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

$$\eqalign{ & \,\,\,\,\,\,1|\overline 1 \,\,\overline {58} \,\,\overline {76} \,(126 \cr & \,\,\,\,\,\,\,\,\,|1 \cr & \,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & \,\,22|\,\,\,\,\,\,58 \cr & \,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,44 \cr & \,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & 246\,|\,\,\,\,\,\,\,14\,76 \cr & \,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,14\,76 \cr & \,\,\,\,\,\,\,\,\,| - - - - - - - \cr & \,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,x \cr & \,\,\,\,\,\,\,\,\,| - - - - - - - \cr & \therefore \sqrt {15876} = 126 \cr} $$

$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,1|\overline 3 \,.\,\,\overline {00} \,\,\overline {00} \,\,\overline {00} \,(1.732 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,|1 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & \,\,\,\,\,\,27|\,\,2\,\,\,00 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,1\,\,\,89 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & \,\,\,343\,|\,\,\,\,\,\,\,\,\,11\,\,00 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,10\,\,29 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - \cr & 3492\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,71\,\,00 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,69\,\,84 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,16 \cr} $$
$$\eqalign{ & \therefore \sqrt 3 \cr & = 1.73 \cr & = \frac{{173}}{{100}} \cr} $$

The squares of numbers having 1 and 9 as the unit's digit end in the digit 1.
$$\eqalign{ & {\text{Such numbers are,}} \cr & 1,9,11,19,21,29,31,39,41,49{\text{ i}}{\text{.e}}{\text{.,}} \cr & {\text{There are 10 such numbers}}{\text{.}} \cr & \therefore {\text{Required percentage}} \cr & = \left( {\frac{{10}}{{50}} \times 100} \right)\% \cr & = 20\% \cr} $$

$$\eqalign{ & {\text{Let the given number be }}x \cr & {\text{Then,}} \cr & \Leftrightarrow {x^2} - 25 = {\left( {x - 25} \right)^2}{\text{ }} \cr & \Leftrightarrow {x^2} - 25 = {x^2} + 625 - 50x \cr & \Leftrightarrow 50x = 650 \cr & \Leftrightarrow x = 13 \cr} $$

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

The first perfect square number after 50 is 64 $$\left( {64 = {8^2}} \right)$$   and the last perfect square number before 1000 is 961 $$\left[ {961 = {{\left( {31} \right)}^2}} \right]$$
So, the perfect squares between 50 and 1000 are the squares of numbers from 8 to 31.
(31 - 8) + 1 = 24
Clearly, these are 24 in number.

Clearly, the man was born between 1800 and 1850 is 1849.
$${\text{And, }}1849 = {\left( {43} \right)^2}$$
So, the man was 43 years old in 1849
$$\eqalign{ & {\text{Thus, he was born in }} \cr & = \left( {1849 - 43} \right) \cr & = 1806 \cr} $$

$$\eqalign{ & \Rightarrow {\text{Q}} = \sqrt {\frac{{{{\left( {8{\text{R}}} \right)}^2}}}{4}} \cr & \Rightarrow {\text{Q}} = \frac{{\sqrt {{{\left( {8{\text{R}}} \right)}^2}} }}{{\sqrt 4 }} \cr & \Rightarrow {\text{Q}} = \frac{{8{\text{R}}}}{2} \cr & \Rightarrow {\text{Q}} = 4{\text{R}} \cr} $$

The smallest such number is 1444 $$\left[ {1444 = {{\left( {38} \right)}^2}} \right]$$
It lies between 1000 and 2000.

$$\eqalign{ & {\text{We have,}} \cr & 1 \times 2 \times 3 \times 4 = 24 \cr & {\text{And }}24 + 1 = 25\left[ {25 = {5^2}} \right] \cr & 2 \times 3 \times 4 \times 5 = 120{\text{ }} \cr & {\text{And 1}}20 + 1 = 121\left[ {121 = {{11}^2}} \right] \cr & 3 \times 4 \times 5 \times 6 = 360{\text{ }} \cr & {\text{And }}360 + 1 = 361\left[ {361 = {{19}^2}} \right] \cr & 4 \times 5 \times 6 \times 7 = 840{\text{ }} \cr & {\text{And }}840 + 1 = 841\left[ {841 = {{29}^2}} \right] \cr & \therefore p = 1 \cr} $$