61.
$$\sqrt {176 + \sqrt {2401} } $$ ÃÂÃÂ ÃÂÃÂ is equal to = ?
(A) 14
(B) 15
(C) 18
(D) 24
Show Answer
Solution:
$$\eqalign{ & {\text{Given expression,}} \cr & = \sqrt {176 + 49} \cr & = \sqrt {225} \cr & = 15 \cr} $$
62.
If $$\sqrt {{3^n}} = 729$$ ÃÂÃÂ , then the value of n is ?
(A) 6
(B) 8
(C) 10
(D) 12
Show Answer
Solution:
$$\eqalign{ & \Leftrightarrow \sqrt {{3^n}} = 729 \cr & \Leftrightarrow \sqrt {{3^n}} = {3^6} \cr & \Leftrightarrow {\left( {\sqrt {{3^n}} } \right)^2} = {\left( {{3^6}} \right)^2} \cr & \Leftrightarrow {3^n} = {3^{12}} \cr & \Leftrightarrow n = 12 \cr} $$
63.
If $$\sqrt x \div \sqrt {441} = 0.02{\text{,}}$$ ÃÂÃÂ ÃÂÃÂ then the value of x is ?
(A) 0.1764
(B) 1.764
(C) 1.64
(D) 2.64
Show Answer
Solution:
$$\eqalign{ & \Leftrightarrow \frac{{\sqrt x }}{{\sqrt {441} }} = 0.02 \cr & \Leftrightarrow \frac{{\sqrt x }}{{21}} = 0.02 \cr & \Leftrightarrow \sqrt x = 0.02 \times 21 \cr & \Leftrightarrow \sqrt x = 0.42 \cr & \Leftrightarrow x = {\left( {0.42} \right)^2} \cr & \Leftrightarrow x = 0.1764{\text{ }} \cr} $$
64.
The square root of $$\left( {{{272}^2} - {{128}^2}} \right)$$ ÃÂÃÂ is = ?
(A) 144
(B) 200
(C) 240
(D) 256
Show Answer
Solution:
$$\eqalign{ & = \sqrt {\left( {{{272}^2} - {{128}^2}} \right)} \cr & = \sqrt {\left( {272 + 128} \right)\left( {272 - 128} \right)} \cr & = \sqrt {400 \times 144} \cr & = \sqrt {57600} \cr & = 240 \cr} $$
65.
If $$\sqrt 5 = 2.236, $$ ÃÂÃÂ then the value of $$\frac{{\sqrt 5 }}{2}$$ $$ - $$ $$\frac{{10}}{{\sqrt 5 }}$$ $$ + $$ $$\sqrt {125} $$ ÃÂÃÂ is equal to :
(A) 5.59
(B) 7.826
(C) 8.944
(D) 10.062
Show Answer
Solution:
$$\eqalign{ & \frac{{\sqrt 5 }}{2} - \frac{{10}}{{\sqrt 5 }} + \sqrt {125} \, \cr & = \frac{{{{\left( {\sqrt 5 } \right)}^2} - 20 + 2\sqrt 5 \times 5\sqrt 5 }}{{2\sqrt 5 }} \cr & = \frac{{5 - 20 + 50}}{{2\sqrt 5 }} \cr & = \frac{{35}}{{2\sqrt 5 }} \times \frac{{\sqrt 5 }}{{\sqrt 5 }} \cr & = \frac{{35\sqrt 5 }}{{10}} \cr & = \frac{{7 \times 2.236}}{2} \cr & = 7 \times 1.118 \cr & = 7.826 \cr} $$
66.
If $$x = 3 + \sqrt 8 ,$$ ÃÂÃÂ then $${x^2} + \frac{1}{{{x^2}}}$$ ÃÂÃÂ is equal to = ?
(A) 30
(B) 34
(C) 36
(D) 38
Show Answer
Solution:
$$\eqalign{ & \because x = 3 + \sqrt 8 \cr & \Rightarrow {x^2} = {\left( {3 + \sqrt 8 } \right)^2} \cr & \Rightarrow {x^2} = {3^2} + {\left( {\sqrt 8 } \right)^2} + 2 \times 3 \times \sqrt 8 \cr & \Rightarrow {x^2} = 9 + 8 + 6\sqrt 8 \cr & \Rightarrow {x^2} = 17 + 12\sqrt 2 \cr} $$ $$\therefore {x^2} + \frac{1}{{{x^2}}}$$ $$ = \left( {17 + 12\sqrt 2 } \right)$$ $$ + \frac{1}{{\left( {17 + 12\sqrt 2 } \right)}}$$ $$ \times \frac{{\left( {17 - 12\sqrt 2 } \right)}}{{\left( {17 - 12\sqrt 2 } \right)}}$$ $$\eqalign{ & = \left( {17 + 12\sqrt 2 } \right) + \frac{{\left( {17 - 12\sqrt 2 } \right)}}{{289 - 288}} \cr & = \left( {17 + 12\sqrt 2 } \right) + \left( {17 - 12\sqrt 2 } \right) \cr & = 17 + 17 \cr & = 34 \cr} $$
67.
$$\left( {\frac{{\sqrt {625} }}{{11}} \times \frac{{14}}{{\sqrt {25} }} \times \frac{{11}}{{\sqrt {196} }}} \right){\kern 1pt} $$ ÃÂÃÂ ÃÂÃÂ is equal to :
Show Answer
Solution:
$$\eqalign{ & {\text{Give}}\,{\text{Expression}} \cr & = \frac{{25}}{{11}} \times \frac{{14}}{5} \times \frac{{11}}{{14}} \cr & = 5 \cr} $$
68.
$${1.5^2} \times \sqrt {0.0225} = ?$$
(A) 0.0375
(B) 0.3375
(C) 3.275
(D) 32.75
Show Answer
Solution:
$$\eqalign{ & = {1.5^2} \times \sqrt {0.0225} \cr & = {1.5^2} \times \sqrt {\frac{{225}}{{10000}}} \cr & = 2.25 \times \frac{{15}}{{100}} \cr & = 2.25 \times 0.15 \cr & = 0.3375 \cr} $$
69.
The approximate value of $$\frac{{3\sqrt {12} }}{{2\sqrt {28} }}$$ÃÂÃÂ $$ \div $$ $$\frac{{2\sqrt {21} }}{{\sqrt {98} }}$$ ÃÂÃÂ is ?
(A) 1.0605
(B) 1.0727
(C) 1.6007
(D) 1.6026
Show Answer
Solution:
$$\eqalign{ & {\text{Given expression,}} \cr & = \frac{{3\sqrt {12} }}{{2\sqrt {28} }} \times \frac{{\sqrt {98} }}{{2\sqrt {21} }} \cr & = \frac{{3\sqrt {4 \times 3} }}{{2\sqrt {4 \times 7} }} \times \frac{{\sqrt {49 \times 2} }}{{2\sqrt {21} }} \cr & = \frac{{6\sqrt 3 }}{{4\sqrt 7 }} \times \frac{{7\sqrt 2 }}{{2\sqrt {21} }} \cr & = \frac{{21\sqrt 6 }}{{4\sqrt {7 \times 21} }} \cr & = \frac{{21\sqrt 6 }}{{28\sqrt 3 }} \cr & = \frac{3}{4}\sqrt 2 \cr & = \frac{3}{4} \times 1.414 \cr & = 3 \times 0.3535 \cr & = 1.0605 \cr} $$
70.
What should come in place of both the question marks in the equation ?
(A) 12
(B) 14
(C) 144
(D) 196
Show Answer
Solution:
$$\eqalign{ & {\text{Let,}} \cr & {\text{ }}\frac{x}{{\sqrt {128} }} = \frac{{\sqrt {162} }}{x} \cr & {\text{Then,}} \cr & \Leftrightarrow {x^2} = \sqrt {128 \times 162} \cr & \Leftrightarrow {x^2} = \sqrt {64 \times 2 \times 18 \times 9} \cr & \Leftrightarrow {x^2} = \sqrt {{8^2} \times {6^2} \times {3^2}} \cr & \Leftrightarrow {x^2} = 8 \times 6 \times 3 \cr & \Leftrightarrow {x^2} = 144 \cr & \Leftrightarrow x = \sqrt {144} \cr & \Leftrightarrow x = 12 \cr} $$