Practice MCQ Questions and Answer on Algebra

Solution:

 $$\frac{{x - {a^2}}}{{b + c}} + \frac{{x - {b^2}}}{{c + a}} + \frac{{x - {c^2}}}{{a + b}} = 4\left( {a + b + c} \right)$$ Note : In such type of question to save your valuable time assume values as per your need which make your calculation easier. Assume a = 1, b = 0, c = 1 $$\eqalign{ & {\text{Make sure there will be no }}\left( {\frac{0}{0}} \right){\text{ form}} \cr & \therefore \frac{{x - 1}}{{1 + 0}} + \frac{{x - 0}}{{1 + 1}} + \frac{{x - 1}}{{1 + 0}} = 4 \cr & \Rightarrow x - 1 + \frac{x}{2} + x - 1 = 4 \times 2 \cr & \Rightarrow x + \frac{x}{2} + x = 8 + 2 \cr & \Rightarrow \frac{{5x}}{2} = 10 \cr & \Rightarrow x = 4 \cr & {\text{Now put values in options take option}} \cr & \left( \text{A} \right),{\left( {a + b + c} \right)^2} = {\left( {1 + 0 + 1} \right)^2} = 4 \cr} $$

Solution:

 $$\eqalign{ & {x^2} - 3x + 1 = 0 \cr & \Rightarrow {x^2} + 1 = 3x \cr & \Rightarrow x + \frac{1}{x} = 3 \cr & {\text{Squaring both sides}} \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 2 = 9 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} = 7 \cr & \therefore {x^2} + x + \frac{1}{x} + \frac{1}{{{x^2}}} \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} + x + \frac{1}{x} \cr & \Rightarrow 7 + 3 \cr & \Rightarrow 10 \cr} $$

Solution:

 $$\eqalign{ & {{\text{7}}^x}{\text{ = }}\frac{1}{{343}} \cr & \Rightarrow {{\text{7}}^x}{\text{ = }}\frac{1}{{{7^3}}} \cr & \Rightarrow {{\text{7}}^x}{\text{ = }}{{\text{7}}^{ - 3}} \cr & \Rightarrow x = - 3 \cr} $$ (If bases are equal then their power are also equal)

Solution:

 $$\eqalign{ & \frac{{4x - 3}}{x} + \frac{{4y - 3}}{y} + \frac{{4z - 3}}{z} = 0 \cr & \Rightarrow \frac{{4x}}{x} - \frac{3}{x} + \frac{{4y}}{y} - \frac{3}{y} + \frac{{4z}}{z} - \frac{3}{z} = 0 \cr & \Rightarrow 4 - \frac{3}{x} + 4 - \frac{3}{y} + 4 - \frac{3}{z} = 0 \cr & \Rightarrow 12 - 3\left( {\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \right) = 0 \cr & \Rightarrow - 3\left( {\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \right) = - 12 \cr & \Rightarrow \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 4 \cr} $$

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:

 $$\eqalign{ & \frac{a}{b}{\text{ = }}\frac{2}{3}{\text{ and }}\frac{b}{c}{\text{ = }}\frac{4}{5}\,\,\left( {{\text{Given}}} \right) \cr & or\,\frac{c}{b} = \frac{5}{4} \cr & \frac{{a + b}}{{b + c}} \cr & = \frac{{b\left( {\frac{a}{b} + 1} \right)}}{{b\left( {\frac{c}{b} + 1} \right)}} \cr & = \frac{{\frac{a}{b} + 1}}{{\frac{c}{b} + 1}} \cr & = \frac{{\left( {\frac{2}{3} + 1} \right)}}{{\left( {\frac{5}{4} + 1} \right)}} \cr & = \frac{{\frac{2 + 3}{3}}}{{\frac{{5 + 4}}{4}}} \cr & = \frac{{5 \times 4}}{{3 \times 9}} \cr & = \frac{{20}}{{27}} \cr & \therefore \frac{{a + b}}{{b + c}} = \frac{{20}}{{27}} \cr & {\bf{Alternate:}} \cr & a{\text{ }}\,\,\,{\text{ }}\,{\text{ }}\,\,\,\,{\text{ }}b{\text{ }}\,\,\,\,{\text{ }}\,\,\,\,{\text{ }}\,\,\,{\text{ }}c \cr & {2_{ \times \left( 4 \right)}}\,\,\,\,\,{\text{ }}{3_{ \times \left( 4 \right)}} \cr & \,{\text{ }}\,{\text{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,{4_{ \times \left( 3 \right)}}{\text{ }}\,\,\,\,\,\,{5_{ \times \left( 3 \right)}} \cr & \overline {\underline {8{\text{ }}\,\,\,\,\,\,\,\,\,\,\,{\text{ }}12{\text{ }}\,\,\,\,\,\,\,\,\,\,{\text{ }}15\,\,\,\,} } \cr & \therefore \frac{{a + b}}{{b + c}} = \frac{{8 + 12}}{{12 + 15}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{20}}{{27}} \cr} $$

Solution:

 $$\eqalign{ & x + \frac{1}{x} = \sqrt 3 \cr & \left( {{\text{Take cube on both sides}}} \right) \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^3} = {\left( {\sqrt 3 } \right)^3} \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3x.\frac{1}{x}\left( {x + \frac{1}{x}} \right) = 3\sqrt 3 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3\sqrt 3 = 3\sqrt 3 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 0 \cr & \therefore {x^6} = - 1 \cr & \therefore {x^{18}} + {x^{12}} + {x^6} + 1 \cr & = {\left( { - 1} \right)^3} + {\left( { - 1} \right)^2} + \left( { - 1} \right) + 1 \cr & = - 1 + 1 - 1 + 1 \cr & = 0 \cr} $$

Solution:

 $$\eqalign{ & 2x + 1 = 0 \cr & \Leftrightarrow x = \frac{{ - 1}}{2}{\text{ }} \cr & 3y - 9 = 0 \cr & \Leftrightarrow y = \frac{9}{3} = 3 \cr & \therefore \left( {x,y} \right) = \left( {\frac{{ - 1}}{2},3} \right) \cr & {\text{Point of intersection}} \cr} $$

Solution:

 $$\eqalign{ & {a^3} + {b^3} + {c^3} = 3abc \cr & {\text{If }}a + b + c = 0 \cr & a = 0.324 \cr & b = 0.221 \cr & c = - 0.545 \cr & \frac{{\left( {0.545} \right)\left( {0.081} \right)\left( {0.51} \right)\left( {5.2} \right)}}{{3abc}} \cr & = - \frac{{0.545 \times 0.081 \times 0.51 \times 5.2}}{{3 \times 0.324 \times 0.221 \times 0.545}} \cr & = - \frac{{81 \times 510 \times 5.2}}{{3 \times 18 \times 18 \times 13 \times 17}} \cr & = - 1 \cr} $$

Solution:

 $$\eqalign{ & a = 2,{\text{ }}b = - 3 \cr & {\text{So,}}27{a^3} - 54{a^2}b + 36a{b^2} - 8{b^3} \cr} $$ $$ = 27 \times {\left( 2 \right)^3} - 54 \times {\left( 2 \right)^2} \times \left( { - 3} \right) + $$       $$36 \times $$ $$\left( 2 \right) \times $$ $${\left( { - 3} \right)^2}$$ $$ -\, 8 \times $$ $${\left( { - 3} \right)^3}$$ $$\eqalign{ & = 27 \times 8 - 54 \times 4 \times - 3 + 36 \times 2 \times 9 - 8 \times - 27 \cr & = 216 + 216 + 648 + 648 \cr & = 1728 \cr} $$