} = 100 \end{aligned}

$\eqalign{ & {\text{Let}}\sqrt {0.0169 \times x} = 1.3 \cr & {\text{Then}},\,0.0169x = {\left( {1.3} \right)^2} = 1.69 \cr & \Rightarrow x = \frac{{1.69}}{{0.0169}} = 100 \cr}$

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