Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

$$\eqalign{ & a + b = 1 \cr & c + d = 1 \cr & a - b = \frac{d}{c} \cr & or\,\,\frac{1}{{a - b}} = \frac{c}{d} \cr & \Rightarrow \frac{{a + b}}{{a - b}} = \frac{c}{d}\left( {\therefore a + b = 1} \right) \cr & {\text{By C & D rule }} \cr & \Rightarrow \frac{{\left( {a + b} \right) + \left( {a - b} \right)}}{{\left( {a + b} \right) - \left( {a - b} \right)}} = \frac{{c + d}}{{c - d}} \cr & \Rightarrow \frac{{2a}}{{2b}} = \frac{{c + d}}{{c - d}} \cr & \Rightarrow \frac{a}{b} = \frac{{c + d}}{{c - d}} \cr} $$
Now multiply & divide by (c + d)
$$\eqalign{ & \frac{a}{b} = \frac{{\left( {c + d} \right)}}{{\left( {c - d} \right)}} \times \frac{{\left( {c + d} \right)}}{{\left( {c + d} \right)}} = \frac{{{{\left( {c + d} \right)}^2}}}{{{c^2} - {d^2}}} \cr & \Rightarrow \frac{a}{b} = \frac{{{{\left( {c + d} \right)}^2}}}{{\left( {{c^2} - {d^2}} \right)}} \cr & \Rightarrow c + d = 1 \cr & \Rightarrow \frac{a}{b} = \frac{1}{{{c^2} - {d^2}}} \cr & \Rightarrow {c^2} - {d^2} = \frac{b}{a} \cr} $$

$$\eqalign{ & x = 3t\,......(i) \cr & y = \frac{1}{2}\left( {t + 1} \right) \cr & x = 2y \cr & \Rightarrow x = 2 \times \frac{1}{2}\left( {t + 1} \right) \cr & \Rightarrow x = t + 1\,......(ii) \cr & \therefore 3t = t + 1 \cr & \left( {{\text{From equation (i) and (ii)}}} \right) \cr & \Rightarrow 2t = 1 \cr & \Rightarrow t = \frac{1}{2} \cr} $$

$$\eqalign{ & {x^2} + \frac{1}{5}x + {a^2} \cr & {{\text{A}}^2} + {\text{2}} \times {\text{AB}} + {{\text{B}}^2} = {\left( {{\text{A}} + {\text{B}}} \right)^2} \cr & {x^2} + 2 \times \frac{1}{{10}} \times x + {a^2} = {\left( {x + \frac{1}{{10}}} \right)^2} \cr & {\text{A}} = x \cr & {\text{B}} = \frac{1}{{10}} \cr & {\text{B}} = a = \frac{1}{{10}} \cr} $$

$$\eqalign{ & \therefore {\text{When}}\left( {x - 1} \right) = 0 \cr & x = 1 \cr & {x^2} + {k_1}x + {k_2}{\text{ }} = 0 \cr & \Rightarrow 1 + {k_1} + {k_2}{\text{ }} = 0 \cr & \Rightarrow {k_1} + {k_2} = - 1\,.......(i) \cr & {\text{When}}\left( {x + 3} \right) = 0 \cr & x = - 3 \cr & 9 - 3{k_1} + {k_2}{\text{ }} = 0 \cr & \Rightarrow - 3{k_1} + {k_2}{\text{ }} = - 9\,.......(ii) \cr & {\text{From equation (i) and (ii)}} \cr & {k_1} = 2,\,\,\,\,\,\,{\text{ }}{k_2} = - 3 \cr} $$

$$\eqalign{ & \frac{{5x}}{{2{x^2} + 5x + 1}} = \frac{1}{3} \cr & \Rightarrow \frac{5}{{\frac{{2{x^2}}}{x} + \frac{{5x}}{x} + \frac{1}{x}}} = \frac{1}{3} \cr & \Rightarrow \frac{5}{{2x + \frac{1}{x} + 5}} = \frac{1}{3} \cr & \Rightarrow 2x + \frac{1}{x} + 5 = 15 \cr & \Rightarrow 2x + \frac{1}{x} = 10 \cr & {\text{Divide by 2 both sides }} \cr & \Rightarrow x + \frac{1}{{2x}} = \frac{{10}}{2} \cr & \Rightarrow 2x + \frac{1}{x} = 5 \cr} $$

$$\eqalign{ & {x^2} + \frac{1}{{{x^2}}} = 83 \cr & {\text{Subtracting 2 from both sides}} \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} - 2 = 83 - 2 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} - 2.x.\frac{1}{x} = 83 - 2 \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^2} = 81 \cr & \Rightarrow x - \frac{1}{x} = 9 \cr & {\text{Take cube on both sides}} \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3{\text{ }}\left( {x - \frac{1}{x}} \right) = 729 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3 \times {\text{9}} = 729 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} = 729 + 27 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} = 756 \cr} $$

$$\eqalign{ & {\left( {a + \frac{1}{a}} \right)^2} = 3 \cr & a + \frac{1}{a} = \sqrt 3 \cr & {\text{Take cube on both sides}} \cr & {\left( {a + \frac{1}{a}} \right)^3} = {\left( {\sqrt 3 } \right)^3} \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3.a.\frac{1}{a}\left( {a + \frac{1}{a}} \right) = 3\sqrt 3 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3\sqrt 3 = 3\sqrt 3 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 0 \cr} $$

$$\eqalign{ & x + y + z = 6 \cr & {x^2} + {y^2} + {z^2} = 20 \cr & \Rightarrow {\left( {x + y + z} \right)^2} = {\left( 6 \right)^2} \cr & \Rightarrow {x^2} + {y^2} + {z^2} + 2\left( {xy + yz + zx} \right) = 36 \cr & \Rightarrow 20 + 2\left( {xy + yz + zx} \right) = 36 \cr & \Rightarrow 2\left( {xy + yz + zx} \right) = 16 \cr & \Rightarrow xy + yz + zx = 8 \cr & \therefore {\text{ }}{x^3} + {y^3} + {z^3} - 3xyz \cr & = \left( {x + y + z} \right)\left( {{\text{ }}{x^2} + {y^2} + {z^2} - xy - zx - yz} \right) \cr & = {x^3} + {y^3} + {z^3} - 3xyz = 6\left( {20 - 8} \right) \cr & = 6\left( {20 - 8} \right) \cr & = 6 \times 12 \cr & = 72 \cr} $$

$$\eqalign{ & x = a - b \cr & y = b - c \cr & z = c - a \cr & \therefore x + y + z \cr & = a - b + b - c + c - a \cr & = 0 \cr & \therefore {x^3} + {y^3} + {z^3} - 3xyz \cr & = \left( {x + y + z} \right)\left( {{\text{ }}{x^2} + {y^2} + {z^2} - xy - zx - yz} \right) \cr & = \left( 0 \right)\left( {{\text{ }}{x^2} + {y^2} + {z^2} - xy - zx - yz} \right) \cr & = 0 \cr} $$

$$\eqalign{ & \left( {x - a} \right)\left( {x - b} \right) = 1{\text{ }} \cr & \Rightarrow \left( {x - a} \right) = \frac{1}{{\left( {x - b} \right)}} \cr & {\text{ }}a - b + 5 = 0 \cr & \Rightarrow {\text{ }}a - b = - 5{\text{ }}\left( {{\text{Given}}} \right) \cr & {\text{Add and subtract }}x \cr & \Rightarrow a - b + x - x = - 5 \cr & \Rightarrow \left( {a - x} \right) + \left( {x - b} \right) = - 5 \cr & \Rightarrow \left( {x - b} \right) - \left( {x - a} \right) = - 5 \cr & \Rightarrow \left( {x - a} \right) - \left( {x - b} \right) = + 5 \cr & \Rightarrow \left( {x - a} \right) - \frac{1}{{\left( {x - a} \right)}} = + 5 \cr & {\text{Taking cube on both sides}} \cr & \Rightarrow {\left( {x - a} \right)^3} - \frac{1}{{{{\left( {x - a} \right)}^3}}} - 3\left( {x - a} \right)\frac{1}{{\left( {x - a} \right)}}\left( {\left( {x - a} \right) - \frac{1}{{\left( {x - a} \right)}}} \right) = {\left( 5 \right)^3} \cr & \Rightarrow {\left( {x - a} \right)^3} - \frac{1}{{{{\left( {x - a} \right)}^3}}} - 3 \times 5 = 125 \cr & \Rightarrow {\left( {x - a} \right)^3} - \frac{1}{{{{\left( {x - a} \right)}^3}}} = 125 + 15 \cr & \Rightarrow {\left( {x - a} \right)^3} - \frac{1}{{{{\left( {x - a} \right)}^3}}} = 140 \cr} $$