Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

$$\eqalign{ & a\left( {x + y} \right) = b\left( {x - y} \right) = 2ab \cr & x + y = 2b \cr & {\text{On squraing both side}} \cr & \Rightarrow {x^2} + {y^2} + 2xy = 4{b^2}\,.....(i) \cr & \Rightarrow b\left( {x - y} \right) = 2ab \cr & \Rightarrow x - y = 2a \cr & {\text{On squraing both side}} \cr & \Rightarrow {x^2} + {y^2} - 2xy = 4{a^2}\,.....(ii) \cr & {\text{Add equation (i) and (ii)}} \cr & \Rightarrow 2\left( {{x^2} + {y^2}} \right) = 4\left( {{a^2} + {b^2}} \right) \cr} $$

$$\eqalign{ & 2x + \frac{1}{{4x}} = 1 \cr & {\text{Dividing by 2 both side}} \cr & \Rightarrow x + \frac{1}{{8x}} = \frac{1}{2} \cr & {\text{Squaring both side }} \cr & \Rightarrow {x^2} + \frac{1}{{64{x^2}}} + 2 \times x \times \frac{1}{{8x}} = \frac{1}{4} \cr & \Rightarrow {x^2} + \frac{1}{{64{x^2}}} + \frac{1}{4} = \frac{1}{4} \cr & \Rightarrow {x^2} + \frac{1}{{64{x^2}}} = \frac{1}{4} - \frac{1}{4} \cr & \Rightarrow {x^2} + \frac{1}{{64{x^2}}} = 0 \cr} $$

$$\eqalign{ & \sqrt x - \sqrt y = 1\,......(i) \cr & \sqrt x + \sqrt y = 17\,.....(ii) \cr & {\text{From equation (i) and (ii)}} \cr & \sqrt x = 9,\sqrt y = 8 \cr & {\text{So, }}\sqrt {xy} = 9 \times 8 = 72 \cr} $$

$$\eqalign{ & \because x = \sqrt 3 + \frac{1}{{\sqrt 3 }} \cr & \Rightarrow x = \frac{{3 + 1}}{{\sqrt 3 }} \cr & \Rightarrow x = \frac{4}{{\sqrt 3 }} \cr & {\text{So, }}\left( {x - \frac{{\sqrt {126} }}{{\sqrt {42} }}} \right)\left( {x - \frac{1}{{x - \frac{{2\sqrt 3 }}{3}}}} \right) \cr & = \left( {\frac{4}{{\sqrt 3 }} - \frac{{\sqrt {126} }}{{\sqrt {42} }}} \right)\left( {\frac{4}{{\sqrt 3 }} - \frac{1}{{\frac{4}{{\sqrt 3 }} - \frac{2}{{\sqrt 3 }}}}} \right) \cr & = \left( {\frac{{4\sqrt {42} - \sqrt {126} \times \sqrt 3 }}{{\sqrt {3 \times } \sqrt {42} }}} \right)\left( {\frac{4}{{\sqrt 3 }} - \frac{1}{{\frac{2}{{\sqrt 3 }}}}} \right) \cr & = \frac{{4\sqrt {42} - 3\sqrt {42} }}{{\sqrt {3 \times } \sqrt {42} }}\left( {\frac{4}{{\sqrt 3 }} - \frac{{\sqrt 3 }}{2}} \right) \cr & = \left( {\frac{{\sqrt {42} }}{{\sqrt {3 \times } \sqrt {42} }}} \right)\left( {\frac{{8 - 3}}{{2\sqrt 3 }}} \right) \cr & = \frac{1}{{\sqrt 3 }} \times \frac{5}{{2\sqrt 3 }} \cr & = \frac{5}{6} \cr} $$

$$\eqalign{ & \frac{{2 + a}}{a} + \frac{{2 + b}}{b} + \frac{{2 + c}}{c} = 4 \cr & \Rightarrow \frac{{2bc + abc + 2ac + abc + 2ab + abc}}{{abc}} = 4 \cr & \Rightarrow \frac{{2\left( {bc + ab + ca} \right)}}{{abc}} + \frac{{3abc}}{{abc}} = 4 \cr & \Rightarrow \frac{{2\left( {bc + ab + ca} \right)}}{{abc}} + 3 = 4 \cr & \Rightarrow \boxed{\frac{{ab + bc + ca}}{{abc}} = \frac{1}{2}} \cr} $$

$$\eqalign{ & x + \frac{1}{x} = 5{\text{ then, }}\frac{{5x}}{{{x^2} + 5x + 1}} \cr & \Rightarrow \frac{5}{{x + 5 + \frac{1}{x}}} \cr & \Rightarrow \frac{5}{{x + \frac{1}{x} + 5}} \cr & \Rightarrow \frac{5}{{5 + 5}} \cr & \Rightarrow \frac{1}{2} \cr} $$

$$\eqalign{ & \frac{1}{a}\left( {{a^2} + 1} \right) = 3 \cr & \Rightarrow a + \frac{1}{a} = 3 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3.a.\frac{1}{a}\left( {a + \frac{1}{a}} \right) = {3^3} \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3\left( 3 \right) = {3^3} \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 27 - 9 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 18 \cr & \Rightarrow \frac{{{a^6} + 1}}{{{a^3}}} = 18 \cr} $$

$$\eqalign{ & {\left( {x - 5} \right)^2} + {\left( {y - 2} \right)^2} + {\left( {z - 9} \right)^2} = 0 \cr & {\text{It is possible only when }} \cr & x - 5 = 0 \cr & x = 5 \cr & y - 2 = 0 \cr & y = 2 \cr & z - 9 = 0 \cr & z = 9 \cr & \therefore x + y - z \cr & = 5 + 2 - 9 \cr & = 7 - 9 \cr & = - 2 \cr} $$

$$\eqalign{ & \therefore {a^3} + {b^3} + {c^3} - 3abc \cr & = \frac{1}{2}\left( {a + b + c} \right)\left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right] \cr & \therefore \frac{{{x^3} + {y^3} + {z^3} - 3xyz}}{{x - y + z}} \cr} $$
$$ = \frac{{\frac{1}{2}\left( {x + y + z} \right)\left[ {{{\left( {x - y} \right)}^2} + {{\left( {y - z} \right)}^2} + {{\left( {z - x} \right)}^2}} \right]}}{{x - y + z}}$$
$$\eqalign{ & = \frac{{\frac{1}{2}\left( {999 + 1000 + 1001} \right)\left( {1 + 1 + 4} \right)}}{{999 - 1000 + 1001}} \cr & = \frac{{\frac{1}{2} \times 6 \times 3000}}{{1000}} \cr & = 9 \cr} $$

$$\eqalign{ & \frac{a}{{1 - 2a}} + \frac{b}{{1 - 2b}} + \frac{c}{{1 - 2c}} = \frac{1}{2} \cr & {\text{Multiply by 2 both side}} \cr & \Rightarrow \frac{{2a}}{{1 - 2a}} + \frac{{2b}}{{1 - 2b}} + \frac{{2c}}{{1 - 2c}} = 1 \cr & {\text{Adding 3 both side}} \cr} $$
$$ \Rightarrow 1 + \frac{{2a}}{{1 - 2a}} + 1 + \frac{{2b}}{{1 - 2b}} + 1 + $$       $$\frac{{2c}}{{1 - 2c}} = $$  $$1 + 3$$
$$ \Rightarrow \frac{1}{{1 - 2a}} + \frac{1}{{1 - 2b}} + \frac{1}{{1 - 2c}} = 4$$