Practice MCQ Questions and Answer on Algebra

Solution:

 $$\eqalign{ & a + \frac{1}{b} = b + \frac{1}{c} = c + \frac{1}{a} \cr & {\text{Put , }}a = \frac{1}{2},b = 2,c = - 1 \cr & \Rightarrow \frac{1}{2} + \frac{1}{2} = 2 - 1 = - 1 + 2 \cr & \Rightarrow 1 = 1 = 1 \cr & abc = \frac{1}{2} \times 2 \times - 1 \cr & \boxed{abc = - 1} \cr & {\text{Again put}} \cr & a = - \frac{1}{2},b = - 2,c = 1 \cr & \Rightarrow a + \frac{1}{b} = b + \frac{1}{c} = c + \frac{1}{a} \cr & \Rightarrow - \frac{1}{2} - \frac{1}{2} = - 2 + 1 = 1 - 2 \cr & \Rightarrow - 1 = - 1 = - 1 \cr & {\text{Equation satisfied}} \cr & \Rightarrow abc = - \frac{1}{2} \times - 2 \times 1 \cr & \boxed{abc = + 1} \cr & {\text{So, }}abc{\text{ can be }} - 1{\text{ and }} + 1 \cr} $$

Solution:

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

Solution:

 $$\eqalign{ & x:y \cr & 7:3{\text{ }} \cr & \therefore {\text{ }}\frac{{xy + {y^2}}}{{{x^2} - {y^2}}} \cr & = \frac{{21 + 9}}{{49 - 9}} \cr & = \frac{{30}}{{40}} \cr & = \frac{3}{4} \cr} $$

Solution:

 $$\eqalign{ & x + y + z = 0,\,\frac{{3{y^2} + {x^2} + {z^2}}}{{2{y^2} - xz}} = ? \cr & {\text{Put }}x = 1,\,y = 1,\,z = - 2 \cr & \Rightarrow \frac{{3{{\left( y \right)}^2} + {x^2} + {z^2}}}{{2{y^2} - xz}} \cr & \Rightarrow \frac{{3{{\left( 1 \right)}^2} + {{\left( 1 \right)}^2} + {{\left( { - 2} \right)}^2}}}{{2{{\left( 1 \right)}^2} - \left( {1 \times \left( { - 2} \right)} \right)}} \cr & \Rightarrow \frac{{4 + 4}}{{2 + 2}} \cr & \Rightarrow \frac{8}{4} \cr & \Rightarrow 2 \cr} $$

Solution:

 $$\eqalign{ & x + \frac{1}{x} = 5 \cr & \therefore \frac{{2x}}{{3{x^2} - 5x + 3}}\,\left( {{\text{Divide by }}x} \right) \cr & = \frac{{\frac{{2x}}{x}}}{{\frac{{3{x^2}}}{x} - \frac{{5x}}{x} + \frac{3}{x}}} \cr & = \frac{2}{{3x + \frac{3}{x} - 5}} \cr & = \frac{2}{{3\left( {x + \frac{1}{x}} \right) - 5}} \cr & = \frac{2}{{3 \times 5 - 5}} \cr & = \frac{2}{{10}} \cr & = \frac{1}{5} \cr} $$

Solution:

 $$\eqalign{ & \Rightarrow x = 3 + \sqrt 8 \cr & \Rightarrow {x^2} = 9 + 8 + 2 \times 3\sqrt 8 \cr & \Rightarrow {x^2} = 17 + 6\sqrt 8 \cr & \Rightarrow \frac{1}{{{x^2}}} = 17 - 6\sqrt 8 \cr & \therefore {x^2} + \frac{1}{{{x^2}}} \cr & = 17 + 6\sqrt 8 + 17 - 6\sqrt 8 \cr & = 34 \cr} $$

Solution:

 $$\eqalign{ & \frac{{{y^2}}}{{2xz}} - \frac{{{x^2}}}{{2yz}} - \frac{{{z^2}}}{{2xy}} \cr & \frac{{{y^3} - {x^3} - {z^3}}}{{2xyz}}\,......\,\left( 1 \right) \cr & x - y + z = 0 \cr & x + z = y \cr & {\text{Cubing both side}} \cr & {\left( {x + z} \right)^3} = {y^3} \cr & {x^3} + {z^3} + 3\left( {x + z} \right)\left( x \right)\left( z \right) = {y^3} \cr & {x^3} + {z^3} + 3\left( y \right)\left( x \right)\left( z \right) = {y^3} \cr & 3xyz = {y^3} - {x^3} - {z^3} \cr & {\text{Put in equation }}\left( 1 \right) \cr & \frac{{3xyz}}{{2xyz}} = \boxed{\frac{3}{2}} \cr} $$

Solution:

 $$\eqalign{ & \frac{{\sqrt 7 - 2}}{{\sqrt 7 + 2}} = a\sqrt 7 + b \cr & {\text{L}}{\text{.H}}{\text{.S }}\frac{{\sqrt 7 - 2}}{{\sqrt 7 + 2}} \times \frac{{\sqrt 7 - 2}}{{\sqrt 7 - 2}} \cr & {\text{ }}\left( {{\text{Rationalisation}}} \right) \cr & = \frac{{{{\left( {\sqrt 7 - 2} \right)}^2}}}{{{{\left( {\sqrt 7 } \right)}^2} - \left( 4 \right)}} \cr & = \frac{{7 + 4 - 4\sqrt 7 }}{{7 - 4}} \cr & = \frac{{11 - 4\sqrt 7 }}{3} \cr & = \frac{{11}}{3} - \frac{4}{3}\sqrt 7 \cr & = - \frac{4}{3}\sqrt 7 + \frac{{11}}{3} \cr & = a\sqrt 7 + b{\text{ }}\left( {{\text{ R}}{\text{.H}}{\text{.S}}{\text{.}}} \right) \cr} $$ (Compare the coefficients of $$\sqrt 7 $$ and constant term) $$\eqalign{ & {\text{a}} = - \frac{4}{3} \cr & b = \frac{{11}}{3} \cr} $$

Solution:

 $$\frac{{{x^2}}}{{{y^2}}} + tx + \frac{{{y^2}}}{4}\left( {{\text{ Given}}} \right)$$ To make it a perfect square it should be in the form $$\eqalign{ & {{\text{A}}^2} \pm 2{\text{AB}} + {{\text{B}}^2} = {\left( {{\text{A}} \pm {\text{B}}} \right)^2} \cr & = {\left( {\frac{x}{y}} \right)^2} \pm tx + {\left( {\frac{y}{2}} \right)^2} \cr & = {{\text{A}}^2} \pm 2{\text{AB}} + {{\text{B}}^2} \cr & {\text{A}} = \frac{x}{y}{\text{, B}} = \frac{y}{2}\,\,\& \,\,{\text{2AB}} = tx \cr & {\text{So, }}tx = \pm 2 \times \frac{x}{y} \times \frac{y}{2} \cr & \Rightarrow tx = \pm x \cr & \Rightarrow t = \pm 1 \cr} $$

Solution:

 $$\eqalign{ & {\text{Let the numbers are a, b}} \cr & {a^2} + {b^2} = 41 \cr & a + b = 9 \cr & \Rightarrow {\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab \cr & \Rightarrow {9^2} = 41 + 2ab \cr & \Rightarrow 81 - 41 = 2ab \cr & \Rightarrow ab = 20 \cr & {\text{Take }} \cr & a = 5 \cr & b = 4 \cr & \Rightarrow {a^3} + {b^3} = {5^3} + {4^3} \cr & \Rightarrow {a^3} + {b^3} = 125 + 64 \cr & \Rightarrow {a^3} + {b^3} = 189 \cr} $$