Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

$$\eqalign{ & {\text{If }}a + b + c = 0 \cr & {\text{Put }}a = 1,{\text{ }}b = 1{\text{ and }}c = - 2 \cr & \therefore \frac{{{a^2} + {b^2} + {c^2}}}{{ab + bc + ca}} \cr & = \frac{{1 + 1 + 4}}{{1 - 2 - 2}} \cr & = \frac{6}{{ - 3}} \cr & = - 2 \cr} $$

$$\eqalign{ & a + b + c = m \cr & \Rightarrow \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0 \cr & \Rightarrow \frac{{ab + bc + ca}}{{abc}} = 0 \cr & \Rightarrow ab + bc + ca = 0 \cr & {\left( {a + b + c} \right)^2} = {\text{ }}{a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ca} \right) \cr & \Rightarrow {m^2} = {\text{ }}{a^2} + {b^2} + {c^2} \cr & \Rightarrow \frac{{{m^2}}}{3} = {\text{ }}\frac{{{a^2} + {b^2} + {c^2}}}{3} \cr} $$

$$\eqalign{ & \left( {\sqrt a + \sqrt b } \right) = 15{\text{ }} \cr & {\text{Square both sides}} \cr & \Rightarrow a + b + 2\sqrt {ab} = 225 \cr & \Rightarrow a + b = 225 - 2\sqrt {ab} \,.....(i) \cr & {\text{ }}\left( {\sqrt a - \sqrt b } \right) = 3 \cr & {\text{Square both sides}} \cr & \Rightarrow a + b - 2\sqrt {ab} = 9 \cr & \Rightarrow a + b = 9 + 2\sqrt {ab} \,.....(ii) \cr & {\text{From (i) and (ii)}} \cr & \Rightarrow 225 - 2\sqrt {ab} = 9 + 2\sqrt {ab} \cr & \Rightarrow 216 = 4\sqrt {ab} \cr & \Rightarrow 54 = \sqrt {ab} \cr & {\text{Divided by 4 on both sides}} \cr & \Rightarrow \frac{{\sqrt {ab} }}{4} = \frac{{54}}{4} \cr & \Rightarrow \frac{{\sqrt {ab} }}{4} = \frac{{27}}{2} \cr} $$

$$\eqalign{ & {\text{Given , }}a + \frac{1}{a} = 3 \cr & {\text{Cube both sides}} \cr & {a^3} + \frac{1}{{{a^3}}} + 3 \times a \times \frac{1}{a}\left( {a + \frac{1}{a}} \right) = {\left( 3 \right)^3} \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3 \times 3 = 27 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 27 - 9 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 18 \cr} $$

$$\eqalign{ & x = \sqrt {a\root 3 \of {ab\sqrt {a\root 3 \of {ab} } } } ....... \propto \cr & {\text{Square both sides}} \cr & {x^2} = a\root 3 \of {abx} {\text{ }}\left( {\therefore \sqrt {a\root 3 \of {ab} } .... \propto } \right) \cr & {\text{Again cube both sides}} \cr & {x^6} = {a^3}abx \cr & {x^5} = {a^4}b \cr & x = \root 5 \of {{a^4}b} \cr} $$

$$\eqalign{ & {\text{Put }}a = b = c = 1 \cr & {\text{Then we have }} \cr & \frac{{m - 3}}{2}{\text{ + }}\frac{{m - 3}}{2}{\text{ + }}\frac{{m - 3}}{2} = 9{\text{ }} \cr & m = 9 \cr} $$
Now, putting values of a, b, c in option,
Only option (C) gives value of m = 9
So, option (C) is correct.

$$\eqalign{ & {a^2} = b + c,{\text{ }}{b^2} = a + c,{\text{ }}{c^2} = b + a \cr & {\text{Taking }}a = 2,{\text{ }}b = 2{\text{ and }}c = 2 \cr & {\text{So,}}{\left( 2 \right)^2} = 2 + 2 \cr & \boxed{4 = 4} \cr & {\text{Now,}}\frac{1}{{a + 1}} + {\text{ }}\frac{1}{{b + 1}} + {\text{ }}\frac{1}{{c + 1}} \cr & {\text{Put }}a = 2,{\text{ }}b = 2{\text{ and }}c = 2 \cr & = \frac{1}{{2 + 1}} + {\text{ }}\frac{1}{{2 + 1}} + {\text{ }}\frac{1}{{2 + 1}} \cr & = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} \cr & = 1 \cr} $$

$$\eqalign{ & a + b = 2c \cr & {\text{Taking }}a = 2,{\text{ }}b = 4{\text{ and }}c = 3 \cr & {\text{So,}}2 + 4 = 2 \times 3 \cr & \boxed{6 = 6} \cr & {\text{Now,}}\frac{a}{{a - c}} + {\text{ }}\frac{c}{{b - c}} \cr & = \frac{2}{{2 - 3}} + {\text{ }}\frac{3}{{4 - 3}} \cr & = \frac{2}{{ - 1}} + \frac{3}{1} \cr & = 1 \cr} $$

$$\eqalign{ & \frac{a}{b} = \frac{1}{2} \cr & {\text{Let }}a = x,{\text{ }}b = 2x \cr & {\text{Then,}}\frac{{\left( {2a - 5b} \right)}}{{\left( {5a + 3b} \right)}} \cr & = \frac{{2x - 10x}}{{5x + 6x}} \cr & {\text{ = }}\frac{{ - 8x}}{{11x}} \cr & = \frac{{ - 8}}{{11}} \cr} $$

$$\eqalign{ & 3{x^2} + 5x + 3 = 0 \cr & 3{x^2} + 3 = - 5x \cr & {\text{Divide by }}3x{\text{ both sides}} \cr & x + \frac{1}{x} = \frac{{ - 5}}{3} \cr & {\text{Then, }}{x^3} + \frac{1}{{{x^3}}} \cr & = {\left( {\frac{{ - 5}}{3}} \right)^3} - 3 \times \left( {\frac{{ - 5}}{3}} \right) \cr & = \frac{{ - 125}}{{27}} + 5 \cr & = \frac{{10}}{{27}} \cr} $$