## Surds and Indices Questions and Answers Part-4

1. $$\frac{{12}}{{3 + \sqrt 5 + 2\sqrt 2 }}$$    is equal to = ?
a) $${\text{1}} - \sqrt 5 + \sqrt 2 + \sqrt {16}$$
b) $${\text{1}} + \sqrt 5 + \sqrt 2 - \sqrt {10}$$
c) $${\text{1}} + \sqrt 5 + \sqrt 2 + \sqrt {10}$$
d) $${\text{1}} - \sqrt 5 - \sqrt 2 + \sqrt {10}$$

Explanation:
\eqalign{ & \frac{{12}}{{3 + \sqrt 5 + 2\sqrt 2 }} \cr & = \frac{{12\left( {3 + \sqrt 5 - 2\sqrt 2 } \right)}}{{\left[ {\left( {3 + \sqrt 5 } \right) + 2\sqrt 2 } \right]\left[ {\left( {3 + \sqrt 5 } \right) - 2\sqrt 2 } \right]}} \cr & = \frac{{12\left( {3 + \sqrt 5 - 2\sqrt 2 } \right)}}{{{{\left( {3 + \sqrt 5 } \right)}^2} - {{\left( {2\sqrt 2 } \right)}^2}}} \cr & = \frac{{12\left( {3 + \sqrt 5 - 2\sqrt 2 } \right)}}{{9 + 5 + 6\sqrt 5 - 8}} \cr & = \frac{{12\left( {3 + \sqrt 5 - 2\sqrt 2 } \right)}}{{6\sqrt 5 + 6}} \cr & = \frac{{2\left( {3 + \sqrt 5 - 2\sqrt 2 } \right)}}{{\sqrt 5 + 1}} \cr & = \frac{{2\left( {3 + \sqrt 5 - 2\sqrt 2 } \right)\left( {\sqrt 5 - 1} \right)}}{{\left( {\sqrt 5 - 1} \right)\left( {\sqrt 5 + 1} \right)}} \cr & = \frac{{2\left( {3\sqrt 5 + 5 - 2\sqrt {10} - 3 - \sqrt 5 + 2\sqrt 2 } \right)}}{{5 - 1}} \cr & = \frac{{2\left( {2\sqrt 5 + 2\sqrt 2 - 2\sqrt {10} + 2} \right)}}{4} \cr & = \frac{{2 \times 2\left( {\sqrt 5 + \sqrt 2 - \sqrt {10} + 1} \right)}}{4} \cr & = \sqrt 5 + \sqrt 2 - \sqrt {10} + 1 \cr & {\text{or,}}\,1 + \sqrt 5 + \sqrt 2 - \sqrt {10} \cr}

2. $$\left[ {8 - {{\left( {\frac{{{4^{\frac{9}{4}}}\sqrt {{{2.2}^2}} }}{{2\sqrt {{2^{ - 2}}} }}} \right)}^{^{\frac{1}{2}}}}} \right]$$     is equal to = ?
a) 32
b) 8
c) 1
d) 0

Explanation:
\eqalign{ & \left[ {8 - {{\left( {\frac{{{4^{\frac{9}{4}}}\sqrt {{{2.2}^2}} }}{{2\sqrt {{2^{ - 2}}} }}} \right)}^{^{\frac{1}{2}}}}} \right] \cr & = \left[ {8 - {{\left( {\frac{{{2^{2 \times \frac{9}{4}}}\sqrt {{2^{1 + 2}}} }}{{2\sqrt {\frac{1}{4}} }}} \right)}^{^{\frac{1}{2}}}}} \right] \cr & = \left[ {8 - {{\left( {\frac{{{2^{\frac{9}{2}}}{{.2}^{\frac{3}{2}}}}}{{2 \times \frac{1}{2}}}} \right)}^{^{\frac{1}{2}}}}} \right] \cr & = \left[ {8 - {{\left( {{2^{\frac{{12}}{2}}}} \right)}^{^{\frac{1}{2}}}}} \right] \cr & = \left[ {8 - \left( {{2^{6 \times \frac{1}{2}}}} \right)} \right] \cr & = \left[ {8 - \left( {{2^3}} \right)} \right] \cr & = \left[ {8 - 8} \right] \cr & = 0 \cr}

3. $$\frac{{3\sqrt 2 }}{{\sqrt 6 + \sqrt 3 }} -$$  $$\frac{{2\sqrt 6 }}{{\sqrt 3 + 1}} +$$  $$\frac{{2\sqrt 3 }}{{\sqrt 6 + 2}}$$  is equal to = ?
a) 3
b) 2
c) 0
d) $$\sqrt 3$$

Explanation:
$$\frac{{3\sqrt 2 }}{{\sqrt 6 + \sqrt 3 }} - \frac{{2\sqrt 6 }}{{\sqrt 3 + 1}} + \frac{{2\sqrt 3 }}{{\sqrt 6 + 2}}$$
$$= \left( {\frac{{3\sqrt 2 }}{{\sqrt 6 + \sqrt 3 }} \times \frac{{\sqrt 6 - \sqrt 3 }}{{\sqrt 6 - \sqrt 3 }}{\text{ }}} \right) -$$      $$\left( {\frac{{2\sqrt 6 }}{{\sqrt 3 + 1}} \times \frac{{\sqrt 3 - 1}}{{\sqrt 3 - 1}}{\text{ }}} \right) +$$     $$\left( {\frac{{2\sqrt 3 }}{{\sqrt 6 + 2}}{\text{ }} \times \frac{{\sqrt 6 - 2}}{{\sqrt 6 - 2}}{\text{ }}} \right)$$
$$= \frac{{3\sqrt 2 \left( {\sqrt 6 - \sqrt 3 } \right)}}{3} -$$    $$\frac{{2\sqrt 6 \left( {\sqrt 3 - 1} \right)}}{2} +$$   $$\frac{{2\sqrt 3 \left( {\sqrt 6 - 2} \right)}}{2}$$
\eqalign{ & = \sqrt {12} - \sqrt 6 - \sqrt {18} + \sqrt 6 + \sqrt {18} - 2\sqrt 3 \cr & = \sqrt {12} - 2\sqrt 3 \cr & = 2\sqrt 3 - 2\sqrt 3 \cr & = 0 \cr}

4. The value of $$\frac{1}{{{{\left( {216} \right)}^{ - \frac{2}{3}}}}}{\text{ + }}$$  $$\frac{1}{{{{\left( {256} \right)}^{ - \frac{3}{4}}}}}{\text{ + }}$$  $$\frac{1}{{{{\left( {32} \right)}^{ - \frac{1}{5}}}}}$$  is = ?
a) 102
b) 105
c) 107
d) 109

Explanation:
\eqalign{ & \frac{1}{{{{\left( {216} \right)}^{ - \frac{2}{3}}}}}{\text{ + }}\frac{1}{{{{\left( {256} \right)}^{ - \frac{3}{4}}}}}{\text{ + }}\frac{1}{{{{\left( {32} \right)}^{ - \frac{1}{5}}}}} \cr & = \frac{1}{{{{\left( {{6^3}} \right)}^{ - \frac{2}{3}}}}}{\text{ + }}\frac{1}{{{{\left( {{4^4}} \right)}^{ - \frac{3}{4}}}}}{\text{ + }}\frac{1}{{{{\left( {{2^5}} \right)}^{ - \frac{1}{5}}}}} \cr & = \frac{1}{{{6^{3 \times \frac{{\left( { - 2} \right)}}{3}}}}}{\text{ + }}\frac{1}{{{4^{4 \times \frac{{\left( { - 3} \right)}}{4}}}}} + \frac{1}{{{2^{5 \times \frac{{\left( { - 1} \right)}}{5}}}}} \cr & = \frac{1}{{{6^{ - 2}}}}{\text{ + }}\frac{1}{{{4^{ - 3}}}}{\text{ + }}\frac{1}{{{2^{ - 1}}}} \cr & = \left( {{6^2} + {4^3} + {2^1}} \right) \cr & = \left( {36 + 64 + 2} \right) \cr & = 102 \cr}

5. $${\left( {48} \right)^{ - \frac{2}{7}}} \times {\left( {16} \right)^{ - \frac{5}{7}}} \times {\left( 3 \right)^{ - \frac{5}{7}}} = ?$$
a) $$\frac{1}{3}$$
b) $$\frac{1}{{48}}$$
c) 1
d) 48

Explanation:
\eqalign{ & {\left( {48} \right)^{ - \frac{2}{7}}} \times {\left( {16} \right)^{ - \frac{5}{7}}} \times {\left( 3 \right)^{ - \frac{5}{7}}} \cr & = {\left( {16 \times 3} \right)^{ - \frac{2}{7}}} \times {\left( {16} \right)^{ - \frac{5}{7}}} \times {\left( 3 \right)^{ - \frac{5}{7}}} \cr & = {\left( {16} \right)^{ - \frac{2}{7}}} \times {\left( 3 \right)^{ - \frac{2}{7}}} \times {\left( {16} \right)^{ - \frac{5}{7}}} \times {\left( 3 \right)^{ - \frac{5}{7}}} \cr & = {\left( {16} \right)^{\left( { - \frac{2}{7} - \frac{5}{7}} \right)}} \times {\left( 3 \right)^{\left( { - \frac{2}{7} - \frac{5}{7}} \right)}} \cr & = {\left( {16} \right)^{\left( { - \frac{7}{7}} \right)}} \times {\left( 3 \right)^{\left( { - \frac{7}{7}} \right)}} \cr & = {\left( {16} \right)^{ - 1}} \times {\left( 3 \right)^{ - 1}} \cr & = \frac{1}{{16}} \times \frac{1}{3} \cr & = \frac{1}{{48}} \cr}.

6. 93 × 62 ÷ 33 = ?
a) 948
b) 972
c) 984
d) 1012

Explanation:
\eqalign{ & \frac{{{9^3} \times {6^2}}}{{{3^3}}} \cr & = \frac{{{{\left( {{3^2}} \right)}^3} \times {{\left( {3 \times 2} \right)}^2}}}{{{3^3}}} \cr & = \frac{{{3^{\left( {3 \times 2} \right)}} \times {3^2} \times {2^2}}}{{{3^3}}} \cr & = \frac{{{3^{\left( {6 + 2} \right)}} \times {2^2}}}{{{3^3}}} \cr & = {3^{\left( {8 - 3} \right)}} \times {2^2} \cr & = {3^5} \times {2^2} \cr & = 243 \times 4 \cr & = 972 \cr}

7. $$\left({\frac{{2+\sqrt 3}}{{2-\sqrt3}}+ \frac{{2 - \sqrt 3}}{{2 + \sqrt 3}} + \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}} \right)$$      Simplifies to :
a) 2 - $$\sqrt 3$$
b) 2 + $$\sqrt 3$$
c) 16 - $$\sqrt 3$$
d) 40 - $$\sqrt 3$$

Explanation:
$$\left( {\frac{{2 + \sqrt 3 }}{{2 - \sqrt 3 }} + \frac{{2 - \sqrt 3 }}{{2 + \sqrt 3 }} + \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}} \right)$$
$$= \left\{ {\frac{{{{\left( {2 + \sqrt 3 } \right)}^2} + {{\left( {2 - \sqrt 3 } \right)}^2}}}{{\left( {2 - \sqrt 3 } \right)\left( {2 + \sqrt 3 } \right)}} + \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}} \times \frac{{\sqrt 3 - 1}}{{\sqrt 3 - 1}}} \right\}$$
$$= \left\{ {\frac{{4 + 3 + 4\sqrt 3 + 4 + 3 - 4\sqrt 3 }}{{4 - 3}} + \frac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{{3 - 1}}} \right\}$$
\eqalign{ & = \left\{ {14 + \frac{{3 + 1 - 2\sqrt 3 }}{2}} \right\} \cr & = 14 + \frac{{2\left( {2 - \sqrt 3 } \right)}}{2} \cr & = 14 + 2 - \sqrt 3 \cr & = 16 - \sqrt 3 \cr}

8. The value of $$\sqrt {\frac{{\left( {\sqrt {12} - \sqrt 8 } \right)\left( {\sqrt 3 + \sqrt 2 } \right)}}{{5 + \sqrt {24} }}}$$       is = ?
a) $$\sqrt 6 - \sqrt 2$$
b) $$\sqrt 6 + \sqrt 2$$
c) $$\sqrt 6 - 2$$
d) $${\text{2}} - \sqrt 6$$

Explanation:
\eqalign{ & \sqrt {\frac{{\left( {\sqrt {12} - \sqrt 8 } \right)\left( {\sqrt 3 + \sqrt 2 } \right)}}{{5 + \sqrt {24} }}} \cr & = \sqrt {\frac{{\sqrt {36} + \sqrt {24} - \sqrt {24} - \sqrt {16} }}{{5 + \sqrt {24} }}} \cr & = \sqrt {\frac{{6 - 4}}{{5 + \sqrt {24} }}} \cr & = \sqrt {\frac{2}{{5 + \sqrt {24} }} \times \frac{{5 - \sqrt {24} }}{{5 - \sqrt {24} }}} \cr & = \sqrt {\frac{{2\left( {5 - \sqrt {24} } \right)}}{{25 - 24 }}} \cr & = \sqrt {2\left( {5 - 2\sqrt 6 } \right)} \cr & = \sqrt {2\left\{ {{{\left( {\sqrt 3 } \right)}^2} + {{\left( {\sqrt 2 } \right)}^2} - 2\sqrt 3 \times \sqrt 2 } \right\}} \cr & = \sqrt {2{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}} \cr & = \sqrt 2 \left( {\sqrt 3 - \sqrt 2 } \right) \cr & = \sqrt 6 - 2 \cr}

9. (19)12 × (19)8 ÷ (19)4 = (19)?
a) 6
b) 8
c) 12
d) None of these

\eqalign{ & \frac{{{{\left( {19} \right)}^{12}} \times {{\left( {19} \right)}^8}}}{{{{\left( {19} \right)}^4}}} \cr & = \frac{{{{19}^{\left( {12 + 8} \right)}}}}{{{{\left( {19} \right)}^4}}} \cr & = \frac{{{{\left( {19} \right)}^{20}}}}{{{{\left( {19} \right)}^4}}} \cr & = {\left( {19} \right)^{\left( {20 - 4} \right)}} \cr & = {\left( {19} \right)^{16}} \cr}
\eqalign{ & {\left( {64} \right)^4} \div {\left( 8 \right)^5} \cr & = {\left( {{8^2}} \right)^4} \div {\left( 8 \right)^5} \cr & = {\left( 8 \right)^{\left( {2 \times 4} \right)}} \div {8^5} \cr & = \frac{{{8^8}}}{{{8^5}}} \cr & = {8^{\left( {8 - 5} \right)}} \cr & = {8^3} \cr}