Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

$$\eqalign{ & a = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}} \cr & b = \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}}{\text{ }} \cr & \therefore a = \frac{1}{b} \cr & a + b = a + \frac{1}{a} \cr & \Rightarrow \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}}{\text{ + }}\frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}} \cr & \Rightarrow \frac{{5 + 1 + 2\sqrt 5 + 5 + 1 - 2\sqrt 5 }}{{{{\left( {\sqrt 5 } \right)}^2} - {{\left( 1 \right)}^2}}} \cr & \Rightarrow \frac{{6 + 2\sqrt 5 + 6 - 2\sqrt 5 }}{{5 - 1}} \cr & \Rightarrow \frac{{12}}{4} \cr & \Rightarrow 3 \cr & \therefore \frac{{{a^2} + ab + {b^2}}}{{{a^2} - ab + {b^2}}} \cr & \Rightarrow \frac{{{a^2} + \frac{1}{{{a^2}}} + ab}}{{{a^2} + \frac{1}{{{a^2}}} - ab}} \cr & \Rightarrow a + \frac{1}{a} = 3 \cr & \Rightarrow {a^2} + \frac{1}{{{a^2}}} \cr & \Rightarrow 9 - 2 \cr & \Rightarrow 7\left( {ab = 1} \right) \cr & \therefore \frac{{{a^2} + \frac{1}{{{a^2}}} + ab}}{{{a^2} + \frac{1}{{{a^2}}} - ab}} \cr & = \frac{{7 + 1}}{{7 - 1}} \cr & = \frac{8}{6} \cr & = \frac{4}{3} \cr} $$

$$\eqalign{ & a = 4.36 \cr & b = 2.39 \cr & c = 1.97 \cr & {\text{ }}a - b - c \cr & = 4.36 - 2.39 - 1.97 \cr & = 0 \cr & {\text{ }}{{\text{a}}^3} - {b^3} - {c^3} - 3abc \cr & = \frac{1}{2}\left( {a - b - c} \right)\left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right] \cr & = 0 \cr} $$

$$\eqalign{ & \frac{{3a + 5b}}{{3a - 5b}} = 5 \cr & \Rightarrow 3a + 5b = 15a - 25b \cr & \Rightarrow 12a = 30b \cr & \Rightarrow 2a = 5b \cr & a:b \cr & 5:2 \cr} $$

$$\eqalign{ & x:y = 3:4 \cr & \frac{{\left( {7x + 3y} \right)}}{{\left( {7x - 3y} \right)}} \cr & = \frac{y}{y}\left( {\frac{{7\frac{x}{y} + 3}}{{7\frac{x}{y} - 3}}} \right) \cr & = \frac{{7 \times \frac{3}{4} + 3}}{{7 \times \frac{3}{4} - 3}} \cr & = \frac{{\frac{{21}}{4} + 3}}{{\frac{{21}}{4} - 3}} \cr & = \frac{{\frac{{21 + 12}}{4}}}{{\frac{{21 - 12}}{4}}} \cr & = \frac{{11}}{3} \,\,or\,\, 11 : 3 \cr} $$

$$\eqalign{ & x + \frac{1}{4}\sqrt x + {a^2} \cr & = {\left( {\sqrt x } \right)^2} + 2 \times \frac{1}{8} \times \sqrt x + {a^2} \cr & \left[ {\left( {{{\text{A}}^2} + {\text{2AB}} + {{\text{B}}^2}} \right) = {{\left( {{\text{A}} + {\text{B}}} \right)}^2}} \right] \cr & {\text{Here, A}} = \sqrt x {\text{ and }} \cr & {\text{B}} = a \cr & {\text{B}} = \frac{1}{8} \cr & \therefore a = \frac{1}{8} \cr} $$

$$\eqalign{ & {\text{ }}{x^{\frac{1}{3}}} = {y^{\frac{1}{4}}}, \cr & \Rightarrow {\text{LCM of 3, 4}} = 12{\text{ }} \cr & \therefore {\text{ }}{\left( {{x^{\frac{1}{3}}}} \right)^{12}} = {\left( {{y^{\frac{1}{4}}}} \right)^{12}} \cr & \Rightarrow {x^4} = {y^3} \cr & {\text{take power '5' on both sides}} \cr & \Rightarrow {\left( {{x^4}} \right)^5} = {\left( {{y^3}} \right)^5} \cr & \Rightarrow {x^{20}} = {y^{15}} \cr} $$

$$\eqalign{ & x = \frac{{\sqrt 3 }}{2} \cr & {\text{or }}1 + x = 1 + \frac{{\sqrt 3 }}{2} \cr & \Rightarrow 1 + x = \frac{{2 + \sqrt 3 }}{2} \cr & \Rightarrow 1 + x = \frac{{2\left( {2 + \sqrt 3 } \right)}}{{2 \times 2}} \cr & \left( {{\text{Divided and multiply by 2}}} \right) \cr & \Rightarrow 1 + x = \frac{{4 + 2\sqrt 3 }}{4} \cr & \Rightarrow 1 + x = \frac{{1 + 3 + 2\sqrt 3 }}{4} \cr & \Rightarrow 1 + x = \frac{{4 + 2\sqrt 3 }}{4} \cr & \Rightarrow 1 + x = \frac{{{{\left( 1 \right)}^2} + {{\left( {\sqrt 3 } \right)}^2} + 2.1.\sqrt 3 }}{4} \cr & \Rightarrow 1 + x = \frac{{{{\left( {1 + \sqrt 3 } \right)}^2}}}{4} \cr & \therefore \sqrt {1 + x} = \frac{{1 + \sqrt 3 }}{2} \cr & \cr & {\bf{Similarly:}} \cr & \sqrt {1 - x} \cr & = \frac{{\sqrt 3 - 1}}{2} \cr & \therefore \frac{{\sqrt {1 + x} }}{{1 + \sqrt {1 + x} }}{\text{ + }}\frac{{\sqrt {1 - x} }}{{1 - \sqrt {1 - x} }} \cr & = \frac{{\frac{{1 + \sqrt 3 }}{2}}}{{1 + \frac{{1 + \sqrt 3 }}{2}}} + \frac{{\frac{{\sqrt 3 - 1}}{2}}}{{1 - \frac{{\sqrt 3 - 1}}{2}}} \cr & = \frac{{1 + \sqrt 3 }}{{3 + \sqrt 3 }} + \frac{{\sqrt 3 - 1}}{{3 - \sqrt 3 }} \cr & = \frac{{1 + \sqrt 3 }}{{\sqrt 3 \left( {\sqrt 3 + 1} \right)}} + \frac{{1 - \sqrt 3 }}{{\sqrt 3 \left( {\sqrt 3 - 1} \right)}} \cr & = \frac{1}{{\sqrt 3 }} + \frac{1}{{\sqrt 3 }} \cr & = \frac{2}{{\sqrt 3 }} \cr} $$

$$\eqalign{ & \Leftrightarrow {x^2} - 4x - 1 = 0 \cr & \Leftrightarrow {x^2} - 1 = 4x \cr & \left( {{\text{divide }}x{\text{ both sides}}} \right) \cr & \Leftrightarrow x - \frac{1}{x} = 4 \cr & \Leftrightarrow {x^2} + \frac{1}{{{x^2}}} - 2 = 16 \cr & \Leftrightarrow {x^2} + \frac{1}{{{x^2}}} = 18 \cr} $$

$$\left( {x + \frac{1}{x}} \right)$$ $$\left( {x - \frac{1}{x}} \right)$$ $$\left( {{x^2} + \frac{1}{{{x^2}}} - 1} \right)$$  $$\left( {{x^2} + \frac{1}{{{x^2}}} + 1} \right)$$
$${\text{ = }}\left( {x + \frac{1}{x}} \right)$$ $$\left( {{x^2} + \frac{1}{{{x^2}}} - 1} \right)$$  $$\left( {x - \frac{1}{x}} \right)$$ $$\left( {{x^2} + \frac{1}{{{x^2}}} + 1} \right)$$
$$\eqalign{ & \because \left( {{\text{A}} + {\text{B}}} \right)\left( {{{\text{A}}^2} - {\text{AB}} + {{\text{B}}^2}} \right) = {{\text{A}}^3} + {{\text{B}}^3} \cr & \because \left( {{\text{A}} - {\text{B}}} \right)\left( {{{\text{A}}^2} + {\text{AB}} + {{\text{B}}^2}} \right) = {{\text{A}}^3} - {{\text{B}}^3} \cr & = \left( {{x^3} + \frac{1}{{{x^3}}}} \right)\left( {{x^3} - \frac{1}{{{x^3}}}} \right) \cr & = \boxed{{x^6} - \frac{1}{{{x^6}}}} \cr} $$

$$\eqalign{ & \Rightarrow {a^{2x + 2}} = 1 \cr & \Rightarrow {a^{2x + 2}} = {a^0} \cr & \Rightarrow 2x + 2 = 0 \cr & \Rightarrow x = - \frac{2}{2} \cr & \Rightarrow x = - 1 \cr} $$