Practice MCQ Questions and Answer on Algebra

Solution:

 a + b + c = 0 Have values a = 1 b = 2 c = - 3 $$ \Rightarrow \left( {\frac{{a + b}}{c} + \frac{{b + c}}{a} + \frac{{c + a}}{b}} \right)$$     $$\left( {\frac{a}{{b + c}} + \frac{b}{{c + a}} + \frac{c}{{a + b}}} \right)$$ $$ \Rightarrow \left( {\frac{{1 + 2}}{{ - 3}} + \frac{{2 - 3}}{1} + \frac{{ - 3 + 1}}{2}} \right)$$     $$\left( {\frac{1}{{2 - 3}} + \frac{2}{{ - 3 + 1}} + \frac{{ - 3}}{{1 + 2}}} \right)$$ $$\eqalign{ & \Rightarrow \left( { - 1 - 1 - 1} \right)\left( { - 1 - 1 - 1} \right) \cr & \Rightarrow - 3 \times - 3 \cr & \Rightarrow 9 \cr} $$

Solution:

 $$\eqalign{ & {\text{ }}x + \frac{1}{x} = 5 \cr & {\text{Take cube on both sides}} \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^3} = {\left( 5 \right)^3} \cr & \Rightarrow {x^3}{\text{ + }}\frac{1}{{{x^3}}} + 3 \times 5 = 125 \cr & \Rightarrow {x^3}{\text{ + }}\frac{1}{{{x^3}}} = 110 \cr & \therefore {\text{Squaring both sides}} \cr & \Rightarrow {\left( {{x^3}{\text{ + }}\frac{1}{{{x^3}}}} \right)^2} = {\left( {110} \right)^2} \cr & \Rightarrow {x^6}{\text{ + }}\frac{1}{{{x^6}}} + 2 = 12100 \cr & \Rightarrow {x^6}{\text{ + }}\frac{1}{{{x^6}}} = 12100 - 2 \cr & \Rightarrow {x^6}{\text{ + }}\frac{1}{{{x^6}}} = 12098 \cr} $$

Solution:

 $$\eqalign{ & {x^2} + \frac{1}{{{x^2}}} = 83 \cr & {\text{Subtracting 2 from both sides}} \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} - 2 = 83 - 2 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} - 2.x.\frac{1}{x} = 83 - 2 \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^2} = 81 \cr & \Rightarrow x - \frac{1}{x} = 9 \cr & {\text{Take cube on both sides}} \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3{\text{ }}\left( {x - \frac{1}{x}} \right) = 729 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3 \times {\text{9}} = 729 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} = 729 + 27 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} = 756 \cr} $$

Solution:

 $$a - \frac{1}{{a - 5}} = 10$$ We can write it is as (subtracting 5 from both sides) $$\eqalign{ & a - 5 - \frac{1}{{a - 5}} = 10 - 5 \cr & {\text{Now, }}\left( {a - 5} \right) - \frac{1}{{\left( {a - 5} \right)}} = 5 \cr} $$ So, take the cube of this equation, $$\eqalign{ & \left[ {{{\left( {a - 5} \right)}^3} - \frac{1}{{{{\left( {a - 5} \right)}^3}}}} \right] = 125 + 3 \times 5 \cr & \left[ {{{\left( {a - 5} \right)}^3} - \frac{1}{{{{\left( {a - 5} \right)}^3}}}} \right] = 140 \cr} $$

Solution:

 $$\eqalign{ & \frac{x}{y}{\text{ = }}\frac{{a + 2}}{{a - 2}} \cr & \frac{{{x^2}}}{{{y^2}}}{\text{ = }}\frac{{{{\left( {a + 2} \right)}^2}}}{{{{\left( {a - 2} \right)}^2}}} \cr} $$ Applying componendo and dividendo $$\eqalign{ & \therefore \frac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}} \cr & = \frac{{{{\left( {a + 2} \right)}^2} - {{\left( {a - 2} \right)}^2}}}{{{{\left( {a + 2} \right)}^2} + {{\left( {a - 2} \right)}^2}}} \cr & = \frac{{8a}}{{2{a^2} + 8}} \cr & = \frac{{4a}}{{{a^2} + 4}} \cr} $$

Solution:

 5x + 9y = 5 . . . . . (i) 125x3 + 729y3 = 120 . . . . . (ii) From equation (i) cubing both sides ⇒ (5x + 9y)3 = 53 ⇒ 125x3 + 729y3 + 3 × 5x × 9y(5x + 9y) = 125 ⇒ 125x3 + 729y3 + 135xy × 5 = 125 ⇒ 120 + 135xy × 5 = 125 ⇒ 135xy × 5 = 5 ⇒ xy = $$\frac{1}{{135}}$$ ∴ product of x & y = $$\frac{1}{{135}}$$

Solution:

 $$\eqalign{ & xy = - 6,\,{x^3} + {y^3} = 19 \cr & {\text{putting }}x = 3,\,y = - 2 \cr & {\text{both condition are satisfied}} \cr & {\text{So, }}\frac{1}{{{x^{ - 1}}}} + \frac{1}{{{y^{ - 1}}}} \cr & = x + y \cr & = 3 - 2 \cr & = 1 \cr} $$

Solution:

 $$\eqalign{ & {a^2} + {b^2} = 5ab \cr & \Rightarrow \frac{{{a^2}}}{{ab}} + \frac{{{b^2}}}{{ab}} = 5 \cr & \Rightarrow \frac{a}{b}{\text{ + }}\frac{b}{a}{\text{ = 5}} \cr & {\text{Squaring the both sides}} \cr & \Rightarrow {\left( {\frac{a}{b}} \right)^2}{\text{ + }}{\left( {\frac{b}{a}} \right)^2} + 2 \times \frac{a}{b} \times \frac{b}{a} = 25 \cr & \Rightarrow \frac{{{a^2}}}{{{b^2}}}{\text{ + }}\frac{{{b^2}}}{{{a^2}}} = 25 - 2 \cr & \Rightarrow \frac{{{a^2}}}{{{b^2}}}{\text{ + }}\frac{{{b^2}}}{{{a^2}}} = 23 \cr} $$

Solution:

 $$\eqalign{ & a*b = a + b + ab \cr & 3*4 \cr & = 3 + 4 + 3 \times 4 \cr & = 19 \cr & 2*3 \cr & = 2 + 3 + 2 \times 3 \cr & = 11 \cr & \therefore 3*4 - 2*3{\text{ }} \cr & = 19 - 11 \cr & = 8 \cr} $$

Solution:

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