Practice MCQ Questions and Answer on Surds and Indices

Solution:

 $$\eqalign{ & \frac{1}{{\sqrt 5 + \sqrt 3 }} \times \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \cr & = \frac{{\sqrt 5 - \sqrt 3 }}{{5 - 3}} \cr & = \frac{{2.236 - 1.732}}{2} \cr & = \frac{{0.504}}{2} \cr & = 0.252 \cr} $$

Solution:

 LCM of 2, 3, 4, 6 is 12 $$\sqrt 2 = $$  $${2^{\frac{1}{2}}} = $$  $${2^{\left( {\frac{1}{2} \times \frac{6}{6}} \right)}} = $$   $${2^{\frac{6}{{12}}}} = $$  $${\left( {{2^6}} \right)^{\frac{1}{{12}}}} = $$  $${\left( {64} \right)^{\frac{1}{{12}}}} = $$   $$\root {12} \of {64} $$ $$\root 6 \of 3 = $$  $${3^{\frac{1}{6}}} = $$  $${3^{\left( {\frac{1}{6} \times \frac{2}{2}} \right)}} = $$   $${3^{\frac{2}{{12}}}} = $$  $${\left( {{3^2}} \right)^{\frac{1}{{12}}}} = $$  $${\left( 9 \right)^{\frac{1}{{12}}}} = $$   $$\root {12} \of 9 $$ $$\root 3 \of 4 = $$  $${4^{\frac{1}{3}}} = $$  $${4^{\left( {\frac{1}{3} \times \frac{4}{4}} \right)}} = $$  $${4^{\frac{4}{{12}}}} = $$  $${\left( {{4^4}} \right)^{\frac{1}{{12}}}} = $$  $${\left( {256} \right)^{\frac{1}{{12}}}} = $$   $$\root {12} \of {256} $$ $$\root 4 \of 5 = $$  $${5^{\frac{1}{4}}} = $$  $${5^{\left( {\frac{1}{4} \times \frac{3}{3}} \right)}} = $$   $${5^{\frac{3}{{12}}}} = $$  $${\left( {{5^3}} \right)^{\frac{1}{{12}}}} = $$  $${\left( {125} \right)^{\frac{1}{{12}}}} = $$   $$\root {12} \of {125} $$ Clearly $$\root {12} \of {256} \,\,i.e.,\root 3 \of 4 $$     is the greatest.

Solution:

 (333 + 333 + 333)(233 + 233) = 6x (3.333)(2.233) = 6x (334)(234) = 6x 634 = 6x x = 34

Solution:

 $$\eqalign{ & a = \frac{1}{{2 - \sqrt 3 }} + \frac{1}{{3 - \sqrt 8 }} + \frac{1}{{4 - \sqrt {15} }} \cr & \Rightarrow \frac{1}{{2 - \sqrt 3 }} \times \frac{{2 + \sqrt 3 }}{{2 + \sqrt 3 }} + \frac{1}{{3 - \sqrt 8 }} \times \frac{{3 + \sqrt 8 }}{{3 + \sqrt 8 }} + \frac{1}{{4 - \sqrt {15} }} \times \frac{{4 + \sqrt {15} }}{{4 + \sqrt {15} }} \cr & \Rightarrow \frac{{2 + \sqrt 3 }}{{4 - 3}} + \frac{{3 + \sqrt 8 }}{{9 - 8}} + \frac{{4 + \sqrt {15} }}{{16 - 15}} \cr & \Rightarrow 2 + \sqrt 3 + 3 + \sqrt 8 + 4 + \sqrt {15} \cr & \Rightarrow 9 + \sqrt 3 + 2\sqrt 2 + \sqrt {15} \cr & a = 9 9 + \sqrt 3 + 2\sqrt 2 + \sqrt {15} 18 \cr & \sqrt 3 = 1.73,\,\sqrt 2 = 1.41,\,\sqrt {15} = 3.9 \cr & \Rightarrow 9 9 + 1.73 + \left( {2 \times 1.41} \right) + 3.9 = 17.4 18 \cr} $$

Solution:

 $$\eqalign{ & {\left( {256} \right)^{\frac{5}{4}}} \cr & = {\left( {{4^4}} \right)^{\frac{5}{4}}} \cr & = {4^{\left( {4 \times \frac{5}{4}} \right)}} \cr & = {4^5} \cr & = 1024 \cr} $$

Solution:

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

Solution:

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

Solution:

 $$\eqalign{ & \root 3 \of 6 ,\,\root 2 \of 5 {\text{ and }}\root 6 \of {12} \cr & {\text{Take LCM of }}\left( {3,\,2{\text{ and }}6} \right) = 6 \cr & {6^{\frac{{1 \times 6}}{3}}},\,{5^{\frac{{1 \times 6}}{2}}},\,{12^{\frac{{1 \times 6}}{6}}} \cr & 36,\,125{\text{ and }}12 \cr & {\text{Clearly, }}\root 2 \of 5 {\text{ is the largest}}{\text{.}} \cr} $$

Solution:

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

Solution:

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