Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

$$\eqalign{ & x:y \cr & 7:3{\text{ }} \cr & \therefore {\text{ }}\frac{{xy + {y^2}}}{{{x^2} - {y^2}}} \cr & = \frac{{21 + 9}}{{49 - 9}} \cr & = \frac{{30}}{{40}} \cr & = \frac{3}{4} \cr} $$

Given [p] means the greatest positive integer less than or [p] equal to p
$$\eqalign{ & \Rightarrow \left[ {\text{p}} \right] = {\text{ p}} \cr & \Rightarrow \left[ -{\text{p}} \right] = {\text{ p}} \cr & \Rightarrow \left[ { - \frac{1}{4}} \right]{\text{ + }}\left[ {4 - \frac{1}{4}} \right] + \left[ 3 \right] \cr & \Rightarrow \frac{1}{4} + 4 - \frac{1}{4} + 3 \cr & \Rightarrow 7 \cr} $$

$$\eqalign{ & \frac{{{{\left( {243} \right)}^{\frac{n}{5}}}{{.3}^{2n + 1}}}}{{{9^n}{{.3}^{n - 1}}}} \cr & = \frac{{{{\left( {{3^5}} \right)}^{\frac{n}{5}}}{{.3}^{2n + 1}}}}{{{3^{2n}}{{.3}^{n - 1}}}} \cr & = \frac{{{3^{n + 2n + 1}}}}{{{3^{2n + n - 1}}}} \cr & = \frac{{{3^{3n + 1}}}}{{{3^{3n - 1}}}} \cr & = {3^{3n + 1 - 3n + 1}} \cr & = {3^2} \cr & = 9 \cr} $$

$$\eqalign{ & x = 0.5 \cr & y = 0.2 \cr & \sqrt {0.6} \times {\left( {3y} \right)^x}{\text{ }} \cr & = \sqrt {0.6} \times {\left( {3 \times 0.2} \right)^{0.5}}{\text{ }} \cr & = \sqrt {0.6} \times \sqrt {0.6} \cr & = 0.6 \cr} $$

$$\eqalign{ & {x^{x\sqrt x }} = {\left( {x\sqrt x } \right)^x} \cr & {x^{x\sqrt x }} = {\left( {{x^{\frac{3}{2}}}} \right)^x} \cr & {x^{x\sqrt x }} = {x^{\frac{3}{2}x}} \cr} $$
(If bases are same then their power is also same)
$$\eqalign{ & \therefore x\sqrt x = \frac{3}{2}x \cr & \Rightarrow \sqrt x = \frac{3}{2} \cr & \Rightarrow x = {\left( {\frac{3}{2}} \right)^2} \cr & \Rightarrow x = \frac{9}{4} \cr} $$

$$\eqalign{ & {a^2} + {b^2} + {c^2} - ab - bc - ca \cr & = \frac{1}{2}\left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right] \cr & = \frac{1}{2}\left[ {{{\left( {7 - 5} \right)}^2} + {{\left( {5 - 3} \right)}^2} + {{\left( {3 - 7} \right)}^2}} \right] \cr & = \frac{1}{2}\left( {4 + 4 + 16} \right) \cr & = \frac{{24}}{2} \cr & = 12 \cr} $$

$$\eqalign{ & {{\text{7}}^x}{\text{ = }}\frac{1}{{343}} \cr & \Rightarrow {{\text{7}}^x}{\text{ = }}\frac{1}{{{7^3}}} \cr & \Rightarrow {{\text{7}}^x}{\text{ = }}{{\text{7}}^{ - 3}} \cr & \Rightarrow x = - 3 \cr} $$
(If bases are equal then their power are also equal)

$$\frac{{\frac{1}{3}.\frac{1}{3}.\frac{1}{3} + \frac{1}{4}.\frac{1}{4}.\frac{1}{4} - 3.\frac{1}{3}.\frac{1}{4}.\frac{1}{5} + \frac{1}{5}.\frac{1}{5}.\frac{1}{5}}}{{\frac{1}{3}.\frac{1}{3} + \frac{1}{4}.\frac{1}{4} + \frac{1}{5}.\frac{1}{5} - \left( {\frac{1}{3}.\frac{1}{4} + \frac{1}{4}.\frac{1}{5} + \frac{1}{5}.\frac{1}{3}} \right)}}$$

A³ + B³ + C³ - 3ABC = (A + B + C)(A² + B² + C² - AB - BC - CA)

$$\therefore \frac{{{{\left( {\frac{1}{3}} \right)}^3} + {{\left( {\frac{1}{4}} \right)}^3} - 3.\frac{1}{3}.\frac{1}{4}.\frac{1}{5} + {{\left( {\frac{1}{5}} \right)}^3}}}{{{{\left( {\frac{1}{3}} \right)}^2} + {{\left( {\frac{1}{4}} \right)}^2} + {{\left( {\frac{1}{5}} \right)}^2} - \frac{1}{3}.\frac{1}{4} - \frac{1}{4}.\frac{1}{5} - \frac{1}{5}.\frac{1}{3}}}$$
$$ = \frac{{\left( {\frac{1}{3} + \frac{1}{4} + \frac{1}{5}} \right)\left[ {{{\left( {\frac{1}{3}} \right)}^2} + {{\left( {\frac{1}{4}} \right)}^2} + {{\left( {\frac{1}{5}} \right)}^2} - \frac{1}{3}.\frac{1}{4} - \frac{1}{4}.\frac{1}{5} - \frac{1}{5}.\frac{1}{3}} \right]}}{{\left[ {{{\left( {\frac{1}{3}} \right)}^2} + {{\left( {\frac{1}{4}} \right)}^2} + {{\left( {\frac{1}{5}} \right)}^2} - \frac{1}{3}.\frac{1}{4} - \frac{1}{4}.\frac{1}{5} - \frac{1}{5}.\frac{1}{3}} \right]}}$$
$$\eqalign{ & = \frac{{20 + 15 + 12}}{{60}} \cr & = \frac{{47}}{{60}} \cr} $$

$$\eqalign{ & {\text{0}}{\text{.13}} \times {{\text{p}}^2} = 13 \cr & \Rightarrow {{\text{p}}^2} = \frac{{13}}{{0.13}} \cr & \Rightarrow {{\text{p}}^2} = \frac{{13}}{{13}} \times 100 \cr & \Rightarrow {{\text{p}}^2} = 100 \cr & \Rightarrow {\text{p}} = 10 \cr} $$

$$\eqalign{ & x = 7 - 4\sqrt 3 \cr & \Rightarrow x = 4 + 3 - 4\sqrt 3 \cr & \Rightarrow x = {\left( 2 \right)^2} + {\left( {\sqrt 3 } \right)^2} - 2 \times 2\sqrt 3 \cr & \Rightarrow {\left( {2 - \sqrt 3 } \right)^2} \cr & \therefore \left[ {{a^2} + {b^2} - 2ab = {{\left( {a - b} \right)}^2}} \right] \cr & \Rightarrow x = {\left( {2 - \sqrt 3 } \right)^2} \cr & \Rightarrow \sqrt x = 2 - \sqrt 3 \cr & \Rightarrow \frac{1}{{\sqrt x }} = \frac{1}{{2 - \sqrt 3 }} \times \frac{{2 + \sqrt 3 }}{{2 + \sqrt 3 }} \cr & \Rightarrow 2 + \sqrt 3 \cr & \therefore \sqrt x {\text{ + }}\frac{1}{{\sqrt x }}{\text{ }} \cr & = 2 - \sqrt 3 + 2 + \sqrt 3 \cr & = 4 \cr} $$