Practice MCQ Questions and Answer on Algebra

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:

 $$\eqalign{ & x \propto \frac{1}{{{y^2}}} \cr & \left( {{\text{Inversely proportional}}} \right) \cr & x = \frac{k}{{{y^2}}} \cr & \left( {{\text{Given}}} \right), \cr & \left( {y = 2} \right){\text{ for }}\left( {x = 1} \right) \cr & \therefore 1 = \frac{k}{{{{\left( 2 \right)}^2}}} \cr & \Rightarrow 1 = \frac{k}{4} \cr & \Rightarrow k = 4 \cr & \therefore {\text{For }}y = 6 \cr & x = \frac{4}{{{{\left( 6 \right)}^2}}} \cr & x = \frac{4}{{36}} \cr & x = \frac{1}{9} \cr} $$

Solution:

 $$\eqalign{ & {x^2} + {y^2} + 1 = 2x \cr & {x^2} - 2x + 1 + {y^2} = 0 \cr & {\left( {x - 1} \right)^2} + {y^2} = 0 \cr & {\text{If }}{{\text{A}}^2} + {{\text{B}}^2} = 0 \cr} $$ [As powers are even it can possible only when A = 0 & B = 0] $$\eqalign{ & \therefore x - 1 = 0 \cr & x = 1 \cr & y = 0 \cr & \therefore {x^3} + {y^5} \cr & = 1 + 0 \cr & = 1 \cr} $$

Solution:

 $$\eqalign{ & x + \frac{{16}}{x} = 8 \cr & {\text{at }}x = 4{\text{ it satisfy}} \cr & {x^2} + \frac{{32}}{{{x^2}}} = {4^2} + \frac{{32}}{{{4^2}}} \cr & = 16 + 2 \cr & = 18 \cr} $$

Solution:

 $$\eqalign{ & {\text{Put }}\left( {x + y + z} \right) = 10 \cr & x = 2 \cr & y = 3 \cr & z = 5 \cr & x\left( {x + y + z} \right) = 20 \cr & \Leftrightarrow 2\left( {10} \right) = 20 \cr & \Leftrightarrow 20 = 20 \cr & {\text{Similarly other will satisfied,}} \cr & {\text{So, value of 2}}\left( {x + y + z} \right) \cr & = 2\left( {10} \right) \cr & = 20 \cr} $$

Solution:

 $$\eqalign{ & {\text{Given, }}x + \frac{1}{x} = \sqrt {13} {\text{ }} \cr & {\text{then ,}}\frac{{3x}}{{\left( {{x^2} - 1} \right)}} \cr & = \frac{3}{{x - \frac{1}{x}}}\,.............(i) \cr & {\text{Now, }}x + \frac{1}{x} = \sqrt {13} \cr & {\text{On squaring both side}} \cr & = {x^2} + \frac{1}{{{x^2}}} \cr & = 13 - 2 \cr & = 11 \cr & = {x^2} + \frac{1}{{{x^2}}} - 2 \cr & = 11 - 2 \cr & = 9 \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^2} = 9 \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^2} = {3^2} \cr & \Rightarrow x - \frac{1}{x} = 3 \cr & {\text{Put this value in equation (i)}} \cr & \Rightarrow \frac{3}{{x - \frac{1}{x}}} \cr & \Rightarrow \frac{3}{3} \cr & \Rightarrow 1 \cr} $$

Solution:

 $$\eqalign{ & x = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}}{\text{ and y}} = \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}} \cr & \therefore x = \frac{1}{y} \cr & \Leftrightarrow xy = 1 \cr & x + y = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}}{\text{ + }}\frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}} \cr & \Rightarrow x + y = \frac{{5 + 1 + 2\sqrt 5 + 5 + 1 - 2\sqrt 5 }}{{5 - 1}} \cr & \Rightarrow x + y = \frac{{12}}{4} \cr & \Rightarrow x + y = 3 \cr & \Rightarrow x + \frac{1}{x} = 3 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} = {\left( 3 \right)^2} - 2 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} = 7 \cr & {\text{Now,}}\frac{{{x^2} + xy + {y^2}}}{{{x^2} - xy + {y^2}}} \cr & = \frac{{{x^2} + {y^2} + xy}}{{{x^2} + {y^2} - xy}} \cr & = \frac{{7 + 1}}{{7 - 1}} \cr & = \frac{8}{6} \cr & = \frac{4}{3} \cr} $$

Solution:

 $$\eqalign{ & m + \frac{1}{{m - 2}} = 4 \cr & \Rightarrow m - 2 + \frac{1}{{m - 2}} = 2 \cr & \left( {{\text{Squaring the both sides}}} \right) \cr} $$ $$ \Rightarrow {\left( {m - 2} \right)^2}{\text{ + }}\frac{1}{{{{\left( {m - 2} \right)}^2}}} + 2 \times $$      $$\left( {m - 2} \right) \times $$  $$\frac{1}{{\left( {m - 2} \right)}}$$   $$ = 4$$ $$ \Rightarrow {\left( {m - 2} \right)^2}{\text{ + }}\frac{1}{{{{\left( {m - 2} \right)}^2}}} = 2$$

Solution:

 $$\eqalign{ & \frac{{144}}{{0.144}} = \frac{{14.4}}{x} \cr & \Rightarrow \frac{{144 \times 1000}}{{144}} = \frac{{144}}{{x \times 10}} \cr & \Rightarrow 1000 = \frac{{144}}{{10x}} \cr & \Rightarrow x = \frac{{144}}{{1000 \times 10}} \cr & \Rightarrow x = \frac{{144}}{{10000}} \cr & \Rightarrow x = 0.0144 \cr} $$

Solution:

 $$\eqalign{ & \sqrt {1 + \frac{x}{{961}}} = \frac{{32}}{{31}} \cr & \left( {{\text{Squaring both sides}}} \right) \cr & \Rightarrow 1 + \frac{x}{{961}} = \frac{{1024}}{{961}} \cr & \Rightarrow \frac{{961 + x}}{{961}} = \frac{{1024}}{{961}} \cr & \Rightarrow x = 1024 - 961 \cr & \Rightarrow x = 63 \cr} $$