Practice MCQ Questions and Answer on Algebra

Solution:

 $$\eqalign{ & {a^3} + {b^3} = 218\,\& \,a + b = 2 \cr & {\left( {a + b} \right)^3} = {\left( 2 \right)^3} \cr & {a^3} + {b^3} + 3\left( {a + b} \right)\left( {ab} \right) = 8 \cr & 218 + 3\left( 2 \right)\left( {ab} \right) = 8 \cr & ab = \frac{{8 - 218}}{6} \cr & ab = \frac{{ - 210}}{6} = - 35 \cr & \sqrt {1 - ab} = \sqrt {1 - \left( { - 35} \right)} \cr & \sqrt {1 - ab} = \sqrt {1 + \left( {35} \right)} \cr & \sqrt {1 - ab} = 6 \cr} $$

Solution:

 $$\eqalign{ & {x^2} - 3x + 1 = 0 \cr & {x^2} + 1 = 3x \cr & {\text{Divide by }}x \cr & \frac{{{x^2}}}{x} + \frac{1}{x} = \frac{{3x}}{x} \cr & \Rightarrow x + \frac{1}{x} = 3 \cr & {\text{Cubing both sides}} \cr & \Rightarrow {x^3}{\text{ + }}\frac{1}{{{x^3}}} + 3x \times \frac{1}{x}\left( {x + \frac{1}{x}} \right) = 27 \cr & \Rightarrow {x^3}{\text{ + }}\frac{1}{{{x^3}}} + 3 \times 3 = 27 \cr & \Rightarrow {x^3}{\text{ + }}\frac{1}{{{x^3}}} = 18 \cr} $$

Solution:

 $$\eqalign{ & {a^3} + {b^3} + {c^3} = 3abc \cr & {\text{If }}a + b + c = 0 \cr & a = 0.324 \cr & b = 0.221 \cr & c = - 0.545 \cr & \frac{{\left( {0.545} \right)\left( {0.081} \right)\left( {0.51} \right)\left( {5.2} \right)}}{{3abc}} \cr & = - \frac{{0.545 \times 0.081 \times 0.51 \times 5.2}}{{3 \times 0.324 \times 0.221 \times 0.545}} \cr & = - \frac{{81 \times 510 \times 5.2}}{{3 \times 18 \times 18 \times 13 \times 17}} \cr & = - 1 \cr} $$

Solution:

 $$\eqalign{ & x \propto \frac{1}{{{y^2}}} \cr & \left( {{\text{Inversely proportional}}} \right) \cr & x = \frac{k}{{{y^2}}} \cr & \left( {{\text{Given}}} \right), \cr & \left( {y = 2} \right){\text{ for }}\left( {x = 1} \right) \cr & \therefore 1 = \frac{k}{{{{\left( 2 \right)}^2}}} \cr & \Rightarrow 1 = \frac{k}{4} \cr & \Rightarrow k = 4 \cr & \therefore {\text{For }}y = 6 \cr & x = \frac{4}{{{{\left( 6 \right)}^2}}} \cr & x = \frac{4}{{36}} \cr & x = \frac{1}{9} \cr} $$

Solution:

 Given, a + b + c = -11 By putting value i.e. a = -4 b = -5 c = -2 Therefore, (a + 4)3 + (b + 5)3 + (c + 2)3 - 3(a + 4)(b + 5)(c + 2) = 0 + 0 + 0 - 0 = 0

Solution:

 $$\left( {1 - \frac{{2xy}}{{{x^2} + {y^2}}}} \right) \div \left( {\frac{{{x^3} - {y^3}}}{{x - y}} - 3xy} \right)$$ $$ = \left( {\frac{{{x^2} + {y^2} - 2xy}}{{{x^2} + {y^2}}}} \right) \div $$    $$\left( {\frac{{{x^3} - {y^3} - 3xy\left( {x - y} \right)}}{{x - y}}} \right)$$ $$\eqalign{ & = \frac{{{{\left( {x - y} \right)}^2}}}{{{x^2} + {y^2}}} \div \frac{{{{\left( {x - y} \right)}^3}}}{{x - y}} \cr & = \frac{{{{\left( {x - y} \right)}^2}}}{{{x^2} + {y^2}}} \div {\left( {x - y} \right)^2} \cr & = \frac{{{{\left( {x - y} \right)}^2}}}{{{x^2} + {y^2}}} \times \frac{1}{{{{\left( {x - y} \right)}^2}}} \cr & = \frac{1}{{{x^2} + {y^2}}} \cr} $$

Solution:

 a = 55 b = 17 c = -72 a + b + c = 55 + 17 - 72 = 0 ∴ a3 + b3 + c3 - 3abc = 0 (a + b + c) = 0 Answer = 0

Solution:

 $$\eqalign{ & x + y + z = 6 \cr & {x^2} + {y^2} + {z^2} = 20 \cr & \Rightarrow {\left( {x + y + z} \right)^2} = {\left( 6 \right)^2} \cr & \Rightarrow {x^2} + {y^2} + {z^2} + 2\left( {xy + yz + zx} \right) = 36 \cr & \Rightarrow 20 + 2\left( {xy + yz + zx} \right) = 36 \cr & \Rightarrow 2\left( {xy + yz + zx} \right) = 16 \cr & \Rightarrow xy + yz + zx = 8 \cr & \therefore {\text{ }}{x^3} + {y^3} + {z^3} - 3xyz \cr & = \left( {x + y + z} \right)\left( {{\text{ }}{x^2} + {y^2} + {z^2} - xy - zx - yz} \right) \cr & = {x^3} + {y^3} + {z^3} - 3xyz = 6\left( {20 - 8} \right) \cr & = 6\left( {20 - 8} \right) \cr & = 6 \times 12 \cr & = 72 \cr} $$

Solution:

 $$a - \frac{1}{{a - 5}} = 10$$ We can write it is as (subtracting 5 from both sides) $$\eqalign{ & a - 5 - \frac{1}{{a - 5}} = 10 - 5 \cr & {\text{Now, }}\left( {a - 5} \right) - \frac{1}{{\left( {a - 5} \right)}} = 5 \cr} $$ So, take the cube of this equation, $$\eqalign{ & \left[ {{{\left( {a - 5} \right)}^3} - \frac{1}{{{{\left( {a - 5} \right)}^3}}}} \right] = 125 + 3 \times 5 \cr & \left[ {{{\left( {a - 5} \right)}^3} - \frac{1}{{{{\left( {a - 5} \right)}^3}}}} \right] = 140 \cr} $$

Solution:

 $$\eqalign{ & x = 3 + 2\sqrt 2 \cr & \Rightarrow x = 2 + 1 + 2\sqrt 2 \cr & \Rightarrow x = {\left( {\sqrt 2 + 1} \right)^2} \cr & \Rightarrow \sqrt x = \sqrt 2 + 1 \cr & \Rightarrow \frac{1}{{\sqrt x }} = \frac{1}{{\sqrt 2 + 1}} \cr & \Rightarrow \frac{1}{{\sqrt x }} = \frac{1}{{\sqrt 2 + 1}} \times \frac{{\sqrt 2 - 1}}{{\sqrt 2 - 1}} \cr & \Rightarrow \frac{1}{{\sqrt x }} = \sqrt 2 - 1 \cr & \therefore \sqrt x - \frac{1}{{\sqrt x }} \cr & = \sqrt 2 + 1 - \left( {\sqrt 2 - 1} \right) \cr & = \sqrt 2 + 1 - \sqrt 2 + 1 \cr & = 2 \cr} $$