111.
ÃÂÃÂ ÃÂÃÂ is equal to :
Solution:
$$\eqalign{ & {\text{Give}}\,{\text{Expression}} \cr & = \frac{{25}}{{11}} \times \frac{{14}}{5} \times \frac{{11}}{{14}} \cr & = 5 \cr} $$
112.
The square root of ÃÂÃÂ ÃÂÃÂ is ?
Solution:
$$\eqalign{ & = \sqrt {\left( {7 + 3\sqrt 5 } \right)\left( {7 - 3\sqrt 5 } \right)} \cr & = \sqrt {{{\left( 7 \right)}^2} - {{\left( {3\sqrt 5 } \right)}^2}} \cr & = \sqrt {49 - 45} \cr & = \sqrt 4 \cr & = 2 \cr} $$
113.
If ÃÂÃÂ then the value of ÃÂÃÂ is = ?
- (A) 0.6122
- (B) 0.8163
- (C) 1.223
- (D) 1.2245
Solution:
$$\eqalign{ & = \frac{{3\sqrt 2 }}{{2\sqrt 3 }} \cr & = \frac{{3\sqrt 2 }}{{2\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }} \cr & = \frac{{3\sqrt 6 }}{{2 \times 3}} \cr & = \frac{{\sqrt 6 }}{2} \cr & = \frac{{2.449}}{2} \cr & = 1.2245 \cr} $$
114.
- (A) 2025
- (B) 2209
- (C) 2304
- (D) 2401
Solution:
$$\eqalign{ & {\text{Let}}, \cr & \sqrt x \times \sqrt {484} = 1034 \cr & {\text{Then}}, \cr & \sqrt x \times 22 = 1034 \cr & \Leftrightarrow \sqrt x = \frac{{1034}}{{22}} = 47 \cr & \Leftrightarrow x = {\left( {47} \right)^2} \cr & \Leftrightarrow x = 2209 \cr} $$
115.
The least number by which 294 must be multiplied to make it a perfect square, is = ?
Solution:
294 = 7 × 7 × 2 × 3 To make it a perfect square, it must be multiplied by 2 × 3 i.e.,6 ∴ Required number = 6
116.
If ÃÂÃÂ and ÃÂÃÂ ÃÂÃÂ then ÃÂÃÂ ÃÂÃÂ is ?
- (A)
- (B) 81
- (C) 87
- (D) None of these
Solution:
$$\eqalign{ & a + b + c = 27\sqrt {29} \cr & \Rightarrow 2c + \frac{3}{2}c + c = 27\sqrt {29} \cr & \Rightarrow \frac{9}{2}c = 27\sqrt {29} \cr & \Rightarrow c = 6\sqrt {29} \cr & \therefore \sqrt {{a^2} + {b^2} + {c^2}} \cr & = \sqrt {{{\left( {a + b + c} \right)}^2} - 2\left( {ab + bc + ca} \right)} \cr & = \sqrt {{{\left( {27\sqrt {29} } \right)}^2} - 2\left( {2c \times \frac{3}{2}c + \frac{3}{2}c \times c + c \times 2c} \right)} \cr & = \sqrt {\left( {729 \times 29} \right) - 2\left( {3{c^2} + \frac{3}{2}{c^2} + 2{c^2}} \right)} \cr & = \sqrt {\left( {729 \times 29} \right) - 2 \times \frac{{13}}{2}{c^2}} \cr & = \sqrt {\left( {729 \times 29} \right) - 13{{\left( {6\sqrt {29} } \right)}^2}} \cr & = \sqrt {29\left( {729 - 468} \right)} \cr & = \sqrt {29 \times 261} \cr & = \sqrt {29 \times 29 \times 9} \cr & = 29 \times 3 \cr & = 87 \cr} $$
117.
What should come in place of both x in the equation
- (A) 12
- (B) 14
- (C) 144
- (D) 196
Solution:
$$\eqalign{ & {\text{Let}}\,\frac{x}{{\sqrt {128} }} = \frac{{\sqrt {162} }}{x} \cr & {\text{Then}}\,{x^2} = \sqrt {128 \times 162} \cr & = \sqrt {64 \times 2 \times 18 \times 9} \cr & = \sqrt {{8^2} \times {6^2} \times {3^2}} \cr & = 8 \times 6 \times 3 \cr & = 144 \cr & \therefore x = \sqrt {144} = 12 \cr} $$