Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

$$\eqalign{ & \frac{{\sqrt 7 - 2}}{{\sqrt 7 + 2}} = a\sqrt 7 + b \cr & {\text{L}}{\text{.H}}{\text{.S }}\frac{{\sqrt 7 - 2}}{{\sqrt 7 + 2}} \times \frac{{\sqrt 7 - 2}}{{\sqrt 7 - 2}} \cr & {\text{ }}\left( {{\text{Rationalisation}}} \right) \cr & = \frac{{{{\left( {\sqrt 7 - 2} \right)}^2}}}{{{{\left( {\sqrt 7 } \right)}^2} - \left( 4 \right)}} \cr & = \frac{{7 + 4 - 4\sqrt 7 }}{{7 - 4}} \cr & = \frac{{11 - 4\sqrt 7 }}{3} \cr & = \frac{{11}}{3} - \frac{4}{3}\sqrt 7 \cr & = - \frac{4}{3}\sqrt 7 + \frac{{11}}{3} \cr & = a\sqrt 7 + b{\text{ }}\left( {{\text{ R}}{\text{.H}}{\text{.S}}{\text{.}}} \right) \cr} $$
(Compare the coefficients of $$\sqrt 7 $$ and constant term)
$$\eqalign{ & {\text{a}} = - \frac{4}{3} \cr & b = \frac{{11}}{3} \cr} $$

$$\eqalign{ & a + \frac{1}{b} = 1{\text{, }}b + \frac{1}{c} = 1{\text{, }}c + \frac{1}{a} = ? \cr & {\text{Put values, }} \cr & a = \frac{1}{2},b = 2,{\text{ }}c = - 1 \cr & c + \frac{1}{a} \cr & = - 1 + \frac{1}{{\left( {\frac{1}{2}} \right)}} \cr & = - 1 + 2 \cr & = 1 \cr & \cr & {\bf{Alternate:}} \cr & \Rightarrow a + \frac{1}{b} = 1 \cr & \Rightarrow a = 1 - \frac{1}{b} = \boxed{\frac{{b - 1}}{b}} \cr & \frac{1}{a} = \frac{b}{{b - 1}} \cr & \Rightarrow b + \frac{1}{c} = 1 \cr & \Rightarrow \frac{1}{c} = 1 - b,\boxed{c = \frac{1}{{1 - b}}} \cr & \therefore c + \frac{1}{a} \cr & = \frac{1}{{1 - b}} + \frac{b}{{b - 1}} \cr & = \frac{1}{{1 - b}} - \frac{b}{{1 - b}} \cr & = \frac{{1 - b}}{{1 - b}} \cr & = 1 \cr} $$

$$\eqalign{ & x + y = 7 \cr & \left( {{\text{cubing both sides}}} \right){\text{ }} \cr & \Rightarrow {\left( {x + y} \right)^3} = {\left( 7 \right)^3} \cr & \Rightarrow {x^3} + {y^3} + 3\left( {x + y} \right)xy = 343 \cr & \Rightarrow {x^3} + {y^3} + 21xy = 343 \cr} $$

$$\eqalign{ & {\text{Given, }}a \otimes b = 2a{\text{ }} \cr & {\text{Where }}a > b \cr & a \otimes b = a + b \cr & {\text{Where }}a < b \cr & a \otimes b = {a^2} \cr & {\text{Where }}a = b \cr & \frac{{\left( {5 \otimes 7} \right) + \left( {4 \otimes 4} \right)}}{{3\left( {5 \otimes 5} \right) - \left( {15 \otimes 11} \right) - 3}}{\text{ }} \cr & {\text{ = }}\frac{{\left( {5 + 7} \right) + {{\left( 4 \right)}^2}}}{{3{{\left( 5 \right)}^2} - \left( {2 \times 15} \right) - 3}} \cr & {\text{ = }}\frac{{12 + 16}}{{75 - 30 - 3}} \cr & {\text{ = }}\frac{{28}}{{42}} \cr & {\text{ = }}\frac{2}{3} \cr} $$

$$\eqalign{ & {\left( {125} \right)^x} = 3125 \cr & \Rightarrow {\left( {{5^3}} \right)^x} = {\left( 5 \right)^5} \cr & \Rightarrow {\left( 5 \right)^{3x}} = {\left( 5 \right)^5} \cr & \Rightarrow 3x = 5 \cr & \Rightarrow x = \frac{5}{3} \cr} $$

$$\eqalign{ & {\text{ }}n + \frac{2}{3}n + \frac{1}{2}n + \frac{1}{7}n = 97 \cr & \Rightarrow \frac{{42n + 28n + 21n + 6n}}{{42}} = 97 \cr & \Rightarrow \frac{{97n}}{{42}} = 97 \cr & \Rightarrow n = 42 \cr} $$

$$\eqalign{ & x - \frac{1}{x} = 4 \cr & \left( {{\text{on squaring }}} \right) \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} - 2 = 16 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} = 18 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 2 - 2 = 18 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 2 = 20 \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^2} = 20 \cr & \Rightarrow x + \frac{1}{x} = \sqrt {20} \cr & \Rightarrow x + \frac{1}{x} = \sqrt {4 \times 5} \cr & \Rightarrow x + \frac{1}{x} = 2\sqrt 5 \cr} $$

$$\eqalign{ & 4{b^2} + \frac{1}{{{b^2}}} = 2 \cr & \Rightarrow {\left( {2b} \right)^2} + {\left( {\frac{1}{b}} \right)^2} + 4 - 4 = 2 \cr & \Rightarrow {\left( {2b + \frac{1}{b}} \right)^2} - 4 = 2 \cr & \Rightarrow {\left( {2b + \frac{1}{b}} \right)^2} = 6 \cr & \Rightarrow 2b + \frac{1}{b} = \sqrt 6 \cr & {\text{Take cube both sides}} \cr & \Rightarrow {\left( {2b + \frac{1}{b}} \right)^3} = {\left( {\sqrt 6 } \right)^3} \cr & \Rightarrow 8{b^3} + \frac{1}{{{b^3}}} + 3 \times 2b \times \frac{1}{b}\left( {2b + \frac{1}{b}} \right) = 6\sqrt 6 \cr & \Rightarrow 8{b^3} + \frac{1}{{{b^3}}} + 6\sqrt 6 = 6\sqrt 6 \cr & \Rightarrow 8{b^3} + \frac{1}{{{b^3}}} = 6\sqrt 6 - 6\sqrt 6 \cr & \Rightarrow 8{b^3} + \frac{1}{{{b^3}}} = 0 \cr} $$

$$\eqalign{ & {\left( {\frac{3}{5}} \right)^3} \times {\left( {\frac{3}{5}} \right)^{ - 6}} = {\left( {\frac{3}{5}} \right)^{2x - 1}} \cr & \Rightarrow {\left( {\frac{3}{5}} \right)^{3 - 6}} = {\left( {\frac{3}{5}} \right)^{2x - 1}} \cr & \Rightarrow - 3 = 2x - 1 \cr & \Rightarrow 2x = - 2 \cr & \Rightarrow x = - 1 \cr} $$

$$\eqalign{ & \frac{{\sqrt {3 + x} + \sqrt {3 - x} }}{{\sqrt {3 + x} - \sqrt {3 - x} }} = \frac{2}{1}{\text{ }} \cr & \Rightarrow \frac{{\sqrt {3 + x} }}{{\sqrt {3 - x} }} = \frac{{2 + 1}}{{2 - 1}} = \frac{3}{1} \cr & \left[ {\frac{{\text{A}}}{{\text{B}}} = \frac{{\text{C}}}{{\text{D}}}} \right] \cr & \left[ {\frac{{{\text{A}} + {\text{B}}}}{{{\text{A}} - {\text{B}}}} = \frac{{{\text{C}} + {\text{D}}}}{{{\text{C}} - {\text{D}}}}} \right] \cr & \Rightarrow \frac{{\sqrt {3 + x} }}{{\sqrt {3 - x} }} = 3{\text{ }} \cr & {\text{Squaring both sides}} \cr & \Rightarrow \frac{{3 + x}}{{3 - x}} = 9 \cr & \Rightarrow 3 + x = 27 - 9x \cr & \Rightarrow 10x = 24 \cr & \Rightarrow x = \frac{{24}}{{10}} \cr & \Rightarrow x = \frac{{12}}{5} \cr} $$