Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

$$\eqalign{ & 2x + 3y - 5z = 18\,......\left( 1 \right) \cr & 3x + 2y + z = 29\,......\left( 2 \right) \cr & x + y + 3z = 17\,......\left( 3 \right) \cr & {\text{Equation}}\left( 2 \right){\text{ and Equation}}\left( 1 \right),{\text{we get}} \cr & 3x + 2y + z = 29 \cr & \underline {2x + 3y - 5z = 18} \to \left( {{\text{Subtracting}}} \right) \cr & x - y + 6z = 11\,......\left( 4 \right) \cr & {\text{Adding Equation}}\left( 4 \right){\text{and Equation}}\left( 3 \right), \cr & 2x + 9z = 28\,......\left( 5 \right) \cr & {\text{And Equation}}\left( 2 \right){\text{and}}\,2 \times {\text{Equation}}\left( 3 \right), \cr & 3x + 2y + z = 29 \cr & \underline {2x + 2y + 6z = 34} \to \left( {{\text{Subtracting}}} \right) \cr & x - 5z = - 5\,......\left( 6 \right) \cr & {\text{Equation}}\left( 5 \right){\text{and}}\,2 \times {\text{Equation}}\left( 6 \right), \cr & 2x + 9z = 28 \cr & \underline {2x - 10z = - 10} \to \left( {{\text{Subtracting}}} \right) \cr & 19z = 38 \cr & \therefore z = 2 \cr & {\text{Now, from equation}}\left( {\text{6}} \right), \cr & x - 5 \times 2 = - 5 \cr & x = 10 - 5 \cr & x = 5 \cr & {\text{And from equation}}\left( {\text{3}} \right), \cr & 5 + y + 3 \times 2 = 17 \cr & y = 17 - 11 \cr & y = 6 \cr & \therefore xy + yz + zx \cr & = 5 \times 6 + 6 \times 2 + 5 \times 2 \cr & = 30 + 12 + 10 \cr & = 52 \cr} $$

$$\eqalign{ & x + y = 4,\,y + z = 5,\,z + x = 6 \cr & {\text{So, }}x + y + z = \frac{{15}}{2} \cr & {\left( {x - y} \right)^2} = {\left( {x + y} \right)^2} - 4xy \cr & = {4^2} - 4 \times 2 \cr & = 8 \cr & {\left( {y - z} \right)^2} = {\left( {y + z} \right)^2} - 4yz \cr & = {5^2} - 4 \times 3 \cr & = 13 \cr & {\left( {z - x} \right)^2} = {\left( {z + x} \right)^2} - 4zx \cr & = {6^2} - 4 \times 4 \cr & = 20 \cr & {x^3} + {y^3} + {z^3} - 3xyz \cr & = \frac{{\left( {x + y + z} \right)}}{2}\left[ {{{\left( {x - y} \right)}^2} + {{\left( {y - z} \right)}^2} + {{\left( {z - x} \right)}^2}} \right] \cr & = \frac{1}{2} \times \frac{{15}}{2}\left[ {8 + 13 + 20} \right] \cr & = \frac{{15 \times 41}}{4} \cr & = 153.75 \cr} $$

x = 2 - p
x + p = 2
Cube both side
(x + p)³ = 2³
x³ + p³ + 3xp(x + p) = 8
x³ + p³ + 2xp × 2 = 8
x³ + 6xp + p³ = 8

$$\eqalign{ & \frac{{{{\left( {{a^2} - 4{b^2}} \right)}^3} + 64{{\left( {{b^2} - 4{c^2}} \right)}^3} + {{\left( {16{c^2} - {a^2}} \right)}^3}}}{{{{\left( {a - 2b} \right)}^3} + {{\left( {2b - 4c} \right)}^3} + {{\left( {4c - a} \right)}^3}}} \cr & {\text{Put }}a = b = c \cr & = \frac{{{{\left( { - 3{a^2}} \right)}^3} + 64{{\left( { - 3{a^2}} \right)}^3} + {{\left( {15{a^2}} \right)}^3}}}{{{{\left( { - a} \right)}^3} + {{\left( { - 2a} \right)}^3} + {{\left( {3a} \right)}^3}}} \cr & = \frac{{{a^6}\left[ { - 27 - 27 \times 64 + {{\left( {15} \right)}^3}} \right]}}{{{a^3}\left[ { - 1 - 8 + 27} \right]}} \cr & = \frac{{3{a^3}\left[ { - 9 - 576 + 1125} \right]}}{{18}} \cr & = \frac{{{a^3} \times 540}}{6} \cr & = 90{a^3} \cr & {\text{From option put }}a = b = c \cr & \left( {\text{A}} \right)\, - 45{a^3} \cr & \left( {\text{B}} \right)\,90{a^3} \cr & {\text{Hence option B is right answer}}{\text{.}} \cr} $$

$$\eqalign{ & k + \frac{1}{k} = 2 \cr & k = - 1 \cr & \frac{{{k^2} + 4k - 2}}{{{k^2} + k - 5}} \cr & = \frac{{{{\left( { - 1} \right)}^2} + 4\left( { - 1} \right) - 2}}{{{{\left( { - 1} \right)}^2} + \left( { - 1} \right) - 5}} \cr & = \frac{{1 - 4 - 2}}{{1 - 1 - 5}} \cr & = \frac{{ - 5}}{{ - 5}} \cr & = 1 \cr} $$

$$\eqalign{ & a + b + c = 2 \cr & ab + bc + ca = 0 \cr & abc = 4 \cr & {a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left[ {\left( {{a^2} + {b^2} + {c^2}} \right) - \left( {ab + bc + ca} \right)} \right] \cr & 28 - 3 \times 4 = 2\left( {{a^2} + {b^2} + {c^2}} \right) \cr & {a^2} + {b^2} + {c^2} = 8 \cr} $$

$$\eqalign{ & x = 2 - {2^{\frac{1}{3}}} + {2^{\frac{2}{3}}} \cr & x - 2 = {2^{\frac{2}{3}}} - {2^{\frac{1}{3}}} \cr & {\left( {x - 2} \right)^3} = {\left( {{2^{\frac{2}{3}}} - {2^{\frac{1}{3}}}} \right)^3} \cr & {x^3} - 8 - 3 \times 2x\left( {x - 2} \right) = 4 - 2 - 3 \times {2^{\frac{2}{3}}} \times {2^{\frac{1}{3}}}\left( {{2^{\frac{2}{3}}} - {2^{\frac{1}{3}}}} \right) \cr & {x^3} - 8 - 6{x^2} + 12x = 4 - 2 - 6\left( {x - 2} \right) \cr & {x^3} - 8 - 6{x^2} + 12x = 2 - 6x + 12 \cr & {x^3} - 6{x^2} + 12x + 6x = 2 + 12 + 8 \cr & {x^3} - 6{x^2} + 18x = 22 \cr} $$

$$\eqalign{ & a + b + c + d = 2 \cr & {\text{Let }}a = b = c = d = \frac{1}{2} \cr & {\text{satisfy the above}} \cr & {\text{so, }}\left( {1 + a} \right)\left( {1 + b} \right)\left( {1 + c} \right)\left( {1 + d} \right) \cr & = \left( {1 + \frac{1}{2}} \right)\left( {1 + \frac{1}{2}} \right)\left( {1 + \frac{1}{2}} \right)\left( {1 + \frac{1}{2}} \right) \cr & = \frac{3}{2} \times \frac{3}{2} \times \frac{3}{2} \times \frac{3}{2} \cr & = \frac{{81}}{{16}} \cr} $$

$$\eqalign{ & 3a = 4b = 6c \cr & \Rightarrow \frac{{3a}}{{12}} = \frac{{4b}}{{12}} = \frac{{6c}}{{12}} \Rightarrow \frac{a}{4} = \frac{b}{3} = \frac{c}{2} = k \cr & \Rightarrow a = 4k,\,b = 3k,\,c = 2k \cr & a + b + c = 27\sqrt {29} \cr & 9k = 27\sqrt {29} \cr & k = 3\sqrt {29} \cr & a = 4 \times 3\sqrt {29} ,\,b = 3 \times 3\sqrt {29} ,\,c = 2 \times 3\sqrt {29} \cr & \sqrt {{a^2} + {b^2} + {c^2}} \cr & = \sqrt {29\left( {144 + 81 + 36} \right)} \cr & = \sqrt {29 \times 261} \cr & = \sqrt {29 \times 29 \times 9} \cr & = 29 \times 3 \cr & = 87 \cr} $$

$$\eqalign{ & \frac{{{x^2} + 1}}{x} = 4\frac{1}{4} \cr & x + \frac{1}{x} = \frac{{17}}{4} \cr & {\text{On cubing both sides,}} \cr & {\left( {x + \frac{1}{x}} \right)^3} = {\left( {\frac{{17}}{4}} \right)^3} \cr & {x^3} + \frac{1}{{{x^3}}} + 3 \times x \times \frac{1}{x}\left( {x + \frac{1}{x}} \right) = \frac{{4913}}{{64}} \cr & {x^3} + \frac{1}{{{x^3}}} + 3\left( {\frac{{17}}{4}} \right) = \frac{{4913}}{{64}} \cr & {x^3} + \frac{1}{{{x^3}}} = \frac{{4913}}{{64}} - \frac{{51}}{4} \cr & {x^3} + \frac{1}{{{x^3}}} = \frac{{4097}}{{64}} \cr} $$