51.
Given 2x = 8y+1 and 9y = 3x-9 , then value of x + y is = ?
- (A) 18
- (B) 21
- (C) 24
- (D) 27
Solution:
$$\eqalign{ & {{\text{2}}^x}{\text{ = }}{{\text{8}}^{y + 1}} \cr & \Leftrightarrow {{\text{2}}^x}{\text{ = }}{\left( {{2^3}} \right)^{y + 1}} = {2^{\left( {3y + 3} \right)}} \cr & \Leftrightarrow x = 3y + 3 \cr & \Leftrightarrow x - 3y = 3.......(i) \cr & {9^y} = {3^{x - 9}} \cr & \Leftrightarrow {\left( {{3^2}} \right)^y}{\text{ = }}{{\text{3}}^{x - 9}} \cr & \Leftrightarrow 2y = x - 9 \cr & \Leftrightarrow x - 2y = 9......({\text{ii}}) \cr & {\text{Subtracting (i) from (ii),}} \cr & {\text{we}}\,{\text{get}}\,y = 6 \cr & {\text{Putting }}y\,{\text{ = 6 in (i),}} \cr & {\text{we get }}x{\text{ = 21}} \cr & \therefore x + y = 21 + 6 = 27 \cr} $$
52.
( - - ) Equals to = ?
- (A) 2 -
- (B) - 2
- (C) 2
- (D) -2
Solution:
$$\eqalign{ & \left( {\sqrt 8 - \sqrt 4 - \sqrt 2 } \right) \cr & = 2\sqrt 2 - 2 - \sqrt 2 \cr & = 2\sqrt 2 - \sqrt 2 - 2 \cr & = \sqrt 2(2 - 1) - 2 \cr & = \sqrt 2 - 2 \cr} $$
53.
Let x = ÃÂÃÂ and y = ÃÂÃÂ ÃÂÃÂ then the value of x2 + y2 is:
Solution:
$$\eqalign{ & x = \root 6 \of {27} - \sqrt {6\frac{3}{4}} \cr & {x^2} = {\left( {{{27}^{\frac{1}{6}}} - \frac{{{{27}^{\frac{1}{2}}}}}{2}} \right)^2} \cr & {x^2} = {27^{\frac{2}{6}}} + \frac{{27}}{4} - \frac{{2 \times {{27}^{\frac{1}{6}}} \times {{27}^{\frac{1}{2}}}}}{2} \cr & {x^2} = 3 + \frac{{27}}{4} - 9 \cr & {x^2} = \frac{3}{4} \cr & y = \frac{{\sqrt {45} + \sqrt {605} + \sqrt {245} }}{{\sqrt {80} + \sqrt {125} }} \cr & y = \frac{{3\sqrt 5 + 11\sqrt 5 + 7\sqrt 5 }}{{4\sqrt 4 + 5\sqrt 5 }} \cr & y = \frac{7}{3} \cr & {y^2} = \frac{{49}}{9} \cr & {x^2} + {y^2} = \frac{3}{4} + \frac{{49}}{9} = \boxed{\frac{{223}}{{36}}} \cr} $$
54.
If ax = b, by = c and cz = a, then the value of xyz is = ?
Solution:
$$\eqalign{ & {a^1} \cr & = {c^z} \cr & = {\left( {{b^y}} \right)^z} \cr & = {b^{yz}} \cr & = {\left( {{a^x}} \right)^{yz}} \cr & = {a^{xyz}} \cr & \Rightarrow xyz = 1 \cr} $$
55.
Simplified from of
Solution:
$$\eqalign{ & {\left[ {{{\left( {\root 5 \of {{x^{ - \frac{3}{5}}}} } \right)}^{ - \frac{5}{3}}}} \right]^{ - 5}} \cr & = {\left[ {{{\left( {{x^{ - \frac{3}{{25}}}}} \right)}^{ - \frac{5}{3}}}} \right]^{ - 5}} \cr & = {\left[ {\left( {{x^{\frac{1}{5}}}} \right)} \right]^{ - 5}} \cr & = {x^{ - \frac{1}{5} \times 5}} \cr & = {x^{ - 1}} \cr & = \frac{1}{x} \cr} $$
56.
The value of ÃÂÃÂ is = ?
- (A) 4
- (B) 9
- (C) 12
- (D) 31.25
Solution:
$$\eqalign{ & {\left( {0.03125} \right)^{ - \frac{2}{5}}} \cr & = {\left[ {{{\left( {0.5} \right)}^5}} \right]^{ - \frac{2}{5}}} \cr & = {0.5^{\left[ {5 \times \left( { - \frac{2}{5}} \right)} \right]}} \cr & = {\left( {0.5} \right)^{ - 2}} \cr & = \frac{1}{{{{\left( {0.5} \right)}^2}}} \cr & = \frac{1}{{0.25}} \cr & = 4 \cr} $$
57.
The Simplified value of
- (A) 2√6
- (B) 2
- (C) √3
- (D) 0
Solution:
$$\eqalign{ & \frac{{\sqrt 6 + 2}}{{\sqrt 2 + \sqrt {2 + \sqrt 3 } }} - \frac{{\sqrt 6 - 2}}{{\sqrt 2 - \sqrt {2 - \sqrt 3 } }} - \frac{{2\sqrt 2 }}{{2 + \sqrt 2 }} \cr & \Rightarrow \frac{{\sqrt 6 + 2}}{{\sqrt 2 + \frac{{\sqrt 3 + 1}}{{\sqrt 2 }}}} - \frac{{\sqrt 6 - 2}}{{\sqrt 2 - \frac{{\sqrt 3 - 1}}{{\sqrt 2 }}}} - \frac{2}{{\sqrt 2 + 1}} \cr & \Rightarrow \frac{{\left( {\sqrt 6 + 2} \right)\sqrt 2 }}{{2 + \sqrt 3 + 1}} - \frac{{\left( {\sqrt 6 - 2} \right)\sqrt 2 }}{{2 - \sqrt 3 + 1}} - \frac{2}{{\sqrt 2 + 1}} \cr & \Rightarrow \frac{{\sqrt 2 }}{{\sqrt 3 }}\left[ {\frac{{\sqrt 6 + 2}}{{\left( {\sqrt 3 + 1} \right)}} - \frac{{\sqrt 6 - 2}}{{\left( {\sqrt 3 - 1} \right)}}} \right] - \frac{2}{{\sqrt 2 + 1}} \times \frac{{\sqrt 2 - 1}}{{\sqrt 2 - 1}} \cr & \Rightarrow \frac{{\sqrt 2 }}{{\sqrt 3 }}\left[ {\frac{{\sqrt {18} + 2\sqrt 3 - \sqrt 6 - 2 - \sqrt {18} + 2\sqrt 3 - \sqrt 6 + 2}}{{3 - 1}}} \right] - \frac{{2\left( {\sqrt 2 - 1} \right)}}{{2 - 1}} \cr & \Rightarrow \frac{{\sqrt 2 }}{{\sqrt 3 }}\left[ {\frac{{2\left( { - \sqrt 6 + 2\sqrt 3 } \right)}}{2}} \right] - 2\left( {\sqrt 2 - 1} \right) \cr & \Rightarrow - \sqrt 3 \times \sqrt 2 \times \frac{{\sqrt 2 }}{{\sqrt 3 }} + 2\sqrt 3 \times \frac{{\sqrt 2 }}{{\sqrt 3 }} - 2\left( {\sqrt 2 - 1} \right) \cr & \Rightarrow - 2 + 2\sqrt 2 - 2\sqrt 2 + 2 \cr & \Rightarrow 0 \cr} $$
58.
If 3(x+y) = 81 and 81(x-y) = 3, then the value of x is = ?
Solution:
$$\eqalign{ & {{\text{3}}^{x + y}}{\text{ = 81}} \cr & {{\text{3}}^{x + y}}{\text{ = }}{{\text{3}}^4} \cr & x + y = 4.....(i) \cr & {\text{8}}{{\text{1}}^{x - y}}{\text{ = 3}} \cr & {3^{4x - 4y}}{\text{ = }}{{\text{3}}^1} \cr & 4x - 4y{\text{ = 1}}....{\text{(ii)}} \cr & {\text{From equation (i) and (ii)}} \cr & 4x - 4y = 1 \cr & 4x + 4y = 16 \cr & 8x = 17 \cr & x = \frac{{17}}{8} \cr} $$
59.
If ÃÂÃÂ ÃÂÃÂ then the value of x is = ?
Solution:
$$\eqalign{ & {{\text{5}}^{\left( {x + 3} \right)}}{\text{ = 2}}{{\text{5}}^{(3x - 4)}} \cr & \Rightarrow {{\text{5}}^{\left( {x + 3} \right)}}{\text{ = }}{\left( {{5^2}} \right)^{(3x - 4)}} \cr & \Rightarrow {{\text{5}}^{\left( {x + 3} \right)}}{\text{ = }}{{\text{5}}^2}^{(3x - 4)} \cr & \Rightarrow {{\text{5}}^{\left( {x + 3} \right)}}{\text{ = }}{{\text{5}}^{(6x - 8)}} \cr & \Rightarrow x + 3 = 6x - 8 \cr & \Rightarrow 5x = 11 \cr & \Rightarrow x = \frac{{11}}{5} \cr} $$
60.
Given = 1.414, the value of ÃÂÃÂ is = ?
- (A) 8.484
- (B) 8.526
- (C) 8.426
- (D) 8.876
Solution:
$$\eqalign{ & \sqrt 2 = 1.414 \cr & \Rightarrow \sqrt 8 {\text{ + 2}}\sqrt {32} - 3\sqrt {128} {\text{ + 4}}\sqrt {50} \cr & \Rightarrow 2\sqrt 2 + 2 \times 4\sqrt 2 - 3 \times 8\sqrt 2 + 4 \times 5\sqrt 2 \cr & \Rightarrow 2\sqrt 2 + 8\sqrt 2 - 24\sqrt 2 + 20\sqrt 2 \cr & \Rightarrow 6\sqrt 2 \cr & \Rightarrow 6 \times 1.414 \cr & \Rightarrow 8.484{\text{ }} \cr} $$