Practice MCQ Questions and Answer on Algebra

Solution:

 $$\eqalign{ & x + y = 15 \cr & \Rightarrow x - 10 = 5 - y \cr & \Rightarrow x - 10 = - \left( {y - 5} \right) \cr & {\text{Take cube on both sides}} \cr & \Rightarrow {\left( {x - 10} \right)^3} = - {\left( {y - 5} \right)^3} \cr & \Rightarrow {\left( {x - 10} \right)^3} + {\left( {y - 5} \right)^3} = 0 \cr} $$

Solution:

 $$\eqalign{ & {\text{ }}x + \frac{1}{x} = 5 \cr & {\text{Take cube on both sides}} \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^3} = {\left( 5 \right)^3} \cr & \Rightarrow {x^3}{\text{ + }}\frac{1}{{{x^3}}} + 3 \times 5 = 125 \cr & \Rightarrow {x^3}{\text{ + }}\frac{1}{{{x^3}}} = 110 \cr & \therefore {\text{Squaring both sides}} \cr & \Rightarrow {\left( {{x^3}{\text{ + }}\frac{1}{{{x^3}}}} \right)^2} = {\left( {110} \right)^2} \cr & \Rightarrow {x^6}{\text{ + }}\frac{1}{{{x^6}}} + 2 = 12100 \cr & \Rightarrow {x^6}{\text{ + }}\frac{1}{{{x^6}}} = 12100 - 2 \cr & \Rightarrow {x^6}{\text{ + }}\frac{1}{{{x^6}}} = 12098 \cr} $$

Solution:

 $${p^3} - {q^3} = \left( {p - q} \right)\left\{ {{p^2} + {q^2} + pq} \right\}$$ $$ \Rightarrow \left( {p - q} \right)\left\{ {{{\left( {p + q} \right)}^2} - xpq} \right\} = $$       $$\left( {p - q} \right)$$ $$\left( {{p^2} + {q^2} + pq} \right)$$ $$\eqalign{ & \Rightarrow {p^2} + {q^2} + 2pq - xpq = {p^2} + {q^2} + pq \cr & \Rightarrow 2pq - pq = xpq \cr & \Rightarrow pq = xpq \cr & \Rightarrow x = 1 \cr} $$

Solution:

 $$\eqalign{ & {\text{Let the numbers are a, b}} \cr & {a^2} + {b^2} = 41 \cr & a + b = 9 \cr & \Rightarrow {\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab \cr & \Rightarrow {9^2} = 41 + 2ab \cr & \Rightarrow 81 - 41 = 2ab \cr & \Rightarrow ab = 20 \cr & {\text{Take }} \cr & a = 5 \cr & b = 4 \cr & \Rightarrow {a^3} + {b^3} = {5^3} + {4^3} \cr & \Rightarrow {a^3} + {b^3} = 125 + 64 \cr & \Rightarrow {a^3} + {b^3} = 189 \cr} $$

Solution:

 $$\eqalign{ & \frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}} + \frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}} + \frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}} + \frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}} \cr & {\text{Put }} \cr & a = 0 \cr & b = 0 \cr & c = 2 \cr & d = 2 \cr & \therefore a + b + c + d = 4 \cr & \Rightarrow 0 + 0 + 2 + 2 = 4 \cr & \Rightarrow 4 = 4\left( {{\text{ satisfy}}} \right) \cr & \frac{1}{{\left( {1 - 0} \right)\left( {1 - 0} \right)\left( {1 - 2} \right)}} + \frac{1}{{\left( {1 - 0} \right)\left( {1 - 2} \right)\left( {1 - 2} \right)}} + \frac{1}{{\left( {1 - 2} \right)\left( {1 - 2} \right)\left( {1 - 0} \right)}} + \frac{1}{{\left( {1 - 2} \right)\left( {1 - 0} \right)\left( {1 - 0} \right)}} \cr & = \frac{1}{{ - 1}} + \left( {\frac{1}{{ + 1}}} \right) + \frac{1}{{ - 1 \times - 1}} + \frac{1}{{ - 1}} \cr & = - 1 + 1 + 1 - 1 \cr & = 0 \cr} $$

Solution:

 $$\eqalign{ & \frac{{2 + a}}{a} + \frac{{2 + b}}{b} + \frac{{2 + c}}{c} = 4 \cr & \Rightarrow \frac{{2bc + abc + 2ac + abc + 2ab + abc}}{{abc}} = 4 \cr & \Rightarrow \frac{{2\left( {bc + ab + ca} \right)}}{{abc}} + \frac{{3abc}}{{abc}} = 4 \cr & \Rightarrow \frac{{2\left( {bc + ab + ca} \right)}}{{abc}} + 3 = 4 \cr & \Rightarrow \boxed{\frac{{ab + bc + ca}}{{abc}} = \frac{1}{2}} \cr} $$

Solution:

 $$\eqalign{ & \sqrt {1 - \frac{{{x^3}}}{{100}}} = \frac{3}{5} \cr & \Rightarrow 1 - \frac{{{x^3}}}{{100}} = {\left( {\frac{3}{5}} \right)^2} \cr & \Rightarrow 1 - \frac{9}{{25}} = \frac{{{x^3}}}{{100}} \cr & \Rightarrow \frac{{16}}{{25}} = \frac{{{x^3}}}{{100}} \cr & \Rightarrow {x^3} = \frac{{16 \times 100}}{{25}} \cr & \Rightarrow {x^3} = 16 \times 4 \cr & \Rightarrow {x^3} = 64 \cr & \Rightarrow \boxed{x = 4} \cr} $$

Solution:

 $$\eqalign{ & \because x = \sqrt 3 + \frac{1}{{\sqrt 3 }} \cr & \Rightarrow x = \frac{{3 + 1}}{{\sqrt 3 }} \cr & \Rightarrow x = \frac{4}{{\sqrt 3 }} \cr & {\text{So, }}\left( {x - \frac{{\sqrt {126} }}{{\sqrt {42} }}} \right)\left( {x - \frac{1}{{x - \frac{{2\sqrt 3 }}{3}}}} \right) \cr & = \left( {\frac{4}{{\sqrt 3 }} - \frac{{\sqrt {126} }}{{\sqrt {42} }}} \right)\left( {\frac{4}{{\sqrt 3 }} - \frac{1}{{\frac{4}{{\sqrt 3 }} - \frac{2}{{\sqrt 3 }}}}} \right) \cr & = \left( {\frac{{4\sqrt {42} - \sqrt {126} \times \sqrt 3 }}{{\sqrt {3 \times } \sqrt {42} }}} \right)\left( {\frac{4}{{\sqrt 3 }} - \frac{1}{{\frac{2}{{\sqrt 3 }}}}} \right) \cr & = \frac{{4\sqrt {42} - 3\sqrt {42} }}{{\sqrt {3 \times } \sqrt {42} }}\left( {\frac{4}{{\sqrt 3 }} - \frac{{\sqrt 3 }}{2}} \right) \cr & = \left( {\frac{{\sqrt {42} }}{{\sqrt {3 \times } \sqrt {42} }}} \right)\left( {\frac{{8 - 3}}{{2\sqrt 3 }}} \right) \cr & = \frac{1}{{\sqrt 3 }} \times \frac{5}{{2\sqrt 3 }} \cr & = \frac{5}{6} \cr} $$

Solution:

 $$\eqalign{ & a = 4.36 \cr & b = 2.39 \cr & c = 1.97 \cr & {\text{ }}a - b - c \cr & = 4.36 - 2.39 - 1.97 \cr & = 0 \cr & {\text{ }}{{\text{a}}^3} - {b^3} - {c^3} - 3abc \cr & = \frac{1}{2}\left( {a - b - c} \right)\left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right] \cr & = 0 \cr} $$

Solution:

 $$\eqalign{ & A = \frac{{0.216 + 0.008}}{{0.36 + 0.04 - 0.12}} \cr & = \frac{{{{\left( {0.6} \right)}^3} + {{\left( {0.2} \right)}^3}}}{{{{\left( {0.6} \right)}^2} + {{\left( {0.2} \right)}^2} - \left( {0.6} \right)\left( {0.2} \right)}} \cr & = \frac{{\left( {0.6 + 0.2} \right)\left\{ {{{\left( {0.6} \right)}^2} + {{\left( {0.2} \right)}^2} - \left( {0.6} \right)\left( {0.2} \right)} \right\}}}{{\left\{ {{{\left( {0.6} \right)}^2} + {{\left( {0.2} \right)}^2} - \left( {0.6} \right)\left( {0.2} \right)} \right\}}} \cr & = 0.8 \cr & B = \frac{{0.729 - 0.027}}{{0.81 + 0.09 + 0.27}} \cr & = \frac{{{{\left( {0.9} \right)}^3} - {{\left( {0.3} \right)}^3}}}{{{{\left( {0.9} \right)}^2} + {{\left( {0.3} \right)}^2} + \left( {0.9} \right)\left( {0.3} \right)}} \cr & = \frac{{\left( {0.9 - 0.3} \right)\left\{ {{{\left( {0.9} \right)}^2} + {{\left( {0.3} \right)}^2} + \left( {0.9} \right)\left( {0.3} \right)} \right\}}}{{\left\{ {{{\left( {0.9} \right)}^2} + {{\left( {0.3} \right)}^2} + \left( {0.9} \right)\left( {0.3} \right)} \right\}}} \cr & = 0.6 \cr & \therefore \,{\left( {{A^2} + {B^2}} \right)^2} \cr & = {\left[ {{{\left( {0.8} \right)}^2} + {{\left( {0.6} \right)}^2}} \right]^2} \cr & = {\left( {0.64 + 0.36} \right)^2} \cr & = {\left( 1 \right)^2} \cr & = 1 \cr} $$