Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

$$\eqalign{ & x = a + \frac{1}{a} \cr & y = a - \frac{1}{a} \cr & \therefore \left( {x + y} \right) = a + \frac{1}{a} + a - \frac{1}{a} = 2a \cr & \therefore \left( {x - y} \right) = a + \frac{1}{a} - a + \frac{1}{a} = \frac{2}{a} \cr & \therefore {x^4}{\text{ + }}{{\text{y}}^4} - 2{x^2}{y^2} \cr & = {\left( {{x^2} - {y^2}} \right)^2} \cr & = {\left[ {\left( {x + y} \right)\left( {x - y} \right)} \right]^2} \cr & = {\left( {2a \times \frac{2}{a}} \right)^2} \cr & = {\left( 4 \right)^2} \cr & = 16 \cr} $$

$$\eqalign{ & p = 101 \cr & \root 3 \of {p\left( {{p^2} - 3p + 3} \right) - 1} \cr & = \root 3 \of {{p^3} - 3{p^2} + 3p - 1} \cr & \therefore \left[ {{{\left( {p - 1} \right)}^3} = {p^3} - {{\left( 1 \right)}^3} - 3p\left( {p - 1} \right)} \right] \cr & = \root 3 \of {{{\left( {p - 1} \right)}^3}} \cr & = p - 1 \cr & = 101 - 1 \cr & = 100{\text{ }} \cr} $$

$$\eqalign{ & 50\% {\text{ of }}\left( {p - q} \right) = 30\% {\text{ of }}\left( {p + q} \right) \cr & \frac{{p - q}}{2} = \frac{3}{{10}}\left( {p + q} \right) \cr & \left[ {50\% = \frac{1}{2}} \right] \cr & \Rightarrow 5\left( {p - q} \right) = 3\left( {p + q} \right) \cr & \Rightarrow 5p - 5q = 3p + 3q \cr & \Rightarrow 2p = 8q \cr & \Rightarrow 1p = 4q \cr & \Rightarrow p:q = 4:1 \cr} $$

$$\eqalign{ & x:y = 2:1 \cr & {\text{then, }}5{x^2} - 13xy + 6{y^2} \cr & \Rightarrow 5 \times 4 - 13 \times 2 \times 1 + 6 \times {1^2} \cr & \Rightarrow 20 - 26 + 6 \cr & \Rightarrow 0 \cr} $$

$$\eqalign{ & \frac{a}{3} = \frac{b}{2} \Rightarrow \frac{a}{b} = \frac{3}{2} \cr & \therefore \frac{{2a + 3b}}{{3a - 2b}} \cr & = \frac{{2 \times 3 + 3 \times 2}}{{3 \times 3 - 2 \times 2}} \cr & = \frac{{6 + 6}}{{9 - 4}} \cr & = \frac{{12}}{5} \cr} $$

$$\eqalign{ & \frac{{2x - y}}{{x + 2y}}\, \times \frac{1}{2}\left( {{\text{Cross multiply}}} \right) \cr & \Rightarrow 4x - 2y = x + 2y \cr & \Rightarrow 3x = 4y \cr & \Rightarrow x:y = 4:3 \cr & \therefore \frac{{3x - y}}{{3x + y}}{\text{ }} \cr & {\text{ = }}\frac{{3 \times 4 - 3}}{{3 \times 4 + 3}} \cr & {\text{ = }}\frac{{12 - 3}}{{12 + 3}} \cr & {\text{ = }}\frac{9}{{15}}{\text{ }} \cr & {\text{ = }}\frac{3}{5} \cr} $$

$$\eqalign{ & x + \frac{1}{x} = 5 \cr & \therefore \frac{{2x}}{{3{x^2} - 5x + 3}}\,\left( {{\text{Divide by }}x} \right) \cr & = \frac{{\frac{{2x}}{x}}}{{\frac{{3{x^2}}}{x} - \frac{{5x}}{x} + \frac{3}{x}}} \cr & = \frac{2}{{3x + \frac{3}{x} - 5}} \cr & = \frac{2}{{3\left( {x + \frac{1}{x}} \right) - 5}} \cr & = \frac{2}{{3 \times 5 - 5}} \cr & = \frac{2}{{10}} \cr & = \frac{1}{5} \cr} $$

$$\eqalign{ & \sqrt {1 - \frac{{{x^3}}}{{100}}} = \frac{3}{5} \cr & \Rightarrow 1 - \frac{{{x^3}}}{{100}} = {\left( {\frac{3}{5}} \right)^2} \cr & \Rightarrow 1 - \frac{9}{{25}} = \frac{{{x^3}}}{{100}} \cr & \Rightarrow \frac{{16}}{{25}} = \frac{{{x^3}}}{{100}} \cr & \Rightarrow {x^3} = \frac{{16 \times 100}}{{25}} \cr & \Rightarrow {x^3} = 16 \times 4 \cr & \Rightarrow {x^3} = 64 \cr & \Rightarrow \boxed{x = 4} \cr} $$

$$\eqalign{ & \sqrt {1 + \frac{x}{9}} = \frac{{13}}{3} \cr & {\text{By option }} \cr & {\text{Put }}x = 160 \cr & \sqrt {1 + \frac{{160}}{9}} = \frac{{13}}{3} \cr & \Rightarrow \sqrt {\frac{{169}}{9}} = \frac{{13}}{3} \cr & \Rightarrow \frac{{13}}{3} = \frac{{13}}{3} \cr & \cr & {\bf{Alternate:}} \cr & {\text{Squaring both sides}} \cr & {\left( {\sqrt {1 + \frac{{x}}{9}} } \right)^2} = {\left( {\frac{{13}}{3}} \right)^2} \cr & \Rightarrow 1 + \frac{x}{9} = \frac{{169}}{9} \cr & \Rightarrow \frac{{9 + x}}{9} = \frac{{169}}{9} \cr & \Rightarrow 9 + x = 169 \cr & \Rightarrow \boxed{x = 160} \cr} $$

$$\eqalign{ & \frac{{4\sqrt 3 + 5\sqrt 2 }}{{\sqrt {48} + \sqrt {18} }} = a + b\sqrt 6 \cr & \Rightarrow \frac{{4\sqrt 3 + 5\sqrt 2 }}{{\sqrt {16 \times 3} + \sqrt {9 \times 2} }} = a + b\sqrt 6 \cr & \Rightarrow \frac{{4\sqrt 3 + 5\sqrt 2 }}{{4\sqrt 3 + 3\sqrt 2 }} = a + b\sqrt 6 \cr & \Rightarrow \frac{{4\sqrt 3 + 5\sqrt 2 }}{{4\sqrt 3 + 3\sqrt 2 }} \times \frac{{4\sqrt 3 - 3\sqrt 2 }}{{4\sqrt 3 - 3\sqrt 2 }} = a + b\sqrt 6 \cr & \Rightarrow \frac{{\left( {4\sqrt 3 + 5\sqrt 2 } \right)\left( {4\sqrt 3 - 3\sqrt 2 } \right)}}{{48 - 18}} = a + b\sqrt 6 \cr & \Rightarrow \frac{{8\sqrt 6 + 18}}{{30}} = a + b\sqrt 6 \cr & \Rightarrow \frac{{8\sqrt 6 }}{{30}} + \frac{{18}}{{30}} = a + b\sqrt 6 \cr & \Rightarrow \frac{4}{{15}}\sqrt 6 + \frac{3}{5} = a + b\sqrt 6 \cr & \Rightarrow \frac{3}{5} + \frac{4}{{15}}\sqrt 6 = a + b\sqrt 6 \cr} $$
By comparing coefficients of rational and irrational parts.
$$\eqalign{ & \Rightarrow a = \frac{3}{5}{\text{ , }}b = \frac{4}{{15}} \cr & \therefore \left( {\frac{3}{5},\frac{4}{{15}}} \right) \cr} $$