Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

$$\eqalign{ & {\text{Let numbers are }} \cr & x = 2r,{\text{ }}y = 3r,{\text{ }}z = 5r \cr & {\text{Sum, }}x + y + z = 2r + 3r + 5r \cr & \Rightarrow x + y + z = 10r \cr & \Rightarrow x + y + z = 80 \cr & r = 8,x = 16,y = 24,z = 40 \cr & {\text{Then, }}z = ax - 8 \cr & \Rightarrow 40 = a \times 16 - 8 \cr & \Rightarrow a = 3 \cr} $$

$$\eqalign{ & \left( {x - 2} \right)\left( {x - p} \right) = {x^2} - ax + 6 \cr & {x^2} - \left( {2 + p} \right)x + 2p = {x^2} - ax + 6 \cr & {\text{Comparision the cofficients}} \cr & 2 + p = a \cr & 2p = 6 \cr & \Leftrightarrow p = 3 \cr & 2 + 3 = a \cr & \Leftrightarrow a = 5 \cr & {\text{Then , }} \cr & p = 3,{\text{ }}a = 5 \cr & a - p = 5 - 3 \cr & \Leftrightarrow a - p = 2 \cr} $$

$$\eqalign{ & {\text{ }}x = \sqrt a + \frac{1}{{\sqrt a }} \cr & {\text{ }}y = \sqrt a - \frac{1}{{\sqrt a }} \cr & {\text{Put }}a = 4 \cr & x = 2 + \frac{1}{2} = \frac{5}{2} \cr & y = 2 - \frac{1}{2} = \frac{3}{2} \cr & {\text{Then, }}{x^4} + {y^4} - 2{x^2}{y^2}{\text{ }} \cr & = {\left( {{x^2} - {y^2}} \right)^2} \cr & = {\left( {\frac{{25}}{4} - \frac{9}{4}} \right)^2} \cr & = {\text{ 16}} \cr} $$

$$\eqalign{ & x = \root 3 \of {{x^2} + 11} - 2 \cr & \Rightarrow x + 2 = \root 3 \of {{x^2} + 11} \cr & \Rightarrow {\text{Taking cube on both side}} \cr & \Rightarrow {\left( {x + 2} \right)^3} = {x^2} + 11 \cr & \Rightarrow {x^3} + 8 + 6x\left( {x + 2} \right) = {x^2} + 11 \cr & \Rightarrow {x^3} + 8 + 6{x^2} + 12x = {x^2} + 11 \cr & \Rightarrow {x^3} + 5{x^2} + 12x = 3 \cr} $$

$$\eqalign{ & {p^2} + \frac{1}{{{p^2}}} = 47 \cr & {\text{On adding 2 both side}} \cr & {p^2} + \frac{1}{{{p^2}}} + 2 = 47 + 2 \cr & \Rightarrow {\left( {p + \frac{1}{p}} \right)^2} = 49 \cr & \Rightarrow \left( {p + \frac{1}{p}} \right) = 7 \cr & \Rightarrow p + \frac{1}{p} = 7 \cr} $$

$$\eqalign{ & {\text{Let,}} \cr & a = {\left( {x - y} \right)^2}, \cr & b = \left( {{x^2} - {y^2}} \right){\text{and}} \cr & c\,\,{\text{be}}\,{\text{the}}\,{\text{third}}\,{\text{proportional}} \cr & {\text{Therefore}}\,\,\,a:b::b:c \cr & i.e.\,\,\,c = \frac{{{b^2}}}{a} \cr & \Rightarrow c = \frac{{{{\left( {{x^2} - {y^2}} \right)}^2}}}{{\left( {x - y} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \frac{{{{\left( {x - y} \right)}^2}{{\left( {x + y} \right)}^2}}}{{\left( {x - y} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \left( {x - y} \right){\left( {x + y} \right)^2} \cr} $$

$$\eqalign{ & x + \sqrt 5 = 5 + \sqrt y \cr & {\text{Put , }}x = 5{\text{ and }}y = 5 \cr & 5 + \sqrt 5 = 5 + \sqrt 5 \cr & {\text{L}}{\text{.H}}{\text{.S}} = {\text{R}}{\text{.H}}{\text{.S}} \cr & \frac{{\sqrt x + y}}{{x + \sqrt y }} \cr & = \frac{{\sqrt 5 + 5}}{{5 + \sqrt 5 }} \cr & = 1 \cr} $$

This is possible only when
$$\eqalign{ & {\left( {x - 3} \right)^2} = 0 \cr & x = 3 \cr & {\left( {y - 4} \right)^2} = 0 \cr & y = 4 \cr & {\left( {z - 5} \right)^2} = 0 \cr & z = 5 \cr & {\text{Then, }}\left( {x + y + z} \right) \cr & = 3 + 4 + 5 \cr & = 12 \cr} $$

$$\eqalign{ & 2x + \frac{1}{{3x}} = 5 \cr & 6{x^2}{\text{ + 1 = 15x}}\,......{\text{(i)}} \cr & {\text{Now,}}\frac{{5x}}{{6{x^2} + 20x + 1}} \cr & = \frac{{5x}}{{6{x^2} + 1 + 20x}} \cr & \left[ {{\text{From equation (i)}}} \right] \cr & = \frac{{5x}}{{15x + 20x}}{\text{ }} \cr & = \frac{{5x}}{{35x}} \cr & = \frac{1}{7} \cr} $$

$$\eqalign{ & a + b = 10{\text{ and }}ab = 21 \cr & \left( {a + b} \right) = 10 \cr & \Rightarrow {a^2} + {b^2} + 2ab = 100 \cr & \Rightarrow {a^2} + {b^2} = 100 - 2ab \cr & \Rightarrow {a^2} + {b^2} = 100 - 2 \times 21 \cr & \Rightarrow {a^2} + {b^2} = 100 - 42 \cr & {a^2} + {b^2} = 58\,.........(i) \cr & {\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab \cr & {\left( {a - b} \right)^2} = 58 - 2 \times 21 \cr & \left[ {{\text{from equation (i)}}} \right] \cr & = {\text{58}} - {\text{42}} \cr & {\left( {a - b} \right)^2} = 16 \cr} $$