11.
$$\sqrt {\sqrt {17956} + \sqrt {24025} } = ?$$
(A) 19
(B) 155
(C) 256
(D) 289
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Solution:
Answer & Solution Answer: Option E Solution: $$\eqalign{ & \sqrt {\sqrt {17956} + \sqrt {24025} } \cr & = \sqrt {134 + 155} \cr & = \sqrt {289} \cr & = 17 \cr} $$
12.
If $$\sqrt 2 = 1.414{\text{,}}$$ ÃÂÃÂ the square root of $$\frac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}}$$ ÃÂÃÂ is nearest to = ?
(A) 0.172
(B) 0.414
(C) 0.586
(D) 1.414
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Solution:
$$\eqalign{ & = \frac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}} \cr & = \frac{{\left( {\sqrt 2 - 1} \right)}}{{\left( {\sqrt 2 + 1} \right)}} \times \frac{{\left( {\sqrt 2 - 1} \right)}}{{\left( {\sqrt 2 - 1} \right)}} \cr & = {\left( {\sqrt 2 - 1} \right)^2} \cr & \therefore \sqrt {\frac{{\sqrt 2 - 1}}{{\sqrt 2 + 1}}} \cr & = \left( {\sqrt 2 - 1} \right) \cr & = \left( {1.414 - 1} \right) \cr & = 0.414 \cr} $$
13.
The square root of $${\text{0}}{\text{.}}\overline {\text{4}} $$ ÃÂÃÂ is ?
(A) $$0.\overline 6 $$
(B) $$0.\overline 7 $$
(C) $$0.\overline 8 $$
(D) $$0.\overline 9 $$
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Solution:
$$\eqalign{ & = \sqrt {{\text{0}}{\text{.}}\overline {\text{4}} {\text{ }}} \cr & = \sqrt {\frac{4}{9}} \cr & = \frac{2}{3} \cr & = 0.666...... \cr & = 0.\overline 6 \cr} $$
14.
The least perfect square number divisible by 3, 4, 5, 6 and 8 is = ?
(A) 900
(B) 1200
(C) 2500
(D) 3600
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Solution:
L.C.M. of 3, 4, 5, 6, 8 is 120 Now 120 = 2 × 2 × 2 × 3 × 5 To make it a perfect square, it must be multiplied by 2 × 3 × 5 So, required number $$\eqalign{ & = {2^2} \times {2^2} \times {3^2} \times {5^2} \cr & = 3600 \cr} $$
15.
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
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Solution:
$$\eqalign{ & \frac{{\sqrt 5 }}{2} - \frac{{10}}{{\sqrt 5 }} + \sqrt {125} \, \cr & = \frac{{{{\left( {\sqrt 5 } \right)}^2} - 20 + 2\sqrt 5 \times 5\sqrt 5 }}{{2\sqrt 5 }} \cr & = \frac{{5 - 20 + 50}}{{2\sqrt 5 }} \cr & = \frac{{35}}{{2\sqrt 5 }} \times \frac{{\sqrt 5 }}{{\sqrt 5 }} \cr & = \frac{{35\sqrt 5 }}{{10}} \cr & = \frac{{7 \times 2.236}}{2} \cr & = 7 \times 1.118 \cr & = 7.826 \cr} $$
16.
The square root of $$\left( {7 + 3\sqrt 5 } \right)$$ÃÂÃÂ $$\left( {7 - 3\sqrt 5 } \right)$$ ÃÂÃÂ is ?
(A) $$\sqrt 5 $$
(B) 2
(C) 4
(D) $$3\sqrt 5 $$
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Solution:
$$\eqalign{ & = \sqrt {\left( {7 + 3\sqrt 5 } \right)\left( {7 - 3\sqrt 5 } \right)} \cr & = \sqrt {{{\left( 7 \right)}^2} - {{\left( {3\sqrt 5 } \right)}^2}} \cr & = \sqrt {49 - 45} \cr & = \sqrt 4 \cr & = 2 \cr} $$
17.
In the equation $$\frac{{4050}}{{\sqrt x }} = 450{\text{,}}$$ ÃÂÃÂ the value of x is = ?
(A) 9
(B) 49
(C) 81
(D) 100
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Solution:
$$\eqalign{ & \frac{{4050}}{{\sqrt x }} = 450 \cr & \Leftrightarrow \sqrt x = \frac{{4050}}{{450}} \cr & \Leftrightarrow \sqrt x = 9 \cr & \Leftrightarrow x = {\left( 9 \right)^2} \cr & \Leftrightarrow x = 81 \cr} $$
18.
If $$\frac{{52}}{x} = \sqrt {\frac{{169}}{{289}}} {\text{,}}$$ ÃÂÃÂ the value of x is = ?
(A) 52
(B) 58
(C) 62
(D) 68
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Solution:
$$\eqalign{ & \Leftrightarrow \frac{{52}}{x}{\text{ = }}\sqrt {\frac{{169}}{{289}}} \cr & \Leftrightarrow \frac{{52}}{x} = \frac{{13}}{{17}} \cr & \Leftrightarrow x = \left( {\frac{{52 \times 17}}{{13}}} \right) \cr & \Leftrightarrow x = 68 \cr} $$
19.
Given $$\sqrt 5 = 2.2361,$$ ÃÂÃÂ $$\sqrt 3 = 1.7321{\text{,}}$$ ÃÂÃÂ then $$\frac{1}{{\sqrt 5 - \sqrt 3 }}$$ ÃÂÃÂ is equal to ?
(A) 1.98
(B) 1.984
(C) 1.9841
(D) 2
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Solution:
$$\eqalign{ & \Rightarrow \frac{1}{{\sqrt 5 - \sqrt 3 }} \cr & = \frac{1}{{\sqrt 5 - \sqrt 3 }} \times \frac{{\left( {\sqrt 5 + \sqrt 3 } \right)}}{{\left( {\sqrt 5 + \sqrt 3 } \right)}} \cr & = \frac{{\left( {\sqrt 5 + \sqrt 3 } \right)}}{{5 - 3}} \cr & = \frac{{\left( {2.2361 + 1.7321} \right)}}{2} \cr & = \frac{{3.9682}}{2} \cr & = 1.9841{\text{ }} \cr} $$
20.
If $$3a = 4b = 6c$$ ÃÂÃÂ and $$a + b + c = 27\sqrt {29} {\text{,}}$$ ÃÂÃÂ ÃÂÃÂ then $$\sqrt {{a^2} + {b^2} + {c^2}} $$ ÃÂÃÂ ÃÂÃÂ is ?
(A) $$3\sqrt {29} $$
(B) 81
(C) 87
(D) None of these
Show Answer
Solution:
$$\eqalign{ & a + b + c = 27\sqrt {29} \cr & \Rightarrow 2c + \frac{3}{2}c + c = 27\sqrt {29} \cr & \Rightarrow \frac{9}{2}c = 27\sqrt {29} \cr & \Rightarrow c = 6\sqrt {29} \cr & \therefore \sqrt {{a^2} + {b^2} + {c^2}} \cr & = \sqrt {{{\left( {a + b + c} \right)}^2} - 2\left( {ab + bc + ca} \right)} \cr & = \sqrt {{{\left( {27\sqrt {29} } \right)}^2} - 2\left( {2c \times \frac{3}{2}c + \frac{3}{2}c \times c + c \times 2c} \right)} \cr & = \sqrt {\left( {729 \times 29} \right) - 2\left( {3{c^2} + \frac{3}{2}{c^2} + 2{c^2}} \right)} \cr & = \sqrt {\left( {729 \times 29} \right) - 2 \times \frac{{13}}{2}{c^2}} \cr & = \sqrt {\left( {729 \times 29} \right) - 13{{\left( {6\sqrt {29} } \right)}^2}} \cr & = \sqrt {29\left( {729 - 468} \right)} \cr & = \sqrt {29 \times 261} \cr & = \sqrt {29 \times 29 \times 9} \cr & = 29 \times 3 \cr & = 87 \cr} $$