Practice MCQ Questions and Answer on Algebra

Solution:

 $$\eqalign{ & {\text{A}}:{\text{B}} = \frac{1}{2}:\frac{3}{8} \cr & \Rightarrow {\text{A}}:{\text{B}} = 8:6 \cr & \Rightarrow {\text{A}}:{\text{B}} = 4:3 \cr & {\text{B}}:{\text{C}} = \frac{1}{3}:\frac{5}{9} \cr & \Rightarrow {\text{B}}:{\text{C}} = 9:15 \cr & \Rightarrow {\text{B}}:{\text{C}} = 3:5 \cr & {\text{C}}:{\text{D}} = \frac{5}{6}:\frac{3}{4} \cr & \Rightarrow {\text{C}}:{\text{D}} = 20:18 \cr & \Rightarrow {\text{C}}:{\text{D}} = 10:9 \cr & {\text{A}}:{\text{ B }}:{\text{C }}:{\text{D}} \cr & 4{\text{ }}:{\text{ }}3 \cr & \,\,\,\,\,\,\,\,\,\,3{\text{ }}:{\text{ }}5 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,10{\text{ }}:{\text{ }}9 \cr & \overline {\underline {{\text{ }}8{\text{ }}:6{\text{ }}:10{\text{ }}:{\text{ }}9{\text{ }}} } \cr} $$

Solution:

 x2 + 8y2 + 12y - 4xy + 9 = 0 x2 + 4y2 - 4xy + 4y2 + 12y + 9 = 0 (x - 2y)2 + (2y + 3)2 = 0 x = 2y 2y = -3 y = $$\frac{{ - 3}}{2}$$ x = -3 (7x + 8y) = 7 × (-3) + 8 × $$\left( {\frac{{ - 3}}{2}} \right)$$ = -21 - 12 = -33

Solution:

 $$\eqalign{ & x + \frac{1}{x} = 5 \cr & {\text{Cubing both sides, we get}} \cr & {x^3} + \frac{1}{{{x^3}}} + 3 \times x \times \frac{1}{x}\left( {x + \frac{1}{x}} \right) = 125 \cr & {x^3} + \frac{1}{{{x^3}}} + 3 \times 5 = 125 \cr & {x^3} + \frac{1}{{{x^3}}} = 110 \cr & {\text{Squaring both sides, we get}} \cr & {x^6} + \frac{1}{{{x^6}}} + 2 = 12100 \cr & {x^6} + \frac{1}{{{x^6}}} = 12098 \cr} $$

Solution:

 $$\eqalign{ & {\text{ }}{x^2}{\text{ + }}\alpha x + \beta = 0 \cr & {\text{Sum of root}} \cr & \alpha + \beta = \frac{{ - \alpha }}{1}\,......(i) \cr & \alpha \beta = \beta \,......(ii) \cr & {\text{From (i) and (ii)}} \cr & {\text{Then, }}\alpha = 1 \cr & {\text{Then, }}\beta = - 2 \cr & {\text{Then value of }} \cr & \Leftrightarrow {\alpha ^3} + {\beta ^3} \cr & = 1 + {\left( { - 2} \right)^3} \cr & = - 7 \cr} $$

Solution:

 $$\eqalign{ & {\text{Given, }}x + \frac{1}{x} = 1 \cr & {\text{Find }}\frac{2}{{{x^2} - x + 2}} = ? \cr & x + \frac{1}{x} = 1 \cr & {x^2} + 1 = x \cr & \left( {{x^2} - x} \right) = - 1 \cr & {\text{Putting value in,}} \cr & = \frac{2}{{{x^2} - x + 2}} \cr & = \frac{2}{{ - 1 + 2}} \cr & = 2 \cr} $$

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:

 $$\eqalign{ & 3{a^2} = {b^2}{\text{ }}\left( {{\text{Given}}} \right) \cr & {\text{ }}\frac{{{{\left( {a + b} \right)}^3} - {{\left( {a - b} \right)}^3}}}{{{{\left( {a + b} \right)}^2} + {{\left( {a - b} \right)}^2}}} \cr} $$   $$ = \frac{{{a^3} + {b^3} + 3ab\left( {a + b} \right)\, - \,\left( {{a^3} - {b^3} - 3ab\left( {a - b} \right)} \right){\text{ }}}}{{{a^2} + {b^2} + 2ab + {\text{ }}{a^2} + {b^2} - 2ab}}$$ $$\eqalign{ & = \frac{{2{b^3}{\text{ + 6}}{{\text{a}}^2}{\text{b }}}}{{2{a^2} + 2{b^2}{\text{ }}}} \cr & = \frac{{{b^3}{\text{ + 3}}{{\text{a}}^2}{\text{b }}}}{{{a^2} + {b^2}{\text{ }}}} \cr & = \frac{{{b^3} + {b^3}{\text{ }}}}{{\frac{{{b^2}}}{3} + {b^2}{\text{ }}}} \cr & = \frac{{2{b^3}}}{{{b^2}\left( {\frac{1}{3} + 1} \right)}} \cr & = \frac{{2b}}{{\frac{4}{3}}} \cr & = \frac{{3b}}{2} \cr} $$

Solution:

 $$\eqalign{ & a + \frac{1}{a} + 1 = 0 \cr & a + \frac{1}{a} = - 1 \cr & {\text{Squaring both sides}} \cr & \Rightarrow {a^2} + \frac{1}{{{a^2}}} + 2 = 1 \cr & \Rightarrow {a^2} + \frac{1}{{{a^2}}} = - 1 \cr & \Rightarrow {a^2} + 1 = - \frac{1}{{{a^2}}}\,.....(i) \cr & \Rightarrow a + \frac{1}{a} = - 1{\text{ }}\left( {{\text{Given}}} \right) \cr & \therefore {a^2} + 1 = - a\,.....(ii) \cr & \Rightarrow - a = \frac{{ - 1}}{{{a^2}}} \cr & {\text{For equation (i) and (ii)}} \cr & {{\text{a}}^3} = 1 \cr & \therefore {{\text{a}}^3} - 1 = 0 \cr & \Rightarrow {a^4} - a = a\left( {{a^3} - 1} \right) \cr & \Rightarrow {a^4} - a = a \times 0 \cr & \Rightarrow {a^4} - a = 0 \cr} $$

Solution:

 $$\eqalign{ & {x^2} - 4x + 1 = 0 \cr & x + \frac{1}{x} = 4 \cr & {x^2} + \frac{1}{{{x^2}}} = 14 \cr & {x^4} + \frac{1}{{{x^4}}} = 194 \cr & {x^9} + {x^7} - 194{x^5} - 194{x^3} \cr & {x^9} + {x^7} - \left( {{x^4} + \frac{1}{{{x^4}}}} \right){x^5} - \left( {{x^4} + \frac{1}{{{x^4}}}} \right){x^3} \cr & = {x^9} + {x^7} - {x^9} - x - {x^7} - \frac{1}{x} \cr & = - \left( {x + \frac{1}{x}} \right) \cr & = - 4 \cr} $$

Solution:

 $$\eqalign{ & x + \frac{1}{x} = 3,\,x \ne 0 \cr & {x^2} + \frac{1}{{{x^2}}} = {3^2} - 2 = 7 \cr & {x^4} + \frac{1}{{{x^4}}} = {7^2} - 2 = 47 \cr & {x^3} + \frac{1}{{{x^3}}} = {3^3} - 3 \times 3 \cr & = 27 - 9 \cr & = 18 \cr & {x^7} + \frac{1}{{{x^7}}} = \left( {{x^4} + \frac{1}{{{x^4}}}} \right)\left( {{x^3} + \frac{1}{{{x^3}}}} \right) - \left( {x + \frac{1}{x}} \right) \cr & = 47 \times 18 - 3 \cr & = 846 - 3 \cr & = 843 \cr} $$