1. $$\left[ {\frac{{{{\left( {0.73} \right)}^3} + {{\left( {0.27} \right)}^3}}}{{{{\left( {0.73} \right)}^2} + {{\left( {0.27} \right)}^2} - \left( {0.73} \right) \times \left( {0.27} \right)}}} \right]$$ simplifies to ?
a) 1
b) 0.4087
c) 0.73
d) 0.27
Explanation:
$$\eqalign{ & {\text{let}} \cr & a = 0.73 \cr & b = 0.27 \cr & = \frac{{{a^3} + {b^3}}}{{{a^2} + {b^2} - ab}} \cr & = \frac{{\left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)}}{{\left( {{a^2} + {b^2} - ab} \right)}} \cr & = \left( {a + b} \right) \cr & = \left( {0.73 + 0.27} \right) \cr & = 1 \cr} $$
2. If $$x = \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }}$$ and $$y = \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }}$$ then $$\left( {x + y} \right)$$ equals ?
a) 8
b) 16
c) $${\text{2}}\sqrt {15} $$
d) $${\text{2}}\left( {\sqrt 5 + \sqrt 3 } \right)$$
Explanation:
$$\eqalign{ & x = \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \cr & x = \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \times \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }} \cr & x = \frac{{{{\left( {\sqrt 5 + \sqrt 3 } \right)}^2}}}{2} \cr & {\text{Similarly}} \cr & y = \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }} \cr & \Rightarrow y = \frac{{{{\left( {\sqrt 5 - \sqrt 3 } \right)}^2}}}{2} \cr & {\text{Now, }}x + y \cr & = \frac{{5 + 3 + 2\sqrt {15} + 5 + 3 - 2\sqrt {15} }}{2} \cr & = \frac{{16}}{2} \cr & = 8 \cr} $$
3. Simplified from of $${\left[ {{{\left( {\root 5 \of {{x^{ - \frac{3}{5}}}} } \right)}^{ - \frac{5}{3}}}} \right]^5}$$ is = ?
a) $$\frac{1}{x}$$
b) x
c) x-5
d) x5
Explanation:
$$\eqalign{ & {\left[ {{{\left( {\root 5 \of {{x^{ - \frac{3}{5}}}} } \right)}^{ - \frac{5}{3}}}} \right]^5} \cr & = {\left[ {{{\left\{ {{{\left( {{x^{ - \frac{3}{5}}}} \right)}^{\frac{1}{5}}}} \right\}}^{ - \frac{5}{3}}}} \right]^5} \cr & = {\left[ {{{\left( {{x^{^{\left\{ {\left( { - \frac{3}{5}} \right) \times \frac{1}{5}} \right\}}}}} \right)}^{ - \frac{5}{3}}}} \right]^5} \cr & = {\left[ {{{\left( {{x^{ - \frac{3}{{25}}}}} \right)}^{ - \frac{5}{3}}}} \right]^5} \cr & = {\left[ {{x^{\left\{ {\left( { - \frac{3}{{25}}} \right) \times \left( { - \frac{5}{3}} \right)} \right\}}}} \right]^5} \cr & = {\left( {{x^{\frac{1}{5}}}} \right)^5} \cr & = {x^{\left( {\frac{1}{5} \times 5} \right)}} \cr & = x \cr} $$
4. What will come in place of both the question marks in the following question :
$$\frac{{{{\left( ? \right)}^{\frac{2}{3}}}}}{{42}} = \frac{5}{{{{\left( ? \right)}^{\frac{1}{3}}}}}$$
a) 10
b) $${\text{10}}\sqrt 2 $$
c) $$\sqrt {20} $$
d) 210
Explanation:
$$\eqalign{ & {\text{Let }}\frac{{{{\left( x \right)}^{\frac{2}{3}}}}}{{42}} = \frac{5}{{{{\left( x \right)}^{\frac{1}{3}}}}} \cr & {x^{\frac{2}{3}}}.{x^{\frac{1}{3}}} = 42 \times 5 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 210 \cr & {x^{\left( {\frac{2}{3} + \frac{1}{3}} \right)}} = 210 \cr & x = 210 \cr} $$
5. The value of $${{\text{5}}^{\frac{1}{4}}} \times {\left( {125} \right)^{0.25}}$$ is = ?
a) $$\sqrt 5 $$
b) 5
c) $${\text{5}}\sqrt 5 $$
d) 25
Explanation:
$$\eqalign{ & {{\text{5}}^{\frac{1}{4}}} \times {\left( {125} \right)^{0.25}} \cr & = {5^{0.25}} \times {\left( {{5^3}} \right)^{0.25}} \cr & = {5^{0.25}} \times {5^{\left( {3 \times 0.25} \right)}} \cr & = {5^{0.25}} \times {5^{0.75}} \cr & = {5^{\left( {0.25 + 0.75} \right)}} \cr & = {5^1} \cr & = 5 \cr} $$
6. $${\left( 6 \right)^4} \div {\left( {36} \right)^3} \times 216 = {6^{\left( {? - 5} \right)}}$$
a) 1
b) 4
c) 6
d) 7
Explanation:
$$\eqalign{ & {\text{Let ,}} \cr & {\left( 6 \right)^4} \div {\left( {36} \right)^3} \times 216 = {6^{\left( {x - 5} \right)}} \cr & {6^{\left( {x - 5} \right)}} = {\left( 6 \right)^4} \div {\left( {{6^2}} \right)^3} \times {6^3} \cr & {6^{\left( {x - 5} \right)}} = {6^4} \div {6^{\left( {2 \times 3} \right)}} \times {6^3} \cr & {6^{\left( {x - 5} \right)}} = {6^4} \div {6^6} \times {6^3} \cr & {6^{\left( {x - 5} \right)}} = {6^{\left( {4 - 6 + 3} \right)}} \cr & {6^{\left( {x - 5} \right)}} = 6 \cr & x - 5 = 1 \cr & x = 6 \cr} $$
7. The value of $${\left( {256} \right)^{\frac{5}{4}}}$$ is = ?
a) 512
b) 984
c) 1024
d) 1032
Explanation:
$$\eqalign{ & {\left( {256} \right)^{\frac{5}{4}}} \cr & = {\left( {{4^4}} \right)^{\frac{5}{4}}} \cr & = {4^{\left( {4 \times \frac{5}{4}} \right)}} \cr & = {4^5} \cr & = 1024 \cr} $$
8. $$\sqrt {2 + \sqrt {2 + \sqrt {2 + ......} } } $$ is equal to ?
a) $$\sqrt 2 $$
b) $${\text{2}}\sqrt 2 $$
c) 2
d) 3
Explanation:
$$\eqalign{ & x = \sqrt {2 + \sqrt {2 + \sqrt {2 + ......} } } \cr & {x^2} = 2 + \sqrt {2 + \sqrt {2 + .......} } \cr & {x^2} = 2 + x \cr & {x^2} - x - 2 = 0 \cr & x\left( {x - 2} \right) + 1\left( {x - 2} \right) = 0 \cr & \left( {x + 1} \right)\left( {x - 2} \right) = 0 \cr & x = 2 \cr} $$
9.The value of $${\text{2}} + \sqrt {0.09} \, - \,\root 3 \of {0.008} \, - \,75\% $$ of 2.80 is = ?
a) 0
b) 0.01
c) -1
d) 0.001
Explanation:
$$\eqalign{ & {\text{2 + }}\sqrt {0.09} - \root 3 \of {0.008} - 75\% \,{\text{of }}2.80 \cr & = 2 + 0.3 - 0.2 - \left( {\frac{3}{4} \times 2.8} \right) \cr & = 2 + 0.3 - 0.2 - 2.10 \cr & = 2.3 - 2.3 \cr & = 0 \cr} $$
10. The value of $${\left( {3 + 2\sqrt 2 } \right)^{ - 3}}$$ $$ + {\left( {3 - 2\sqrt 2 } \right)^{ - 3}} = ?$$
a) 189
b) 180
c) 108
d) 198
Explanation:
$$\eqalign{ & {\left( {3 + 2\sqrt 2 } \right)^{ - 3}} + {\left( {3 - 2\sqrt 2 } \right)^{ - 3}} \cr & = {\left( {\frac{1}{{3 + 2\sqrt 2 }}} \right)^3} + {\left( {\frac{1}{{3 - 2\sqrt 2 }}} \right)^3} \cr} $$
$$ = {\left( {\frac{1}{{\left( {3 - 2\sqrt 2 } \right)}} \times \frac{{3 + 2\sqrt 2 }}{{3 + 2\sqrt 2 }}} \right)^3} + $$ $${\left( {\frac{1}{{\left( {3 + 2\sqrt 2 } \right)}} \times \frac{{3 - 2\sqrt 2 }}{{3 - 2\sqrt 2 }}} \right)^3}$$
$$\eqalign{ & = {\left( {\frac{{3 - 2\sqrt 2 }}{{9 - 8}}} \right)^3} + {\left( {\frac{{3 + 2\sqrt 2 }}{{9 - 8}}} \right)^3} \cr & = {\left( {3 - 2\sqrt 2 } \right)^3} + {\left( {3 + 2\sqrt 2 } \right)^3} \cr & a = 3 - 2\sqrt 2 \cr & b = 3 + 2\sqrt 2 \cr & \left[ { {a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)} \right] \cr & = \left( {3 - 2\sqrt 2 + 3 + 2\sqrt 2 } \right)\left( {17 + 17 - 1} \right) \cr & = \left( 6 \right)\left( {33} \right) \cr & = 198 \cr} $$