Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

$$\eqalign{ & \frac{{3.6 \times 1.62 + 0.48 \times 3.6}}{{1.8 \times 0.8 + 10.8 \times 0.3 - 2.16}} \cr & = \frac{{\frac{1}{{1000}}\left[ {36 \times 162 + 48 \times 36} \right]}}{{\frac{1}{{100}}\left[ {18 \times 8 + 108 \times 3 - 216} \right]}} \cr & = \frac{1}{{10}}\left[ {\frac{{36\left( {162 + 48} \right)}}{{18\left( {8 + 18 - 12} \right)}}} \right] \cr & = \frac{1}{{10}}\left[ {\frac{{36 \times 210}}{{18 \times 14}}} \right] \cr & = 3 \cr} $$

By putting a = 8, b = 2
(8 + 2)² - 2(8 + 2) = 80
8 × 2 = 16
So, 3a - 19b
= 3 × 8 - 19 × 2
= -14

$$\eqalign{ & {x^2} - 3x + 1 = 0 \cr & {x^2} + 1 = 3x \cr & x + \frac{1}{x} = 3 \cr & \Rightarrow \frac{{\left( {{x^4} + \frac{1}{{{x^2}}}} \right)}}{{\left( {{x^2} + 5x + 1} \right)}} \cr & = \frac{{x\left( {{x^3} + \frac{1}{{{x^3}}}} \right)}}{{\left( {{x^2} + 1 + 5x} \right)}} \cr & = \frac{{x\left[ {{3^3} - 3 \times 3} \right]}}{{\left( {3x + 5x} \right)}} \cr & = \frac{{x\left[ {18} \right]}}{{\left( {8x} \right)}} \cr & = \frac{9}{4} \cr} $$

a + b + c = 8 and ab + bc + ca = 12
a³ + b³ + c³ - 3abc = ?
Let c = 0
a + b = 8, ab = 12
a³ + b³ = (a + b)(a² + b² - ab)
a³ + b³ = (a + b)[(a + b)² - 3ab]
a³ + b³ = 8[8² - 3 × 12]
a³ + b³ = 224

$$\eqalign{ & a - \frac{1}{a} = b\,........\left( {\text{i}} \right) \cr & b - \frac{1}{b} = c\,........\left( {{\text{ii}}} \right) \cr & c - \frac{1}{c} = a\,........\left( {{\text{iii}}} \right) \cr & {\text{Add Equation }}\left( {\text{i}} \right),\,\left( {{\text{ii}}} \right){\text{ and}}\left( {{\text{iii}}} \right) \cr & a + b + c - \left[ {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right] = a + b + c \cr & \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0\,........\left( {{\text{iv}}} \right) \cr & \Rightarrow a - \frac{1}{a} = b \cr & {\text{Squaring both sides}} \cr & {a^2} + \frac{1}{{{a^2}}} - 2 = {b^2}\,........\left( {\text{v}} \right) \cr & \Rightarrow b - \frac{1}{b} = c \cr & {b^2} + \frac{1}{{{b^2}}} - 2 = {c^2}\,........\left( {{\text{vi}}} \right) \cr & \Rightarrow c + \frac{1}{c} = a \cr & {c^2} + \frac{1}{{{c^2}}} - 2 = {a^2}\,........\left( {{\text{vii}}} \right) \cr & {\text{Add Equation }}\left( {\text{v}} \right),\,\left( {{\text{vi}}} \right){\text{ and}}\left( {{\text{vii}}} \right) \cr & {a^2} + {b^2} + {c^2} + \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} - 2 - 2 - 2 = {a^2} + {b^2} + {c^2} \cr & \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} = 6\,........\left( {{\text{viii}}} \right) \cr & \Rightarrow \left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right) \cr & = \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} + \frac{2}{{ab}} + \frac{2}{{bc}} + \frac{2}{{ca}} \cr & {0^2} = 6 + 2\left[ {\frac{1}{{ab}} + \frac{1}{{bc}} + \frac{1}{{ca}}} \right] \cr & \frac{1}{{ab}} + \frac{1}{{bc}} + \frac{1}{{ca}} = \frac{{ - 6}}{2} = - 3 \cr} $$

x + y + z = 22
xy + yz + zx = 35
(x + y + z)² = x² + y² + z² + 2(xy + yz + zx)
(22)² = x² + y² + z² + 2 × 35
484 - 70 = x² + y² + z²
x² + y² + z² = 414
(x - y)² + (y - z)² + (z - x)²
= 2(x² + y² + z² - xy - yz - zx)
= 2(414 - 35)
= 2 × 379
= 758

$$\eqalign{ & {x^2} + \frac{1}{{{x^2}}} = \frac{{31}}{9} \cr & x + \frac{1}{x} = \sqrt {\frac{{31}}{9} + 2} \cr & x + \frac{1}{x} = \frac{7}{3} \cr & {x^3} + \frac{1}{{{x^3}}} = {n^3} - 3n \cr & {x^3} + \frac{1}{{{x^3}}} = {\left( {\frac{7}{3}} \right)^3} - 3 \times \frac{7}{3} \cr & {x^3} + \frac{1}{{{x^3}}} = \frac{{343}}{{27}} - 7 \cr & {x^3} + \frac{1}{{{x^3}}} = \frac{{343 - 7 \times 27}}{{27}} \cr & {x^3} + \frac{1}{{{x^3}}} = \frac{{154}}{{27}} \cr} $$

$$\eqalign{ & {x^4} + \frac{1}{{{x^4}}} = 194, \cr & {x^4} + \frac{1}{{{x^4}}} + 2 = 194 + 2 \cr & {x^2} + \frac{1}{{{x^2}}} + 2 = 16 \cr & x + \frac{1}{x} = 4 \cr & {x^2} + 1 = 4x \cr & {x^2} - 4x = - 1 \cr & {x^2} - 4x + 1 = 0 \cr & {x^2} - 4x + 4 - 3 = 0 \cr & {x^2} - 4x + 4 = 3 \cr & {\left( {x - 2} \right)^2} = 3 \cr} $$

$$\eqalign{ & {\left[ {a + \frac{1}{a}} \right]^2} - 2\left[ {a - \frac{1}{a}} \right] = 12 \cr & {\left[ {a - \frac{1}{a}} \right]^2} + 4 - 2\left[ {a - \frac{1}{a}} \right] = 12 \cr & {\left[ {a - \frac{1}{a}} \right]^2} - 2\left[ {a - \frac{1}{a}} \right] - 8 = 0 \cr & {\text{Let }}a - \frac{1}{a} = x \cr & \therefore \,{x^2} - 2x - 8 = 0 \cr & {x^2} - 4x + 2x - 8 = 0 \cr & \left( {x - 4} \right)\left( {x + 2} \right) = 0 \cr & x = 4,\,x = - 2 \cr & a - \frac{1}{a} = - 2,\,a - \frac{1}{a} = 4 \cr & \therefore \,a + \frac{1}{a} = \sqrt {{4^2} + 4} \cr & a + \frac{1}{a} = 2\sqrt 5 \cr & a - \frac{1}{a} = 4 \cr & \therefore \,2a = 2\sqrt 5 + 4 \cr & a = 2 + \sqrt 5 \cr} $$

$$\eqalign{ & \sqrt x - \frac{1}{{\sqrt x }} = \sqrt 5 \cr & {\text{Square both side}} \cr & x + \frac{1}{x} - 2.x.\frac{1}{x} = 5 \cr & x + \frac{1}{x} = 5 + 2 \cr & {\text{Square both side}} \cr & {x^2} + \frac{1}{{{x^2}}} + 2.x.\frac{1}{x} = 49 \cr & {x^2} + \frac{1}{{{x^2}}} = 47 \cr} $$