1. If 32x-y = 3x+y = $$\sqrt {27} {\text{,}}$$ the value of y is = ?
a) $$\frac{1}{2}$$
b) $$\frac{1}{4}$$
c) $$\frac{3}{2}$$
d) $$\frac{3}{4}$$
Explanation:
$$\eqalign{ & {{\text{3}}^{2x - y}}{\text{ = }}{{\text{3}}^{x + y}}{\text{ = }}\sqrt {{3^3}} = {3^{\frac{3}{2}}} \cr & 2x - y = \frac{3}{2}and \,\,x + y = \frac{3}{2} \cr & 3x = \frac{3}{2} + \frac{3}{2} = 3 \cr & x = 1 \cr & y = \left( {\frac{3}{2} - 1} \right) = \frac{1}{2} \cr} $$
2. $$\frac{{\sqrt {10 + \sqrt {25 + \sqrt {108 + \sqrt {154 + \sqrt {225} } } } } }}{{\root 3 \of 8 }} $$ = ?
a) 8
b) 4
c) $$\frac{1}{2}$$
d) 2
Explanation:
$$\eqalign{ & \frac{{\sqrt {10 + \sqrt {25 + \sqrt {108 + \sqrt {154 + \sqrt {225} } } } } }}{{\root 3 \of 8 }} \cr & \frac{{\sqrt {10 + \sqrt {25 + \sqrt {108 + \sqrt {169} } } } }}{2} \cr & \frac{{\sqrt {10 + \sqrt {25 + \sqrt {121} } } }}{2} \cr & \frac{{\sqrt {10 + \sqrt {36} } }}{2} \cr & \frac{{\sqrt {16} }}{2} \cr & \Rightarrow \frac{4}{2} \cr & \Rightarrow 2 \cr} $$
3. $$\frac{{{6^2} + {7^2} + {8^2} + {9^2} + {{10}^2}}}{{\sqrt {7 + 4\sqrt 3 } - \sqrt {4 + 2\sqrt 3 } }}$$ is equal to = ?
a) 330
b) 355
c) 305
d) 366
Explanation:
$$\eqalign{ & \frac{{{6^2} + {7^2} + {8^2} + {9^2} + {{10}^2}}}{{\sqrt {7 + 4\sqrt 3 } - \sqrt {4 + 2\sqrt 3 } }} \cr & \frac{{{6^2} + {7^2} + {8^2} + {9^2} + {{10}^2}}}{{\sqrt {{{\left( {2 + \sqrt 3 } \right)}^2}} - \sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} }} \cr & \frac{{{6^2} + {7^2} + {8^2} + {9^2} + {{10}^2}}}{{2 + \sqrt 3 - \sqrt 3 - 1}} \cr & {6^2} + {7^2} + {8^2} + {9^2} + {10^2} \cr & 36 + 49 + 64 + 81 + 100 \cr & 330 \cr} $$
4.Given 2x = 8y+1 and 9y = 3x-9 , then value of x + y is = ?
a) 18
b) 21
c) 24
d) 27
Explanation:
$$\eqalign{ & {{\text{2}}^x}{\text{ = }}{{\text{8}}^{y + 1}} \cr & \Leftrightarrow {{\text{2}}^x}{\text{ = }}{\left( {{2^3}} \right)^{y + 1}} = {2^{\left( {3y + 3} \right)}} \cr & x = 3y + 3 \cr & x - 3y = 3.......(i) \cr & {9^y} = {3^{x - 9}} \cr & \Leftrightarrow {\left( {{3^2}} \right)^y}{\text{ = }}{{\text{3}}^{x - 9}} \cr & 2y = x - 9 \cr & x - 2y = 9......({\text{ii}}) \cr & {\text{Subtracting (i) from (ii),}} \cr & {\text{we}}\,{\text{get}}\,y = 6 \cr & {\text{Putting }}y\,{\text{ = 6 in (i),}} \cr & {\text{we get }}x{\text{ = 21}} \cr & x + y = 21 + 6 = 27 \cr} $$
5. What are the values of x and y that satisfy the equation, $${{\text{2}}^{0.7x}}{\text{.}}{{\text{3}}^{ - 1.25y}}{\text{ = }}\frac{{8\sqrt 6 }}{{27}}{\text{ ?}}$$
a) x = 2.5, y = 6
b) x = 3, y = 5
c) x = 3, y = 4
d) x = 5, y = 2
Explanation:
$$\eqalign{ & {{\text{2}}^{0.7x}}{\text{.}}{{\text{3}}^{ - 1.25y}}{\text{ = }}\frac{{8\sqrt 6 }}{{27}} \cr & \Leftrightarrow \frac{{{{\text{2}}^{0.7x}}}}{{{{\text{3}}^{ 1.25y}}}}{\text{ = }}\frac{{{2^3}{{.2}^{\frac{1}{2}}}{{.3}^{\frac{1}{2}}}}}{{{3^3}}} \cr & \Leftrightarrow \frac{{{2^{\left( {3 + \frac{1}{2}} \right)}}}}{{{3^{\left( {3 - \frac{1}{2}} \right)}}}} = \frac{{{2^{\frac{7}{2}}}}}{{{2^{\frac{5}{2}}}}} = \frac{{{2^{3.5}}}}{{{3^{2.5}}}} \cr & 0.7x = 3.5 \Rightarrow x = \frac{{3.5}}{{0.7}}{\text{ = 5}} \cr & {\text{and }}1.25y = 2.5 \cr & \Rightarrow y = \frac{{2.5}}{{1.25}} = 2 \cr} $$
6. Evaluate : $$\sqrt {20} + \sqrt {12} + \root 3 \of {729} \,\, - $$ $$\frac{4}{{\sqrt 5 - \sqrt 3 }} \,- $$ $$\sqrt {81} = ?$$
a) $$\sqrt 2 $$
b) $$\sqrt 3 $$
c) 0
d) $$2\sqrt 2 $$
Explanation:
$$\sqrt {20} + \sqrt {12} + \root 3 \of {729} - \frac{4}{{\sqrt 5 - \sqrt 3 }} - \sqrt {81} $$
$$ = 2\sqrt 5 + 2\sqrt 3 + 9\, - $$ $$\left( {\frac{4}{{\sqrt 5 - \sqrt 3 }} \times \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }}} \right)$$ $$\, - \,9$$
$$\eqalign{ & = 2\sqrt 5 + 2\sqrt 3 + 9 - \left( {\frac{{4\left( {\sqrt 5 + \sqrt 3 } \right)}}{2}} \right) - 9 \cr & = 2\sqrt 5 + 2\sqrt 3 + 9 - 2\sqrt 5 - 2\sqrt 3 - 9 \cr & = 0 \cr} $$
7. If $$\frac{{4 + 3\sqrt 3 }}{{\sqrt {7 + 4\sqrt 3 } }} = A + \sqrt B {\text{,}}$$ then B - A is = ?
a) -13
b) $${\text{2}}\sqrt {13} $$
c) 13
d) $${\text{3}}\sqrt 3 - \sqrt 7 $$
Explanation:
$$\eqalign{ & \frac{{4 + 3\sqrt 3 }}{{\sqrt {7 + 4\sqrt 3 } }}{\text{ = }}A + \sqrt B \cr & \Rightarrow \sqrt {7 + 4\sqrt 3 } \cr & \Rightarrow \sqrt {{2^2} + {{\left( {\sqrt 3 } \right)}^2} + 2 \times 2\sqrt 3 } \cr & \Rightarrow \sqrt {{{\left( {2 + \sqrt 3 } \right)}^2}} \cr & \Rightarrow \left( {2 + \sqrt 3 } \right) \cr & \Rightarrow \frac{{4 + 3\sqrt 3 }}{{2 + \sqrt 3 }}{\text{ = }}A + \sqrt B \cr & \frac{{4 + 3\sqrt 3 }}{{2 + \sqrt 3 }} \times \frac{{2 - \sqrt 3 }}{{2 - \sqrt 3 }} = A + \sqrt B \cr & \frac{{\left( {4 + 3\sqrt 3 } \right)\left( {2 - \sqrt 3 } \right)}}{{4 - 3}} = A + \sqrt B \cr & 8 - 4\sqrt 3 + 6\sqrt 3 - 9 = A + \sqrt B \cr & 2\sqrt 3 - 1 = A + \sqrt B \cr & A = - 1\,\,\& \,\,\sqrt B = 2\sqrt 3 \cr & B = 2\sqrt 3 \times 2\sqrt 3 = 12 \cr & {\text{So}},B - A = 12 - \left( { - 1} \right) = 13 \cr} $$
8. Given that 100.48 = x, 100.70 = y and xz = y2 then the value of z is close to = ?
a) 1.45
b) 1.88
c) 2.9
d) 3.7
Explanation:
$$\eqalign{ & {x^z}{\text{ = }}{y^2}{\text{ }} \cr & {\left( {{{10}^{0.48}}} \right)^z} = {\left( {{{10}^{0.70}}} \right)^2} \cr & {10^{\left( {0.48z} \right)}} = {10^{\left( {2 \times 0.70} \right)}} = {10^{1.40}} \cr & 0.48z = 1.40 \cr & z = \frac{{140}}{{48}} = \frac{{35}}{{12}} \cr & z = 2.9\left( {{\text{approx}}} \right) \cr} $$
9. If m and n are whole numbers such that mn = 121, then the value of (m - 1)n+1 is = ?
a) 1
b) 10
c) 121
d) 1000
Explanation:
$$\eqalign{ & {\text{We know that}} \cr & {\text{1}}{{\text{1}}^2} = 121 \cr & {\text{Putting}} \cr & m = 11\& n = 2 \cr & {\text{we get}} \cr & {\left( {m - 1} \right)^{n + 1}} \cr & = {\left( {11 - 1} \right)^{\left( {2 + 1} \right)}} \cr & = {10^3} \cr & = 1000 \cr} $$
10. 1 + (3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1) is equal to =
a) $$\frac{{{3^{64}} - 1}}{2}$$
b) $$\frac{{{3^{64}} + 1}}{2}$$
c) 364 - 1
d) 364 + 1
Explanation:
$$1 + \left( {3 + 1} \right)$$ $$\left( {{3^2} + 1} \right)$$ $$\left( {{3^4} + 1} \right)$$ $$\left( {{3^8} + 1} \right)$$ $$\left( {{3^{16}} + 1} \right)$$ $$\left( {{3^{32}} + 1} \right)$$
$$ = 1 + \frac{1}{2}\left[ {\left( {3 - 1} \right)\left( {3 + 1} \right)\left( {{3^2} + 1} \right)\left( {{3^4} + 1} \right)\left( {{3^8} + 1} \right)\left( {{3^{16}} + 1} \right)\left( {{3^{32}} + 1} \right)} \right]$$
$$ = 1 + \frac{1}{2}\left[ {\left( {{3^2} - 1} \right)\left( {{3^2} + 1} \right)\left( {{3^4} + 1} \right)\left( {{3^8} + 1} \right)\left( {{3^{16}} + 1} \right)\left( {{3^{32}} + 1} \right)} \right]$$
$$ = 1 + \frac{1}{2}\left[ {\left( {{3^4} - 1} \right)\left( {{3^4} + 1} \right)\left( {{3^8} + 1} \right)\left( {{3^{16}} + 1} \right)\left( {{3^{32}} + 1} \right)} \right]$$
$$\eqalign{ & = 1 + \frac{1}{2}\left[ {\left( {{3^8} - 1} \right)\left( {{3^8} + 1} \right)\left( {{3^{16}} + 1} \right)\left( {{3^{32}} + 1} \right)} \right] \cr & = 1 + \frac{1}{2}\left[ {\left( {{3^{16}} - 1} \right)\left( {{3^{16}} + 1} \right)\left( {{3^{32}} + 1} \right)} \right] \cr & = 1 + \frac{1}{2}\left[ {\left( {{3^{32}} - 1} \right)\left( {{3^{32}} + 1} \right)} \right] \cr & = 1 + \frac{1}{2}\left[ {\left( {{3^{64}} + 1} \right)} \right] \cr & = \frac{{2 + {3^{64}} - 1}}{2} \cr & = \frac{{{3^{64}} + 1}}{2} \cr} $$