Practice MCQ Questions and Answer on Surds and Indices

Solution:

 $$\eqalign{ & x = \sqrt {a\root 3 \of {ab\sqrt {a\root 3 \of {ab} } } } ..... \propto \cr & {\text{Square both side}} \cr & {x^2} = a\,\root 3 \of {ab\,x} \,\,\,\,\,\left( {\because \sqrt {a\root 3 \of {ab} } ..... \propto } \right) \cr & {\text{Again cube both sides}} \cr & {x^6} = {a^3}ab\,x \cr & {x^5} = {a^4}b \cr & x = \root 5 \of {{a^4}b} \cr} $$

Solution:

 Given exp. = $$ = \frac{1}{{\left( {1 + \frac{{{x^b}}}{{{x^a}}} + \frac{{{x^c}}}{{{x^a}}}} \right)}} + $$   $$\frac{1}{{\left( {1 + \frac{{{x^a}}}{{{x^b}}} + \frac{{{x^c}}}{{{x^b}}}} \right)}} + $$   $$\frac{1}{{\left( {1 + \frac{{{x^b}}}{{{x^c}}} + \frac{{{x^a}}}{{{x^c}}}} \right)}}$$ $$ = \frac{{{x^a}}}{{\left( {{x^a} + {x^b} + {x^c}} \right)}} + $$   $$\frac{{{x^b}}}{{\left( {{x^a} + {x^b} + {x^c}} \right)}} + $$   $$\frac{{{x^c}}}{{\left( {{x^a} + {x^b} + {x^c}} \right)}}$$ $$\eqalign{ & = \frac{{ {{x^a} + {x^b} + {x^c}} }}{{ {{x^a} + {x^b} + {x^c}} }} \cr & = 1 \cr} $$

Solution:

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

Solution:

 $$\eqalign{ & \frac{1}{{\sqrt {11 - 2\sqrt {30} } }} \cr & = \frac{1}{{\sqrt {6 + 5 - 2 \times \sqrt 6 \times \sqrt 5 } }} \cr & = \frac{1}{{\sqrt {{{\left( {\sqrt 6 } \right)}^2} + {{\left( {\sqrt 5 } \right)}^2} - 2 \times \sqrt 6 \times \sqrt 5 } }} \cr & = \frac{1}{{\sqrt {{{\left( {\sqrt 6 - \sqrt 5 } \right)}^2}} }} \cr & = \frac{1}{{\sqrt 6 - \sqrt 5 }} \cr & = \frac{{\left( {\sqrt 6 + \sqrt 5 } \right)}}{{\left( {\sqrt 6 - \sqrt 5 } \right)\left( {\sqrt 6 + \sqrt 5 } \right)}} \cr & = \sqrt 6 + \sqrt 5 \cr & \frac{3}{{\sqrt {7 - 2\sqrt {10} } }} \cr & = \frac{3}{{\sqrt {5 + 2 - 2 \times \sqrt 5 \times \sqrt 2 } }} \cr & = \frac{3}{{\sqrt 5 - \sqrt 2 }} \cr & = \frac{{3 \times \left( {\sqrt 5 + \sqrt 2 } \right)}}{{\left( {\sqrt 5 - \sqrt 2 } \right)\left( {\sqrt 5 + \sqrt 2 } \right)}} \cr & = \frac{{3\left( {\sqrt 5 + \sqrt 2 } \right)}}{{5 - 2}} \cr & = \sqrt 5 + \sqrt 2 \cr & \frac{4}{{\sqrt {8 + 4\sqrt 3 } }} \cr & = \frac{4}{{\sqrt {8 + 2\sqrt {12} } }} \cr & = \frac{4}{{\sqrt {6 + 2 + 2 \times \sqrt 6 \times \sqrt 2 } }} \cr & = \frac{4}{{\sqrt {{{\left( {\sqrt 6 + \sqrt 2 } \right)}^2}} }} \cr & = \frac{{4 \times \left( {\sqrt 6 - \sqrt 2 } \right)}}{{\left( {\sqrt 6 + \sqrt 2 } \right)\left( {\sqrt 6 - \sqrt 2 } \right)}} \cr & = \frac{{4\left( {\sqrt 6 - \sqrt 2 } \right)}}{{6 - 2}} \cr & = \sqrt 6 - \sqrt 2 \cr & \therefore {\text{Expression}} \cr & = \left( {\sqrt 6 + \sqrt 5 } \right) - \left( {\sqrt 5 + \sqrt 2 } \right) - \left( {\sqrt 6 - \sqrt 2 } \right) \cr & = \sqrt 6 + \sqrt 5 - \sqrt 5 - \sqrt 2 - \sqrt 6 + \sqrt 2 \cr & = 0 \cr} $$

Solution:

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

Solution:

 $$\eqalign{ & {\left( {10} \right)^{150}} \div {\left( {10} \right)^{146}} = \frac{{{{10}^{150}}}}{{{{10}^{146}}}} \cr & = {10^{150 - 146}} \cr & = {10^4} \cr & = 10000 \cr} $$

Solution:

 $$\eqalign{ & {\left[ {{{\left( {\root 5 \of {{x^{ - \frac{3}{5}}}} } \right)}^{ - \frac{5}{3}}}} \right]^5} \cr & = {\left[ {{{\left\{ {{{\left( {{x^{ - \frac{3}{5}}}} \right)}^{\frac{1}{5}}}} \right\}}^{ - \frac{5}{3}}}} \right]^5} \cr & = {\left[ {{{\left( {{x^{^{\left\{ {\left( { - \frac{3}{5}} \right) \times \frac{1}{5}} \right\}}}}} \right)}^{ - \frac{5}{3}}}} \right]^5} \cr & = {\left[ {{{\left( {{x^{ - \frac{3}{{25}}}}} \right)}^{ - \frac{5}{3}}}} \right]^5} \cr & = {\left[ {{x^{\left\{ {\left( { - \frac{3}{{25}}} \right) \times \left( { - \frac{5}{3}} \right)} \right\}}}} \right]^5} \cr & = {\left( {{x^{\frac{1}{5}}}} \right)^5} \cr & = {x^{\left( {\frac{1}{5} \times 5} \right)}} \cr & = x \cr} $$

Solution:

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

Solution:

 $$\eqalign{ & x = \root 6 \of {27} - \sqrt {6\frac{3}{4}} \cr & {x^2} = {\left( {{{27}^{\frac{1}{6}}} - \frac{{{{27}^{\frac{1}{2}}}}}{2}} \right)^2} \cr & {x^2} = {27^{\frac{2}{6}}} + \frac{{27}}{4} - \frac{{2 \times {{27}^{\frac{1}{6}}} \times {{27}^{\frac{1}{2}}}}}{2} \cr & {x^2} = 3 + \frac{{27}}{4} - 9 \cr & {x^2} = \frac{3}{4} \cr & y = \frac{{\sqrt {45} + \sqrt {605} + \sqrt {245} }}{{\sqrt {80} + \sqrt {125} }} \cr & y = \frac{{3\sqrt 5 + 11\sqrt 5 + 7\sqrt 5 }}{{4\sqrt 4 + 5\sqrt 5 }} \cr & y = \frac{7}{3} \cr & {y^2} = \frac{{49}}{9} \cr & {x^2} + {y^2} = \frac{3}{4} + \frac{{49}}{9} = \boxed{\frac{{223}}{{36}}} \cr} $$

Solution:

 $$\eqalign{ & \frac{{343 \times 49}}{{216 \times 16 \times 81}} \cr & = \frac{{{7^3} \times {7^2}}}{{{6^3} \times {2^4} \times {3^4}}} \cr & = \frac{{{7^{\left( {3 + 2} \right)}}}}{{{6^3} \times {{\left( {2 \times 3} \right)}^4}}} \cr & = \frac{{{7^5}}}{{{6^3} \times {6^4}}} \cr & = \frac{{{7^5}}}{{{6^{\left( {3 + 4} \right)}}}} \cr & = \frac{{{7^5}}}{{{6^7}}} \cr} $$