Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

$$\eqalign{ & \frac{{5.35 \times 5.35 \times 5.35 + 3.65 \times 3.65 \times 3.65}}{{53.5 \times 53.5 + 36.5 \times 36.5 - 53.5 \times 36.5}} \cr & = \frac{{{{5.35}^3} + {{3.65}^3}}}{{{{53.5}^2} + {{36.5}^2} - 53.5 \times 36.5}} \cr & = \frac{{\left( {5.35 + 3.65} \right)\left[ {{{\left( {5.35} \right)}^2} + {{\left( {3.65} \right)}^2} - 5.35 \times 3.65} \right]}}{{{{10}^2}\left[ {{{\left( {5.35} \right)}^2} + {{\left( {3.65} \right)}^2} - 5.35 \times 3.65} \right]}} \cr & = \frac{9}{{{{10}^2}}} \cr & = 0.09 \cr} $$

$$\eqalign{ & f\left( x \right) = 2{x^3} + a{x^2} + bx - 2 \cr & x + 2 = 0 \cr & x = - 2 \cr & f\left( { - 2} \right) = 2{\left( { - 2} \right)^3} + a{\left( { - 2} \right)^2} + b\left( { - 2} \right) - 2 = 0 \cr & \Rightarrow - 16 + 4a - 2b - 2 = 0 \cr & \Rightarrow 4a - 2b - 18 = 0 \cr & {\text{Now go through option - }} \cr & {\text{from option A}} \cr & 4 \times 3 - 2\left( { - 3} \right) - 18 = 0 \cr & 0 = 0{\text{ satisfy}} \cr} $$

$$\eqalign{ & {x^2} - 3x - 1 = 0 \cr & x\left( {x - 3 - \frac{1}{x}} \right) = 0 \cr & x - \frac{1}{x} = 3 \cr & {x^2} + \frac{1}{{{x^2}}} = 11 \cr & \frac{{\left( {{x^2} + 8x - 1} \right)}}{{{x^3} + \frac{1}{x}}} \cr & = \frac{{x\left( {\frac{{x - 1}}{{x + 8}}} \right)}}{{x\left( {\frac{{{x^2} + 1}}{{{x^2}}}} \right)}} \cr & = \frac{{3 + 8}}{{11}} \cr & = 1 \cr} $$

$$\eqalign{ & a + b + c = 3 \cr & {\text{Put }}a = 4,\,b = - 1,\,c = 0 \cr & \frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)}} + \frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)}} + \frac{1}{{\left( {1 - c} \right)\left( {1 - a} \right)}} \cr & = \frac{1}{{ - 3 \times 2}} + \frac{1}{{2 \times 1}} - \frac{1}{3} \cr & = \frac{1}{2} - \frac{1}{3} - \frac{1}{6} \cr & = \frac{0}{6} \cr & = 0 \cr} $$

$$\eqalign{ & A = \frac{{{x^8} - 1}}{{{x^4} + 1}} = \frac{{\left( {{x^4} + 1} \right)\left( {{x^4} - 1} \right)}}{{\left( {{x^4} + 1} \right)}} = {x^4} - 1 \cr & B = \frac{{{y^4} - 1}}{{{y^2} + 1}} = \frac{{\left( {{y^2} + 1} \right)\left( {{y^2} - 1} \right)}}{{\left( {{y^2} + 1} \right)}} = {y^2} - 1 \cr & x = 2,\,\,y = 9 \cr & A = {x^4} - 1 = {\left( 2 \right)^4} - 1 = 15 \cr & B = {y^2} - 1 = {\left( 9 \right)^2} - 1 = 80 \cr & {A^2} + 2AB + A{B^2} \cr & = {\left( {15} \right)^2} + 2 \times 15 \times 80 + 15 \times {\left( {80} \right)^2} \cr & = 15\left( {15 + 160 + 6400} \right) \cr & = 15 \times \left( {6575} \right) \cr & = 98625 \cr} $$

$$\eqalign{ & {x^2} - 12x + 33 = 0 \cr & {x^2} - 12x + 36 - 3 = 0 \cr & {\left( {x - 6} \right)^2} - 3 = 0 \cr & x = \sqrt 3 + 6 \cr & {\left( {x - 4} \right)^2} + \frac{1}{{{{\left( {x - 4} \right)}^2}}} \cr & \therefore {\left( {\sqrt 3 + 2} \right)^2} + \frac{1}{{{{\left( {\sqrt 3 + 2} \right)}^2}}} \cr & = {\left( {\sqrt 3 + 2} \right)^2} + {\left( {\sqrt 3 - 2} \right)^2} \cr & = 2\left( {{{\sqrt 3 }^2} + {2^2}} \right) \cr & = 2 \times 7 \cr & = 14 \cr} $$

a + b = 8
a + a²b + b + ab² = 128
(a + b) + ab(a + b) = 128
(1 + ab)(a + b) = 128
1 + ab = 16
ab = 15
a³ + b³
= (a + b)³ - 3ab(a + b)
= 512 - 3 × 15 × 8
= 512 - 360
= 152

x² + 16 = -4x
x³ - 64
= x³ - (4)³
= (x - 4)(x² + 16 + 4x)
= (x - 4)(-4x + 4x)
= 0

x + y = 1
put y = 0
x = 1
x³ + 3xy + y³
= 1³ + 0 + 0
= 1

$$\eqalign{ & \frac{{{x^4} - \frac{{27}}{{{x^2}}}}}{{{x^2} - 3x - 3}} \cr & = \frac{{x\left( {{x^3} - \frac{{27}}{{{x^3}}}} \right)}}{{x\left( {x - 3 - \frac{3}{x}} \right)}}.....\left( {\text{i}} \right) \cr & {x^3} - \frac{{27}}{{{x^3}}} \cr & = {6^3} + 3 \times 3 \times 6 \cr & = 216 + 54 \cr & = 270 \cr & \frac{{{x^3} - \frac{{27}}{{{x^3}}}}}{{x - \frac{3}{x} - 3}} = \frac{{270}}{{6 - 3}} = 90 \cr} $$