Surds and Indices Questions and Answers Part-3

1. The value of $${\left( {\sqrt 8 } \right)^{\frac{1}{3}}}$$  is = ?
a) 2
b) 4
c) $$\sqrt 2 $$
d) 8

Answer: c
Explanation:
$$\eqalign{ & {\left( {\sqrt 8 } \right)^{\frac{1}{3}}} \cr & = {\left( {{8^{\frac{1}{2}}}} \right)^{\frac{1}{3}}} \cr & = {8^{\left( {\frac{1}{2} \times \frac{1}{3}} \right)}} \cr & = {8^{\frac{1}{6}}} \cr & = {\left( {{2^3}} \right)^{\frac{1}{6}}} \cr & = {2^{\left( {3 \times \frac{1}{6}} \right)}} \cr & = {2^{\frac{1}{2}}} \cr & = \sqrt 2 \cr} $$

2. The value of $${\left( {\frac{{32}}{{243}}} \right)^{ - \frac{4}{5}}}$$   is = ?
a) $$\frac{4}{9}$$
b) $$\frac{9}{4}$$
c) $$\frac{{16}}{{81}}$$
d) $$\frac{{81}}{{16}}$$

Answer: d
Explanation:
$$\eqalign{ & {\left( {\frac{{32}}{{243}}} \right)^{ - \frac{4}{5}}} \cr & {\text{ = }}{\left\{ {{{\left( {\frac{2}{3}} \right)}^5}} \right\}^{ - \frac{4}{5}}} \cr & {\text{ = }}{\left( {\frac{2}{3}} \right)^{5 \times \frac{{\left( { - 4} \right)}}{5}}} \cr & {\text{ = }}{\left( {\frac{2}{3}} \right)^{\left( { - 4} \right)}} \cr & {\text{ = }}{\left( {\frac{3}{2}} \right)^4} \cr & {\text{ = }}\frac{{{3^4}}}{{{2^4}}} \cr & {\text{ = }}\frac{{81}}{{16}}{\text{ }} \cr} $$

3. The value of $${\text{2}}{{\text{7}}^{ - \frac{2}{3}}}$$ lies between = ?
a) 0 and 1
b) 0 and 2
c) 2 and 3
d) 3 and 4

Answer: a
Explanation:
$$\eqalign{ & {\text{ 2}}{{\text{7}}^{ - \frac{2}{3}}} \cr & = {\left( {{3^3}} \right)^{ - \frac{2}{3}}} \cr & = {3^{\left[ {3 \times \left( { - \frac{2}{3}} \right)} \right]}} \cr & = {3^{ - 2}} \cr & = \frac{1}{{{3^2}}} \cr & = \frac{1}{9} \cr & {\text{Clearly}},\,\,0 < \frac{1}{9} < 1 \cr} $$

4. The value of $$\frac{1}{{\sqrt {3.25} + \sqrt {2.25} }}$$    $$ +\, \frac{1}{{\sqrt {4.25} + \sqrt {3.25} }}$$    $$ +\, \frac{1}{{\sqrt {5.25} + \sqrt {4.25} }}$$    $$ +\, \frac{1}{{\sqrt {6.25} + \sqrt {5.25} }}$$    is = ?
a) 1.00
b) 1.25
c) 1.50
d) 2.25

Answer: a
Explanation:
$$\eqalign{ & \frac{1}{{\sqrt {3.25} + \sqrt {2.25} }} \times \frac{{\sqrt {3.25} - \sqrt {2.25} }}{{\sqrt {3.25} - \sqrt {2.25} }} \cr & = \frac{{\sqrt {3.25} - \sqrt {2.25} }}{{3.25 - 2.25}} \cr & = \sqrt {3.25} - \sqrt {2.25} \,......(i) \cr & {\text{Similarly}} \cr & \frac{1}{{\sqrt {4.25} + \sqrt {3.25} }} \cr & = \sqrt {4.25} - \sqrt {3.25} \,.......(ii) \cr & \frac{1}{{\sqrt {5.25} + \sqrt {4.25} }} \cr & = \sqrt {5.25} - \sqrt {4.25} \,.......(iii) \cr & \frac{1}{{\sqrt {6.25} + \sqrt {5.25} }} \cr & = \sqrt {6.25} - \sqrt {5.25} \,.......(iv) \cr & {\text{Now}}\,{\text{add}}\,{\text{all}}\,{\text{them}} \cr & \sqrt {3.25} - \sqrt {2.25} + \sqrt {4.25} - \sqrt {3.25} + \sqrt {5.25} - \sqrt {4.25} + \sqrt {6.25} - \sqrt {5.25} \cr & = \sqrt {6.25} - \sqrt {2.25} \cr & = 2.5 - 1.5 \cr & = 1 \cr} $$

5. $$\frac{{{3^0} + {3^{ - 1}}}}{{{3^{ - 1}} - {3^0}}}$$   is simplified to = ?
a) -2
b) -1
c) 1
d) 2

Answer: a
Explanation:
$$\eqalign{ & \frac{{{3^0} + {3^{ - 1}}}}{{{3^{ - 1}} - {3^0}}} \cr & = \frac{{1 + \frac{1}{3}}}{{\frac{1}{3} - 1}} \cr & = \frac{{\frac{4}{3}}}{{ - \frac{2}{3}}} \cr & = - 2 \cr} $$

6. Simplify : $$\left( {\frac{{\frac{3}{{2 + \sqrt 3 }} - \frac{2}{{2 - \sqrt 3 }}}}{{2 - 5\sqrt 3 }}} \right) = ?$$
a) $$\frac{1}{2} - 5\sqrt 3 $$
b) 2 - $$5\sqrt 3 $$
c) 1
d) 0

Answer: c
Explanation:
$$\eqalign{ & \frac{{\frac{3}{{2 + \sqrt 3 }} - \frac{2}{{2 - \sqrt 3 }}}}{{2 - 5\sqrt 3 }} \cr & = \frac{{\frac{{3\left( {2 - \sqrt 3 } \right) - 2\left( {2 + \sqrt 3 } \right)}}{{\left( {2 + \sqrt 3 \,} \right)\left( {2 - \sqrt 3 } \right)}}}}{{2 - 5\sqrt 3 }} \cr & = \frac{{6 - 3\sqrt 3 - 4 - 2\sqrt 3 }}{{\left( {2 + \sqrt 3 } \right)\left( {2 - \sqrt 3 } \right)\left( {2 - 5\sqrt 3 } \right)}} \cr & = \frac{{2 - 5\sqrt 3 }}{{2 - 5\sqrt 3 }} \cr & = 1 \cr} $$

7. ($$\sqrt 8$$ - $$\sqrt 4 $$ - $$\sqrt 2 $$) Equals to = ?
a) 2 - $$\sqrt 2 $$
b) $$\sqrt 2 $$ - 2
c) 2
d) -2

Answer: b
Explanation:
$$\eqalign{ & \left( {\sqrt 8 - \sqrt 4 - \sqrt 2 } \right) \cr & = 2\sqrt 2 - 2 - \sqrt 2 \cr & = 2\sqrt 2 - \sqrt 2 - 2 \cr & = \sqrt 2 - 2 \cr} $$

8. $${\left( {64} \right)^{ - \frac{2}{3}}} \times {\left( {\frac{1}{4}} \right)^{ - 2}}$$    is equal to ?
a) 1
b) 2
c) $$\frac{1}{2}$$
d) $$\frac{1}{{16}}$$

Answer: a
Explanation:
$$\eqalign{ & {\text{6}}{{\text{4}}^{ - \frac{2}{3}}} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & = {\left( {{4^3}} \right)^{ - \frac{2}{3}}} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & = {4^{ - 2}} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & = {\left( {\frac{1}{4}} \right)^2} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & = {\left( {\frac{1}{4}} \right)^{2 - 2}} \cr & = {\left( {\frac{1}{4}} \right)^0} \cr & = 1 \cr} $$

9. The value of $$\left( {\frac{{{9^2} \times {{18}^4}}}{{{3^{16}}}}} \right)$$   is = ?
a) $$\frac{3}{2}$$
b) $$\frac{4}{9}$$
c) $$\frac{{16}}{{81}}$$
d) $$\frac{{32}}{{243}}$$

Answer: c
Explanation:
$$\eqalign{ & \left( {\frac{{{9^2} \times {{18}^4}}}{{{3^{16}}}}} \right) \cr & = \frac{{{9^2} \times {{\left( {9 \times 2} \right)}^4}}}{{{3^{16}}}} \cr & = \frac{{{{\left( {{3^2}} \right)}^2} \times {{\left( {{3^2}} \right)}^4} \times {2^4}}}{{{3^{16}}}} \cr & = \frac{{{3^4} \times {3^8} \times {2^4}}}{{{3^{16}}}} \cr & = \frac{{{3^{\left( {4 + 8} \right)}} \times {2^4}}}{{{3^{16}}}} \cr & = \frac{{{3^{12}} \times {2^4}}}{{{3^{16}}}} \cr & = \frac{{{2^4}}}{{{3^{\left( {16 - 12} \right)}}}} \cr & = \frac{{{2^4}}}{{{3^4}}} \cr & = \frac{{16}}{{81}} \cr} $$

10.$$\left[ {{4^3} \times {5^4}} \right] \div {4^5} = ?$$
a) 29.0825
b) 30.0925
c) 35.6015
d) 39.0625

Answer: d
Explanation:
$$\eqalign{ & \frac{{{4^3} \times {5^4}}}{{{4^5}}} \cr & = \frac{{{5^4}}}{{{4^{\left( {5 - 3} \right)}}}} \cr & = \frac{{{5^4}}}{{{4^2}}} \cr & = \frac{{625}}{{16}} \cr & = 39.0625 \cr} $$