Square Root And Cube Root - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & {\text{Give}}\,{\text{Expression}} \cr & = \frac{{25}}{{11}} \times \frac{{14}}{5} \times \frac{{11}}{{14}} \cr & = 5 \cr} $$

$$\eqalign{ & {\text{Let}}\sqrt {0.0169 \times x} = 1.3 \cr & {\text{Then}},\,0.0169x = {\left( {1.3} \right)^2} = 1.69 \cr & \Rightarrow x = \frac{{1.69}}{{0.0169}} = 100 \cr} $$

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

A number ending in 8 can never be a perfect square.

$$\eqalign{ & \,\,\,\,\,\,\,\,2|\overline 6 \overline {40} \overline {09} \,(\,\,253 \cr & \,\,\,\,\,\,\,\,\,\,\,|4 \cr & \,\,\,\,\,\,\,\,\,\,\,| - - - - - \cr & \,\,\,\,45\,|240 \cr & \,\,\,\,\,\,\,\,\,\,\,|225 \cr & \,\,\,\,\,\,\,\,\,\,\,| - - - - - \cr & 503\,\,|\,\,\,\,\,\,1509 \cr & \,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,1509 \cr & \,\,\,\,\,\,\,\,\,\,\,| - - - - - \cr & \,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,x \cr & \,\,\,\,\,\,\,\,\,\,\,| - - - - - \cr & \therefore \sqrt {64009} = 253 \cr} $$

$$\eqalign{ & \,\,\,\,\,2|4\,\,\overline {12} \,\,\overline {09} \,(203 \cr & \,\,\,\,\,\,\,\,\,|4 \cr & \,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & \,\,40|\,\,\,\,\,\,\,\,12 \cr & \,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,0 \cr & \,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & \,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,12\,09 \cr & 403|\,\,\,\,\,\,\,\,\,12\,09 \cr & \,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & \,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,x \cr & \,\,\,\,\,\,\,\,\,| - - - - - - - - \, \cr & \therefore \sqrt {41209} = 203 \cr} $$

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

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

$$\eqalign{ & \sqrt {\sqrt {17956} + \sqrt {24025} } \cr & = \sqrt {134 + 155} \cr & = \sqrt {289} \cr & = 17 \cr} $$

Sum of prime numbers greater than 4 but less than 16
$$\eqalign{ & = \left( {5 + 7 + 11 + 13} \right) \cr & = 36 \cr & \therefore \frac{1}{4} \times 36 \cr & = 9 \cr & = {3^2} \cr} $$