## Surds and Indices Questions and Answers Part-10

1. (3x - 2y) : (2x + 3y) = 5 : 6, then one of the value of $${\left( {\frac{{\root 3 \of x + \root 3 \of y }}{{\root 3 \of x - \root 3 \of y }}} \right)^2}{\text{ is = ?}}$$
a) $$\frac{1}{{25}}$$
b) 5
c) $$\frac{1}{5}$$
d) 25

Explanation:
\eqalign{ & \frac{{\left( {3x - 2y} \right)}}{{\left( {3x + 2y} \right)}} = \frac{5}{6} \cr & \Rightarrow 18x - 12y = 10x + 15y \cr & \Rightarrow 8x = 27y \cr & \Rightarrow \frac{x}{y} = \frac{{27}}{8} \cr & \Rightarrow {\left( {\frac{{\root 3 \of x + \root 3 \of y }}{{\root 3 \of x - \root 3 \of y }}} \right)^2} \cr & {\left( {\frac{{\root 3 \of {27} + \root 3 \of 8 }}{{\root 3 \of {27} - \root 3 \of 8 }}} \right)^2} \cr & {\left( {\frac{{3 + 2}}{{3 - 2}}} \right)^2} \cr & {\left( 5 \right)^2} \cr & 25 \cr}

2. The exponential form of $$\sqrt {\sqrt 2 \times \sqrt 3 } {\text{ is = ?}}$$
a) $${{\text{6}}^{ - \frac{1}{2}}}$$
b) $${{\text{6}}^{\frac{1}{2}}}$$
c) $${{\text{6}}^{\frac{1}{4}}}$$
d) 6

Explanation:
\eqalign{ & {\text{The exponential form of }} \cr & \sqrt {\sqrt 2 \times \sqrt 3 } \cr & = \sqrt {{6^{\frac{1}{2}}}} \cr & = {\left( {{6^{\frac{1}{2}}}} \right)^{\frac{1}{2}}} \cr & = {6^{\frac{1}{4}}} \cr}

3. The quotient when 10100 is divided by 575 is
a) 225 × 1075
b) 1025
c) 275
d) 275 × 1025

Explanation:
\eqalign{ & {\text{Expression,}} \cr & {\text{ = }}\frac{{{{\left( {10} \right)}^{100}}}}{{{{\left( 5 \right)}^{75}}}} \cr & = \frac{{{{\left( {2 \times 5} \right)}^{100}}}}{{{{\left( 5 \right)}^{75}}}} \cr & = \frac{{{{\left( 2 \right)}^{100}} \times {{\left( 5 \right)}^{100}}}}{{{{\left( 5 \right)}^{75}}}} \cr & = {2^{100}} \times \frac{{{5^{100}}}}{{{5^{75}}}} \cr & = {2^{100}} \times {5^{\left( {100 - 75} \right)}}.....\left[ {\because \frac{{{a^m}}}{{{a^n}}} = {a^{m - n}}} \right] \cr & = {2^{100}} \times {5^{25}} \cr & = {2^{25}} \times {5^{25}} \times {2^{75}}.....\left[ {\because {a^m} \times {a^n} = {a^{m + n}}} \right] \cr & = {\left( {10} \right)^{25}} \times {2^{75}}.....\left[ {\because {a^m} \times {b^m} = a{b^m}} \right] \cr}

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

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

5. 21? × 216.5 = 2112.4
a) 18.9
b) 4.4
c) 5.9
d) 13.4

Explanation:
\eqalign{ & {21^?} \times {21^{6.5}} = {21^{12.4}}.....\left[ {\because {a^m} \times {a^n} = {a^{m + n}}} \right] \cr & \Rightarrow {21^{? + 6.5}} = {21^{12.4}} \cr & \Rightarrow ? + 6.5 = 12.4 \cr & \Rightarrow ? = 12.4 - 6.5 \cr & \Rightarrow ? = 5.9 \cr}

6. $${25^{2.7}} \times {5^{4.2}} \div {5^{5.4}} = {25^?}$$
a) 1.6
b) 1.7
c) 3.2
d) None of these

Explanation:
\eqalign{ & {\text{Let }}{25^{2.7}} \times {5^{4.2}} \div {5^{5.4}} = {25^x} \cr & {\text{Then, }}{25^{2.7}} \times {5^{(4.2 - 5.4)}} = {25^x} \cr & {25^{2.7}} \times {5^{( - 1.2)}} = {25^x} \cr & {25^{2.7}} \times \frac{1}{{{5^{1.2}}}} = {25^x} \cr & \frac{{{{25}^{2.7}}}}{{{{\left( {{5^2}} \right)}^{0.6}}}} = {25^x} \cr & \frac{{{{\left( {25} \right)}^{2.7}}}}{{{{\left( {25} \right)}^{0.6}}}} = {25^x} \cr & {25^x} = {25^{\left( {2.7 - 0.6} \right)}} = {25^{2.1}} \cr & x = 2.1 \cr}

7. $${8^{2.4}} \times {2^{3.7}} \div {\left( {16} \right)^{1.3}} = {2^?}$$
a) 4.8
b) 5.7
c) 5.8
d) 7.1

Explanation:
\eqalign{ & {\text{Let }}{8^{2.4}} \times {2^{3.7}} \div {\left( {16} \right)^{1.3}} = {2^x} \cr & {\text{Then,}}{\left( {{2^3}} \right)^{2.4}} \times {2^{3.7}} \div {\left( {{2^4}} \right)^{1.3}} = {2^x} \cr & {2^{\left( {3 \times 2.4} \right)}} \times {2^{3.7}} \div {2^{\left( {4 \times 1.3} \right)}} = {2^x} \cr & {2^{7.2}} \times {2^{3.7}} \div {2^{5.2}} = {2^x} \cr & {2^x} = {2^{\left( {7.2 + 3.7 - 5.2} \right)}} \cr & {2^x} = {2^{5.7}} \cr & x = 5.7 \cr}

8. If 3(x+y) = 81 and 81(x-y) = 3, then the value of x is = ?
a) 42
b) $$\frac{{15}}{8}$$
c) $$\frac{{17}}{8}$$
d) 39

Explanation:
\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}

9.Simplified from of $${\left[ {{{\left( {\root 5 \of {{x^{ - \frac{3}{5}}}} } \right)}^{ - \frac{5}{3}}}} \right]^{ - 5}} = ?$$
a) x5
b) x-5
c) x
d) $$\frac{1}{x}$$

\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}
10. Find the value of x in the expression : $$\root 4 \of {3x + 1} = 2$$
\eqalign{ & \root 4 \of {3x + 1} = 2 \cr & {\left( {\root 4 \of {3x + 1} } \right)^4} = {2^4} \cr & {\left( {3x + 1} \right)^{4 \times \frac{1}{4}}} = 16 \cr & 3x = 15 \cr & x = 5 \cr}