Quantitative Aptitude
ALGEBRA MCQs
Basic Algebraic Identities Of School Algebra & Elementary Surds
Total Questions : 1010
| Page 95 of 101 pages
Answer: Option B. -> $${\text{3}}\sqrt 3 $$
$$\eqalign{
& x = \sqrt 3 - \frac{1}{{\sqrt 3 }}{\text{ and }}y = \sqrt 3 + \frac{1}{{\sqrt 3 }} \cr
& \Rightarrow \frac{{{x^2}}}{y} + \frac{{{y^2}}}{x} \cr
& = \frac{{{x^3} + {y^3}}}{{xy}} \cr
& = \frac{{\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right)}}{{xy}} \cr
& \therefore x + y \cr
& = \sqrt 3 - \frac{1}{{\sqrt 3 }} + \sqrt 3 + \frac{1}{{\sqrt 3 }} \cr
& = 2\sqrt 3 \cr
& \therefore xy \cr
& = \sqrt 3 - \frac{1}{{\sqrt 3 }} \times \sqrt 3 + \frac{1}{{\sqrt 3 }} \cr
& = 3 - \frac{1}{3} \cr
& = \frac{8}{3} \cr
& \Rightarrow \frac{{\left( {x + y} \right)\left( {{x^2} + {y^2} + 2xy - 2xy - xy} \right)}}{{xy}} \cr
& \Rightarrow \frac{{\left( {x + y} \right)\left( {{{\left( {x + y} \right)}^2} - 3xy} \right)}}{{xy}} \cr
& \Rightarrow \frac{{2\sqrt 3 \left( {{{\left( {2\sqrt 3 } \right)}^2} - 3 \times \frac{8}{3}} \right)}}{{\frac{8}{3}}} \cr
& \Rightarrow \frac{{2\sqrt 3 \left( {12 - 8} \right)}}{{\frac{8}{3}}} \cr
& \Rightarrow \frac{{2 \times 3\sqrt 3 \left( 4 \right)}}{8} \cr
& \Rightarrow 3\sqrt 3 \cr} $$
$$\eqalign{
& x = \sqrt 3 - \frac{1}{{\sqrt 3 }}{\text{ and }}y = \sqrt 3 + \frac{1}{{\sqrt 3 }} \cr
& \Rightarrow \frac{{{x^2}}}{y} + \frac{{{y^2}}}{x} \cr
& = \frac{{{x^3} + {y^3}}}{{xy}} \cr
& = \frac{{\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right)}}{{xy}} \cr
& \therefore x + y \cr
& = \sqrt 3 - \frac{1}{{\sqrt 3 }} + \sqrt 3 + \frac{1}{{\sqrt 3 }} \cr
& = 2\sqrt 3 \cr
& \therefore xy \cr
& = \sqrt 3 - \frac{1}{{\sqrt 3 }} \times \sqrt 3 + \frac{1}{{\sqrt 3 }} \cr
& = 3 - \frac{1}{3} \cr
& = \frac{8}{3} \cr
& \Rightarrow \frac{{\left( {x + y} \right)\left( {{x^2} + {y^2} + 2xy - 2xy - xy} \right)}}{{xy}} \cr
& \Rightarrow \frac{{\left( {x + y} \right)\left( {{{\left( {x + y} \right)}^2} - 3xy} \right)}}{{xy}} \cr
& \Rightarrow \frac{{2\sqrt 3 \left( {{{\left( {2\sqrt 3 } \right)}^2} - 3 \times \frac{8}{3}} \right)}}{{\frac{8}{3}}} \cr
& \Rightarrow \frac{{2\sqrt 3 \left( {12 - 8} \right)}}{{\frac{8}{3}}} \cr
& \Rightarrow \frac{{2 \times 3\sqrt 3 \left( 4 \right)}}{8} \cr
& \Rightarrow 3\sqrt 3 \cr} $$
Answer: Option D. -> 0
$$\eqalign{
& x + y = 15 \cr
& \Rightarrow x - 10 = 5 - y \cr
& \Rightarrow x - 10 = - \left( {y - 5} \right) \cr
& {\text{Take cube on both sides}} \cr
& \Rightarrow {\left( {x - 10} \right)^3} = - {\left( {y - 5} \right)^3} \cr
& \Rightarrow {\left( {x - 10} \right)^3} + {\left( {y - 5} \right)^3} = 0 \cr} $$
$$\eqalign{
& x + y = 15 \cr
& \Rightarrow x - 10 = 5 - y \cr
& \Rightarrow x - 10 = - \left( {y - 5} \right) \cr
& {\text{Take cube on both sides}} \cr
& \Rightarrow {\left( {x - 10} \right)^3} = - {\left( {y - 5} \right)^3} \cr
& \Rightarrow {\left( {x - 10} \right)^3} + {\left( {y - 5} \right)^3} = 0 \cr} $$
Question 943. If a + b + c + d = 4, then find the value of $$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$$ + $$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$$ + $$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$$ + $$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$$ is?
Answer: Option A. -> 0
$$\eqalign{
& \frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}} + \frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}} + \frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}} + \frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}} \cr
& {\text{Put }} \cr
& a = 0 \cr
& b = 0 \cr
& c = 2 \cr
& d = 2 \cr
& \therefore a + b + c + d = 4 \cr
& \Rightarrow 0 + 0 + 2 + 2 = 4 \cr
& \Rightarrow 4 = 4\left( {{\text{ satisfy}}} \right) \cr
& \frac{1}{{\left( {1 - 0} \right)\left( {1 - 0} \right)\left( {1 - 2} \right)}} + \frac{1}{{\left( {1 - 0} \right)\left( {1 - 2} \right)\left( {1 - 2} \right)}} + \frac{1}{{\left( {1 - 2} \right)\left( {1 - 2} \right)\left( {1 - 0} \right)}} + \frac{1}{{\left( {1 - 2} \right)\left( {1 - 0} \right)\left( {1 - 0} \right)}} \cr
& = \frac{1}{{ - 1}} + \left( {\frac{1}{{ + 1}}} \right) + \frac{1}{{ - 1 \times - 1}} + \frac{1}{{ - 1}} \cr
& = - 1 + 1 + 1 - 1 \cr
& = 0 \cr} $$
$$\eqalign{
& \frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}} + \frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}} + \frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}} + \frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}} \cr
& {\text{Put }} \cr
& a = 0 \cr
& b = 0 \cr
& c = 2 \cr
& d = 2 \cr
& \therefore a + b + c + d = 4 \cr
& \Rightarrow 0 + 0 + 2 + 2 = 4 \cr
& \Rightarrow 4 = 4\left( {{\text{ satisfy}}} \right) \cr
& \frac{1}{{\left( {1 - 0} \right)\left( {1 - 0} \right)\left( {1 - 2} \right)}} + \frac{1}{{\left( {1 - 0} \right)\left( {1 - 2} \right)\left( {1 - 2} \right)}} + \frac{1}{{\left( {1 - 2} \right)\left( {1 - 2} \right)\left( {1 - 0} \right)}} + \frac{1}{{\left( {1 - 2} \right)\left( {1 - 0} \right)\left( {1 - 0} \right)}} \cr
& = \frac{1}{{ - 1}} + \left( {\frac{1}{{ + 1}}} \right) + \frac{1}{{ - 1 \times - 1}} + \frac{1}{{ - 1}} \cr
& = - 1 + 1 + 1 - 1 \cr
& = 0 \cr} $$
Answer: Option B. -> $$\frac{1}{2}$$
$$\eqalign{
& x - \frac{1}{x} = 1{\text{ }} \cr
& \Rightarrow \frac{{{x^4} - \frac{1}{{{x^2}}}}}{{3{x^2} + 5x - 3}} \cr
& {\text{Divide and multiply by }}x \cr
& \Rightarrow \frac{{\frac{{{x^4}}}{x} - \frac{1}{{{x^3}}}}}{{\frac{{3{x^2}}}{x} + \frac{{5x}}{x} - \frac{3}{x}}} \cr
& \Rightarrow \frac{{{x^3} - \frac{1}{{{x^3}}}}}{{3x + \frac{3}{x} + 5}} \cr
& \Rightarrow \frac{{{x^3} - \frac{1}{{{x^3}}}}}{{3\left( {x - \frac{1}{x}} \right) + 5}} \cr
& \Rightarrow x - \frac{1}{x} = 1{\text{ }} \cr
& {\text{Take cube on both sides}} \cr
& \Rightarrow {\left( {x - \frac{1}{x}} \right)^3} = {\left( 1 \right)^3}{\text{ }} \cr
& \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3\left( {x - \frac{1}{x}} \right) = 1 \cr
& \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3\left( 1 \right) = 1 \cr
& \Rightarrow {x^3} - \frac{1}{{{x^3}}} = 4 \cr
& \Rightarrow \frac{{{x^3} - \frac{1}{{{x^3}}}}}{{3\left( {x - \frac{1}{x}} \right) + 5}} \cr
& \Rightarrow \frac{4}{{3 \times 1 + 5}} \cr
& \Rightarrow \frac{4}{8} \cr
& \Rightarrow \frac{1}{2} \cr} $$
$$\eqalign{
& x - \frac{1}{x} = 1{\text{ }} \cr
& \Rightarrow \frac{{{x^4} - \frac{1}{{{x^2}}}}}{{3{x^2} + 5x - 3}} \cr
& {\text{Divide and multiply by }}x \cr
& \Rightarrow \frac{{\frac{{{x^4}}}{x} - \frac{1}{{{x^3}}}}}{{\frac{{3{x^2}}}{x} + \frac{{5x}}{x} - \frac{3}{x}}} \cr
& \Rightarrow \frac{{{x^3} - \frac{1}{{{x^3}}}}}{{3x + \frac{3}{x} + 5}} \cr
& \Rightarrow \frac{{{x^3} - \frac{1}{{{x^3}}}}}{{3\left( {x - \frac{1}{x}} \right) + 5}} \cr
& \Rightarrow x - \frac{1}{x} = 1{\text{ }} \cr
& {\text{Take cube on both sides}} \cr
& \Rightarrow {\left( {x - \frac{1}{x}} \right)^3} = {\left( 1 \right)^3}{\text{ }} \cr
& \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3\left( {x - \frac{1}{x}} \right) = 1 \cr
& \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3\left( 1 \right) = 1 \cr
& \Rightarrow {x^3} - \frac{1}{{{x^3}}} = 4 \cr
& \Rightarrow \frac{{{x^3} - \frac{1}{{{x^3}}}}}{{3\left( {x - \frac{1}{x}} \right) + 5}} \cr
& \Rightarrow \frac{4}{{3 \times 1 + 5}} \cr
& \Rightarrow \frac{4}{8} \cr
& \Rightarrow \frac{1}{2} \cr} $$
Answer: Option B. -> (a + b + c)3 = 27abc
$$\eqalign{
& {\text{ }}{a^{\frac{1}{3}}} + {b^{\frac{1}{3}}} + {c^{\frac{1}{3}}} = 0 \cr
& \Rightarrow {\text{ }}{a^{\frac{1}{3}}} + {b^{\frac{1}{3}}} = - {c^{\frac{1}{3}}} \cr
& {\text{Take cube on both sides}} \cr
& \Rightarrow {\left( {{\text{ }}{a^{\frac{1}{3}}} + {b^{\frac{1}{3}}}} \right)^3} = {\left( { - {c^{\frac{1}{3}}}} \right)^3} \cr
& \Rightarrow {\text{ }}a + b + 3{a^{\frac{1}{3}}}{b^{\frac{1}{3}}}\left( {{\text{ }}{a^{\frac{1}{3}}} + {b^{\frac{1}{3}}}} \right) = - c \cr
& \Rightarrow {\text{ }}a + b + 3{a^{\frac{1}{3}}}{b^{\frac{1}{3}}}\left( {{\text{ }} - {c^{\frac{1}{3}}}} \right) = - c \cr
& \Rightarrow a + b + c = 3{a^{\frac{1}{3}}}{b^{\frac{1}{3}}}{c^{\frac{1}{3}}} \cr
& {\text{Again taking cube }} \cr
& \Rightarrow {\left( {a + b + c} \right)^3} = 27abc \cr} $$
$$\eqalign{
& {\text{ }}{a^{\frac{1}{3}}} + {b^{\frac{1}{3}}} + {c^{\frac{1}{3}}} = 0 \cr
& \Rightarrow {\text{ }}{a^{\frac{1}{3}}} + {b^{\frac{1}{3}}} = - {c^{\frac{1}{3}}} \cr
& {\text{Take cube on both sides}} \cr
& \Rightarrow {\left( {{\text{ }}{a^{\frac{1}{3}}} + {b^{\frac{1}{3}}}} \right)^3} = {\left( { - {c^{\frac{1}{3}}}} \right)^3} \cr
& \Rightarrow {\text{ }}a + b + 3{a^{\frac{1}{3}}}{b^{\frac{1}{3}}}\left( {{\text{ }}{a^{\frac{1}{3}}} + {b^{\frac{1}{3}}}} \right) = - c \cr
& \Rightarrow {\text{ }}a + b + 3{a^{\frac{1}{3}}}{b^{\frac{1}{3}}}\left( {{\text{ }} - {c^{\frac{1}{3}}}} \right) = - c \cr
& \Rightarrow a + b + c = 3{a^{\frac{1}{3}}}{b^{\frac{1}{3}}}{c^{\frac{1}{3}}} \cr
& {\text{Again taking cube }} \cr
& \Rightarrow {\left( {a + b + c} \right)^3} = 27abc \cr} $$
Answer: Option A. -> (a - 2b - 4)(a - 2b + 2)
$$\eqalign{
& {a^2} + 4{b^2} + 4b - 4ab - 2a - 8 \cr
& = {a^2} - 4ab + 4{b^2} - 2a + 4b - 8 \cr
& = {\left( {a - 2b} \right)^2} - 2\left( {a - 2b} \right) - 8 \cr
& {\text{Put }} t = a - 2b \cr
& = {t^2} - 2t - 8 \cr
& = {t^2} - 4t + 2t - 8 \cr
& = t\left( {t - 4} \right) + 2\left( {t - 4} \right) \cr
& = \left( {t + 2} \right)\left( {t - 4} \right) \cr
& = \left( {a - 2b - 4} \right)\left( {a - 2b + 2} \right) \cr
& \left( {{\text{Put the value of assume }}t} \right) \cr} $$
$$\eqalign{
& {a^2} + 4{b^2} + 4b - 4ab - 2a - 8 \cr
& = {a^2} - 4ab + 4{b^2} - 2a + 4b - 8 \cr
& = {\left( {a - 2b} \right)^2} - 2\left( {a - 2b} \right) - 8 \cr
& {\text{Put }} t = a - 2b \cr
& = {t^2} - 2t - 8 \cr
& = {t^2} - 4t + 2t - 8 \cr
& = t\left( {t - 4} \right) + 2\left( {t - 4} \right) \cr
& = \left( {t + 2} \right)\left( {t - 4} \right) \cr
& = \left( {a - 2b - 4} \right)\left( {a - 2b + 2} \right) \cr
& \left( {{\text{Put the value of assume }}t} \right) \cr} $$
Answer: Option D. -> 0
$$\frac{1}{{{a^2} + ax + {x^2}}}$$ $$ - $$ $$\frac{1}{{{a^2} - ax + {x^2}}}$$ $$ + $$ $$\frac{1}{{{a^4} + {a^2}{x^2} + {x^4}}}$$
$$ = \frac{{{a^2} - ax + {x^2} - {a^2} - ax - {x^2}}}{{\left( {{a^2} + {x^2} + ax} \right)\left( {{a^2} + {x^2} - ax} \right)}} + $$ $$\frac{{2ax}}{{{a^4} + {a^2}{x^2} + {x^4}}}$$
$$\eqalign{
& = \frac{{ - 2ax}}{{{{\left( {{a^2} + {x^2}} \right)}^2} - {{\left( {ax} \right)}^2}}} + \frac{{2ax}}{{{a^4} + {x^4} + {a^2}{x^2}}} \cr
& = \frac{{ - 2ax}}{{{a^4} + {x^4} + 2{a^2}{x^2} - {a^2}{x^2}}} + \frac{{2ax}}{{{a^4} + {x^4} + {a^2}{x^2}}} \cr
& = \frac{{ - 2ax}}{{{a^4} + {x^4} + {a^2}{x^2}}} + \frac{{2ax}}{{{a^4} + {x^4} + {a^2}{x^2}}} \cr
& = \frac{{ - 2ax + 2ax}}{{{a^4} + {x^4} + 2{a^2}{x^2} - {a^2}{x^2}}} \cr
& = 0 \cr} $$
$$\frac{1}{{{a^2} + ax + {x^2}}}$$ $$ - $$ $$\frac{1}{{{a^2} - ax + {x^2}}}$$ $$ + $$ $$\frac{1}{{{a^4} + {a^2}{x^2} + {x^4}}}$$
$$ = \frac{{{a^2} - ax + {x^2} - {a^2} - ax - {x^2}}}{{\left( {{a^2} + {x^2} + ax} \right)\left( {{a^2} + {x^2} - ax} \right)}} + $$ $$\frac{{2ax}}{{{a^4} + {a^2}{x^2} + {x^4}}}$$
$$\eqalign{
& = \frac{{ - 2ax}}{{{{\left( {{a^2} + {x^2}} \right)}^2} - {{\left( {ax} \right)}^2}}} + \frac{{2ax}}{{{a^4} + {x^4} + {a^2}{x^2}}} \cr
& = \frac{{ - 2ax}}{{{a^4} + {x^4} + 2{a^2}{x^2} - {a^2}{x^2}}} + \frac{{2ax}}{{{a^4} + {x^4} + {a^2}{x^2}}} \cr
& = \frac{{ - 2ax}}{{{a^4} + {x^4} + {a^2}{x^2}}} + \frac{{2ax}}{{{a^4} + {x^4} + {a^2}{x^2}}} \cr
& = \frac{{ - 2ax + 2ax}}{{{a^4} + {x^4} + 2{a^2}{x^2} - {a^2}{x^2}}} \cr
& = 0 \cr} $$
Answer: Option C. -> $$\frac{5}{2}$$
$$\eqalign{
& x + \frac{1}{x} = 3 \cr
& \frac{{3{x^2} - 4x + 3}}{{{x^2} - x + 1}} \cr
& = \frac{{\frac{{3{x^2}}}{x} - \frac{{4x}}{x} + \frac{3}{x}}}{{\frac{{{x^2}}}{x} - \frac{x}{x} + \frac{1}{x}}} \cr
& = \frac{{3\left( {x + \frac{1}{x}} \right) - 4}}{{\left( {x + \frac{1}{x}} \right) - 1}} \cr
& = \frac{{3 \times 3 - 4}}{{3 - 1}} \cr
& = \frac{{9 - 4}}{2} \cr
& = \frac{5}{2} \cr} $$
$$\eqalign{
& x + \frac{1}{x} = 3 \cr
& \frac{{3{x^2} - 4x + 3}}{{{x^2} - x + 1}} \cr
& = \frac{{\frac{{3{x^2}}}{x} - \frac{{4x}}{x} + \frac{3}{x}}}{{\frac{{{x^2}}}{x} - \frac{x}{x} + \frac{1}{x}}} \cr
& = \frac{{3\left( {x + \frac{1}{x}} \right) - 4}}{{\left( {x + \frac{1}{x}} \right) - 1}} \cr
& = \frac{{3 \times 3 - 4}}{{3 - 1}} \cr
& = \frac{{9 - 4}}{2} \cr
& = \frac{5}{2} \cr} $$
Answer: Option B. -> 10
$$\because $$ x = 11
x5 - 12x4 + 12x3 - 12x2 + 12x - 1
= x5 - 11x4 - x4 + 11x3 + x3 - 11x2 - x2 + 11x + x - 1
= 115 - 11.114 - 114 + 11.113 + 113 - 11.112 - 112 + 11.11 + 11 - 1
= 0 - 0 + 0 + 0 + 11 - 1
= 10
$$\because $$ x = 11
x5 - 12x4 + 12x3 - 12x2 + 12x - 1
= x5 - 11x4 - x4 + 11x3 + x3 - 11x2 - x2 + 11x + x - 1
= 115 - 11.114 - 114 + 11.113 + 113 - 11.112 - 112 + 11.11 + 11 - 1
= 0 - 0 + 0 + 0 + 11 - 1
= 10
Answer: Option D. -> 3
$$\eqalign{
& {x^2} + {y^2} + {z^2} - xy - yz - zx \cr
& = \frac{1}{2}\left[ {{{\left( {x - y} \right)}^2} + {{\left( {y - z} \right)}^2} + {{\left( {z - x} \right)}^2}} \right] \cr} $$
$$ = \frac{1}{2}$$ $$\left[ {{{\left( {997 - 998} \right)}^2} + {{\left( {998 - 999} \right)}^2} + {{\left( {999 - 997} \right)}^2}} \right]$$
$$\eqalign{
& = \frac{1}{2}\left( {1 + 1 + 4} \right) \cr
& = 3 \cr} $$
$$\eqalign{
& {x^2} + {y^2} + {z^2} - xy - yz - zx \cr
& = \frac{1}{2}\left[ {{{\left( {x - y} \right)}^2} + {{\left( {y - z} \right)}^2} + {{\left( {z - x} \right)}^2}} \right] \cr} $$
$$ = \frac{1}{2}$$ $$\left[ {{{\left( {997 - 998} \right)}^2} + {{\left( {998 - 999} \right)}^2} + {{\left( {999 - 997} \right)}^2}} \right]$$
$$\eqalign{
& = \frac{1}{2}\left( {1 + 1 + 4} \right) \cr
& = 3 \cr} $$
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