Practice MCQ Questions and Answer on Square Root and Cube Root

Solution:

 $$\eqalign{ & {\left( {11} \right)^2} = 121{\text{ }} \cr & {\text{And }} \cr & {\left( {17} \right)^2} = 289 \cr} $$ So, the perfect squares between 120 and 300 are the squares of numbers from 11 to 17. Clearly, these are 7 in number.

Solution:

 $$\eqalign{ & \Rightarrow 3\sqrt 5 + \sqrt {125} = 17.88 \cr & \Rightarrow {\text{ }}3\sqrt 5 + \sqrt {25 \times 5} = 17.88 \cr & \Rightarrow {\text{ }}3\sqrt 5 + 5\sqrt 5 = 17.88 \cr & \Rightarrow {\text{ }}8\sqrt 5 = 17.88 \cr & \Rightarrow \sqrt 5 = 2.235 \cr & \therefore \sqrt {80} + 6\sqrt 5 \cr & = \sqrt {16 \times 5} + 6\sqrt 5 \cr & = 4\sqrt 5 + 6\sqrt 5 \cr & = 10\sqrt 5 \cr & = \left( {10 \times 2.235} \right) \cr & = 22.35 \cr} $$

Solution:

 $$\eqalign{ & \frac{{\sqrt 5 }}{2} - \frac{{10}}{{\sqrt 5 }} + \sqrt {125} \, \cr & = \frac{{{{\left( {\sqrt 5 } \right)}^2} - 20 + 2\sqrt 5 \times 5\sqrt 5 }}{{2\sqrt 5 }} \cr & = \frac{{5 - 20 + 50}}{{2\sqrt 5 }} \cr & = \frac{{35}}{{2\sqrt 5 }} \times \frac{{\sqrt 5 }}{{\sqrt 5 }} \cr & = \frac{{35\sqrt 5 }}{{10}} \cr & = \frac{{7 \times 2.236}}{2} \cr & = 7 \times 1.118 \cr & = 7.826 \cr} $$

Solution:

 Given expression, $$ = \frac{1}{{\left( {\sqrt 9 - \sqrt 8 } \right)}} \times \frac{{\left( {\sqrt 9 + \sqrt 8 } \right)}}{{\left( {\sqrt 9 + \sqrt 8 } \right)}}$$     $$ - \frac{1}{{\left( {\sqrt 8 - \sqrt 7 } \right)}}$$   $$ \times \frac{{\left( {\sqrt 8 + \sqrt 7 } \right)}}{{\left( {\sqrt 8 + \sqrt 7 } \right)}}$$   $$ + \frac{1}{{\left( {\sqrt 7 - \sqrt 6 } \right)}}$$   $$ \times \frac{{\left( {\sqrt 7 + \sqrt 6 } \right)}}{{\left( {\sqrt 7 + \sqrt 6 } \right)}}$$   $$ - \frac{1}{{\left( {\sqrt 6 - \sqrt 5 } \right)}}$$   $$ \times \frac{{\left( {\sqrt 6 + \sqrt 5 } \right)}}{{\left( {\sqrt 6 + \sqrt 5 } \right)}}$$   $$ + \frac{1}{{\left( {\sqrt 5 - \sqrt 4 } \right)}}$$   $$ \times \frac{{\left( {\sqrt 5 + \sqrt 4 } \right)}}{{\left( {\sqrt 5 + \sqrt 4 } \right)}}$$ $$ = \frac{{\left( {\sqrt 9 + \sqrt 8 } \right)}}{{\left( {9 - 8} \right)}} - \frac{{\left( {\sqrt 8 + \sqrt 7 } \right)}}{{\left( {8 - 7} \right)}}$$     $$ + \frac{{\left( {\sqrt 7 + \sqrt 6 } \right)}}{{\left( {7 - 6} \right)}}$$   $$ - \frac{{\left( {\sqrt 6 + \sqrt 5 } \right)}}{{\left( {6 - 5} \right)}}$$   $$ + \frac{{\left( {\sqrt 5 + \sqrt 4 } \right)}}{{\left( {5 - 4} \right)}}$$ $$ = \left( {\sqrt 9 + \sqrt 8 } \right) - \left( {\sqrt 8 + \sqrt 7 } \right)$$     $$ + \left( {\sqrt 7 + \sqrt 6 } \right)$$   $$ - \left( {\sqrt 6 + \sqrt 5 } \right)$$   $$ + \left( {\sqrt 5 + \sqrt 4 } \right)$$ $$ = \left( {\sqrt 9 + \sqrt 4 } \right)$$ $$ = 3 + 2$$ $$ = 5$$

Solution:

 $$\eqalign{ & {\text{Given expression,}} \cr & = \frac{{3\sqrt {12} }}{{2\sqrt {28} }} \times \frac{{\sqrt {98} }}{{2\sqrt {21} }} \cr & = \frac{{3\sqrt {4 \times 3} }}{{2\sqrt {4 \times 7} }} \times \frac{{\sqrt {49 \times 2} }}{{2\sqrt {21} }} \cr & = \frac{{6\sqrt 3 }}{{4\sqrt 7 }} \times \frac{{7\sqrt 2 }}{{2\sqrt {21} }} \cr & = \frac{{21\sqrt 6 }}{{4\sqrt {7 \times 21} }} \cr & = \frac{{21\sqrt 6 }}{{28\sqrt 3 }} \cr & = \frac{3}{4}\sqrt 2 \cr & = \frac{3}{4} \times 1.414 \cr & = 3 \times 0.3535 \cr & = 1.0605 \cr} $$

Solution:

 The number of digits of the square root of a perfect square number of n digits is $$\eqalign{ & {\text{(i)}}\frac{n}{2}{\text{, if n is even}} \cr & {\text{(ii)}}\frac{{n + 1}}{2}{\text{, if n is odd}} \cr & {\text{Here, }}n = 12 \cr & {\text{So, required number of digits}} \cr & = \frac{n}{2} \cr & = \frac{{12}}{2} \cr & = 6{\text{ }} \cr} $$

Solution:

 $$\eqalign{ & \Leftrightarrow {{\text{n}}^2} = {\left( {{\text{25}}} \right)^{64}} \times {\left( {64} \right)^{25}} \cr & \Leftrightarrow {{\text{n}}^2} = {\left( {{{\text{5}}^2}} \right)^{64}} \times {\left( {{2^6}} \right)^{25}} \cr & \Leftrightarrow {{\text{n}}^2} = {5^{128}} \times {2^{150}} \cr & \Leftrightarrow {{\text{n}}^2} = {5^{128}} \times {2^{128}} \times {2^{22}} \cr & \Leftrightarrow n = {5^{64}} \times {2^{64}} \times {2^{11}} \cr & \Leftrightarrow n = {\left( {5 \times 2} \right)^{64}} \times {2^{11}} \cr & \Leftrightarrow n = {10^{64}} \times 2048 \cr} $$ ∴ Sum of digits of n = 2 + 0 + 4 + 8 = 14

Solution:

 $$\eqalign{ & \Leftrightarrow \sqrt {{3^n}} = 729 \cr & \Leftrightarrow \sqrt {{3^n}} = {3^6} \cr & \Leftrightarrow {\left( {\sqrt {{3^n}} } \right)^2} = {\left( {{3^6}} \right)^2} \cr & \Leftrightarrow {3^n} = {3^{12}} \cr & \Leftrightarrow n = 12 \cr} $$

Solution:

 $$\eqalign{ & 0.000326 = \frac{{326}}{{{{10}^6}}} \cr & \,\,\,\,1|\overline 3 \,\,\overline {26} \,\,(18 \cr & \,\,\,\,\,\,\,|\,\,1 \cr & \,\,\,\,\,\,\,| - - - - - - \cr & 28|\,\,\,2\,26 \cr & \,\,\,\,\,\,\,|\,\,\,2\,24 \cr & \,\,\,\,\,\,\,| - - - - - - \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2 \cr} $$ ∴ Required number to be subtracted $$\eqalign{ & = \frac{2}{{{{10}^6}}} \cr & = 0.000002 \cr} $$

Solution:

 $$\therefore \sqrt {123454321} = 11111$$