Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

$$\eqalign{ & x + \frac{1}{{4x}} = \frac{3}{2} \cr & {\text{Multiply by 2 both sides}} \cr & \therefore 2x + \frac{1}{{2x}} = 3 \cr & {\text{Take cube both sides}} \cr & \Rightarrow {\left( {2x + \frac{1}{{2x}}} \right)^3} = {\left( 3 \right)^3} \cr & \Rightarrow {\text{8}}{x^3}{\text{ + }}\frac{1}{{8{x^3}}} + 3.2x.\frac{1}{{2x}}\left( {2x{\text{ + }}\frac{1}{{2x}}} \right) = 27 \cr & \Rightarrow {\text{8}}{x^3}{\text{ + }}\frac{1}{{8{x^3}}} + 3\left( 3 \right) = 27 \cr & \Rightarrow {\text{8}}{x^3}{\text{ + }}\frac{1}{{8{x^3}}} = 18 \cr} $$

$$\eqalign{ & 3x + \frac{1}{{2x}} = 5 \cr & \Rightarrow {\text{Multiply both sides by }}\frac{2}{3} \cr & \therefore 3x \times \frac{2}{3} + \frac{1}{2}x \times \frac{2}{3} = 5 \times \frac{2}{3} \cr & \Rightarrow 2x + \frac{1}{{3x}} = \frac{{10}}{3} \cr & \therefore {\text{Taking cube on both sides}} \cr & \Rightarrow {\text{8}}{x^3}{\text{ + }}\frac{1}{{27{x^3}}} + 3.2x.\frac{1}{{3x}}\left( {{\text{ 2}}x{\text{ + }}\frac{1}{{3x}}} \right) = {\left( {\frac{{10}}{3}} \right)^3} \cr & \Rightarrow {\text{8}}{x^3}{\text{ + }}\frac{1}{{27{x^3}}} + 2\left( {\frac{{10}}{3}} \right) = \left( {\frac{{1000}}{{27}}} \right) \cr & \Rightarrow {\text{8}}{x^3}{\text{ + }}\frac{1}{{27{x^3}}} = \frac{{1000}}{{27}} - \frac{{20}}{3} \cr & \Rightarrow {\text{8}}{x^3}{\text{ + }}\frac{1}{{27{x^3}}} = \frac{{1000 - 180}}{{27}} \cr & \Rightarrow {\text{8}}{x^3}{\text{ + }}\frac{1}{{27{x^3}}} = \frac{{820}}{{27}} \cr & \Rightarrow {\text{8}}{x^3}{\text{ + }}\frac{1}{{27{x^3}}} = 30\frac{{10}}{{27}} \cr} $$

$$\eqalign{ & x + y = z \cr & x + y - z = 0 \cr & {\text{If }}a + b + c = 0 \cr & {\text{then }}{a^3} + {b^3} + {c^3} - 3abc = 0 \cr & \Rightarrow {x^3} + {y^3} - {z^3} = - 3xyz \cr & \therefore {x^3} + {y^3} - {z^3} + 3xyz = 0 \cr & \Rightarrow 3xyz - 3xyz = 0 \cr} $$

$$\eqalign{ & {a^3} - {b^3} - {c^3} - 3abc = 0 \cr & \therefore a - b - c = 0 \cr & \Rightarrow a = b + c \cr} $$

$$\eqalign{ & a = 2.361 \cr & b = 3.263 \cr & c = 5.624 \cr & a + b - c = 0 \cr & 2.361 + 3.263 - 5.624 = 0 \cr & 0 = 0 \cr & \therefore {a^3} + {b^3} - {c^3} + 3abc \cr & \Rightarrow 0 \cr} $$

$$\eqalign{ & p = 124 \cr & \root 3 \of {p\left( {{p^2} + 3p + 3} \right) + 1} \cr & = \root 3 \of {{p^3} + 3{p^2} + 3p + 1} \cr & = \root 3 \of {{{\left( {p + 1} \right)}^3}} \cr & = \root 3 \of {{{\left( {125} \right)}^3}} \cr & = 125 \cr} $$

$$\eqalign{ & x + \frac{1}{x} = 2 \cr & \left( {{\text{Assume }}x = 1{\text{, so, }}1 + 1 = 2} \right) \cr & {x^{17}}{\text{ + }}\frac{1}{{{x^{19}}}} \cr & = {\left( 1 \right)^{17}} + \frac{1}{{{{\left( 1 \right)}^{19}}}} \cr & = 1 + 1 \cr & = 2 \cr} $$

$$\eqalign{ & x:y = 3:4 \cr & \therefore \frac{{5x - 2y}}{{7x + 2y}} \cr & = \frac{{5 \times 3 - 2 \times 4}}{{7 \times 3 + 2 \times 4}} \cr & = \frac{{15 - 8}}{{21 + 8}} \cr & = \frac{7}{{29}} \cr} $$

$$\eqalign{ & \frac{{2p}}{{{p^2} - 2p + 1}} = \frac{1}{4} \cr & \Rightarrow \frac{{\frac{{2p}}{p}}}{{\frac{{{p^2}}}{p} - \frac{{2p}}{p} + \frac{1}{p}}} = \frac{1}{4} \cr & \Rightarrow \frac{2}{{p + \frac{1}{p} - 2}} = \frac{1}{4} \cr & \Rightarrow p + \frac{1}{p} - 2 = 8 \cr & \Rightarrow p + \frac{1}{p} = 10 \cr} $$

$$\eqalign{ & {\text{Let , }} \cr & a = 1 \cr & b = 180 \cr & c = 1 \cr & \therefore {a^3}b = abc = 180\left( {{\text{Satisfied}}} \right) \cr & \therefore c = 1 \cr} $$