Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

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

$$\eqalign{ & {4^{4x + 1}} = \frac{1}{{64}} \cr & \Rightarrow {4^{4x + 1}} = \frac{1}{{{{\left( 4 \right)}^3}}} \cr & \Rightarrow {4^{4x + 1}} = {\left( 4 \right)^{ - 3}} \cr & \Rightarrow 4x + 1 = - 3 \cr & \Rightarrow 4x = - 4 \cr & \Rightarrow x = - 1 \cr} $$

$$\eqalign{ & {\left( {\sqrt 5 } \right)^7} \div {\left( {\sqrt 5 } \right)^5}{\text{ = }}{{\text{5}}^{\text{P}}} \cr & \Rightarrow \frac{{{{\left( {\sqrt 5 } \right)}^7}}}{{{{\left( {\sqrt 5 } \right)}^5}}} = {{\text{5}}^{\text{P}}} \cr & \Rightarrow {\left( {\sqrt 5 } \right)^2} = {{\text{5}}^{\text{P}}} \cr & \Rightarrow {{\text{5}}^{\text{1}}} = {\text{ }}{{\text{5}}^{\text{P}}} \cr & \Rightarrow \boxed{{\text{P}} = 1} \cr} $$

$$\eqalign{ & 1.5a = 0.04b \cr & \frac{a}{b} = \frac{{0.04}}{{1.5}} = \frac{4}{{100}} \times \frac{{10}}{{15}} = \frac{2}{{75}} \cr & {\text{Let }}a = 2x,{\text{ }}b = 75x \cr & \therefore \frac{{b - a}}{{b + a}} = \frac{{75x - 2x}}{{75x + 2x}} = \frac{{73}}{{77}} \cr & \cr & {\bf{Alternate:}} \cr & \frac{a}{b} = \frac{{0.04}}{{1.5}} \cr & \therefore \frac{{b - a}}{{b + a}} = \frac{{1.5 - 0.04}}{{1.5 + 0.04}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{1.46}}{{1.54}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{73}}{{77}} \cr} $$

$$\eqalign{ & \frac{{{x^2} - x + 1}}{{{x^2} + x + 1}} = \frac{3}{2}\left( {{\text{Given}}} \right) \cr & \Rightarrow \frac{{x\left\{ {\left( {x + \frac{1}{x}} \right) - 1} \right\}}}{{x\left\{ {\left( {x + \frac{1}{x}} \right) + 1} \right\}}} = \frac{3}{2} \cr & \Rightarrow \frac{{\left( {x + \frac{1}{x}} \right) - 1}}{{\left( {x + \frac{1}{x}} \right) + 1}} = \frac{3}{2} \cr & \,\,\,\,\,\,\,\,\,\,{\text{Let }}\left( {x + \frac{1}{x} = y} \right) \cr & \Rightarrow \frac{{y - 1}}{{y + 1}} = \frac{3}{2} \cr & \Rightarrow 2\left( {y - 1} \right) = 3\left( {y + 1} \right) \cr & \Rightarrow 2y - 2 = 3y + 3 \cr & \Rightarrow y = - 2 - 3 \cr & \Rightarrow y = - 5 \cr & \therefore x + \frac{1}{x} = - 5 \cr & {\text{ }} \cr} $$

$$\eqalign{ & \Rightarrow x = 3 + \sqrt 8 \cr & \Rightarrow {x^2} = 9 + 8 + 2 \times 3\sqrt 8 \cr & \Rightarrow {x^2} = 17 + 6\sqrt 8 \cr & \Rightarrow \frac{1}{{{x^2}}} = 17 - 6\sqrt 8 \cr & \therefore {x^2} + \frac{1}{{{x^2}}} \cr & = 17 + 6\sqrt 8 + 17 - 6\sqrt 8 \cr & = 34 \cr} $$

$$\eqalign{ & x = 5 + 2\sqrt 6 \cr & \Leftrightarrow x = 3 + 2 + 2\sqrt 3 \times \sqrt 2 \cr & \Leftrightarrow x = {\left( {\sqrt 3 } \right)^2} + {\left( {\sqrt 2 } \right)^2} + 2\sqrt 3 \times \sqrt 2 \cr & \Leftrightarrow x = {\left( {\sqrt 3 + \sqrt 2 } \right)^2} \cr & \Leftrightarrow \sqrt x = \sqrt 3 + \sqrt 2 \cr & {\text{Similarly,}} \cr & \Leftrightarrow \frac{1}{{\sqrt x }} = \sqrt 3 - \sqrt 2 \cr & \therefore \left( {\sqrt x + \frac{1}{{\sqrt x }}} \right) \cr & = \sqrt 3 + \sqrt 2 + \sqrt 3 - \sqrt 2 \cr & = 2\sqrt 3 \cr} $$

$$\eqalign{ & x + \frac{9}{x} = 6 \cr & {\text{Take value of x}} \cr & {\text{Let }}x = 3 \cr & 3 + \frac{9}{3} = 6{\text{ }}\left( {{\text{Proved}}} \right) \cr & {\text{So, }}x = 3 \cr & \therefore {x^2} + \frac{9}{{{x^2}}} \cr & = 9 + \frac{9}{9} \cr & = 10 \cr & \cr & {\bf{Alternate:}} \cr & x + \frac{9}{x} = 6 \cr & {\text{On squaring,}} \cr & \Rightarrow {\left( {x + \frac{9}{x}} \right)^2} = 36 \cr & \Rightarrow {x^2} + \frac{{81}}{{{x^2}}} + 2 \times x \times \frac{9}{x} = 36 \cr & \Rightarrow {x^2} + \frac{{81}}{{{x^2}}} - 18 = 0 \cr & \Rightarrow {\left( {x - \frac{9}{x}} \right)^2} = 0 \cr & \Rightarrow x = \frac{9}{x} \cr & \Rightarrow {x^2} = 9 \cr & {\text{Hence, }}\left( {{x^2} + \frac{9}{{{x^2}}}} \right) \cr & = \left( {9 + \frac{9}{9}} \right) \cr & = 10 \cr} $$

$$\eqalign{ & x + \frac{1}{x} = 3{\text{ }}\left( {{\text{Given}}} \right) \cr & \frac{{{x^3} + \frac{1}{x}}}{{{x^2} - x + 1}}{\text{ }}\left( {{\text{Divide by }}x} \right) \cr & \Rightarrow \frac{{\frac{{{x^3}}}{x} + \frac{1}{{{x^2}}}}}{{\frac{{{x^2}}}{x} - \frac{x}{x} + \frac{1}{x}}} \cr & \Rightarrow \frac{{{x^2} + \frac{1}{{{x^2}}}}}{{x - 1 + \frac{1}{x}}} \cr & \Rightarrow \frac{{{x^2} + \frac{1}{{{x^2}}}}}{{x + \frac{1}{x} - 1}} \cr & \therefore x + \frac{1}{x} = 3 \cr & \therefore {x^2} + \frac{1}{{{x^2}}} = 9 - 2 = 7 \cr & \therefore \frac{{{x^2} + \frac{1}{{{x^2}}}}}{{x + \frac{1}{x} - 1}} = \frac{7}{{3 - 1}} = \frac{7}{2} \cr} $$

$$\eqalign{ & a + \frac{1}{a} + 1 = 0 \cr & a + \frac{1}{a} = - 1 \cr & {\text{Squaring both sides}} \cr & \Rightarrow {a^2} + \frac{1}{{{a^2}}} + 2 = 1 \cr & \Rightarrow {a^2} + \frac{1}{{{a^2}}} = - 1 \cr & \Rightarrow {a^2} + 1 = - \frac{1}{{{a^2}}}\,.....(i) \cr & \Rightarrow a + \frac{1}{a} = - 1{\text{ }}\left( {{\text{Given}}} \right) \cr & \therefore {a^2} + 1 = - a\,.....(ii) \cr & \Rightarrow - a = \frac{{ - 1}}{{{a^2}}} \cr & {\text{For equation (i) and (ii)}} \cr & {{\text{a}}^3} = 1 \cr & \therefore {{\text{a}}^3} - 1 = 0 \cr & \Rightarrow {a^4} - a = a\left( {{a^3} - 1} \right) \cr & \Rightarrow {a^4} - a = a \times 0 \cr & \Rightarrow {a^4} - a = 0 \cr} $$