Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

$$\eqalign{ & 999x + 888y = 1332 \cr & \underline {888x + 999y = 555} \cr & 1887\left( {x + y} \right) = 1887 \cr & \Rightarrow x + y = \frac{{1887}}{{1887}} \cr & \Rightarrow x + y = 1 \cr} $$

$$\eqalign{ & {\text{According to the question,}} \cr & {a^2} + {b^2} + {c^2} = ab + bc + ca \cr & {\text{Put }}a = 1 \cr & b = 1 \cr & c = 1 \cr & \therefore {a^2} + {b^2} + {c^2} = ab + bc + ca \cr & \Rightarrow {1^2} + {1^2} + {1^2} = 1 \times 1 + 1 \times 1 + 1 \times 1 \cr & \Rightarrow 1 + 1 + 1 = 1 + 1 + 1 \cr & \Rightarrow 3 = 3{\text{ }}\left( {{\text{Satisfy}}} \right) \cr & \therefore \frac{{a + c}}{b} \cr & = \frac{{1 + 1}}{1} \cr & = 2 \cr} $$

$$\eqalign{ & {\text{Let}} \cr & a = 0 \cr & b = 1 \cr & \Rightarrow {a^3} + {b^3} - ab - {\left( {{a^2} - {b^2}} \right)^2} \cr & \Rightarrow 0 + 1 - 0 - {\left( {0 - 1} \right)^2} \cr & \Rightarrow 1 - 1 \cr & \Rightarrow 0 \cr} $$

$$\eqalign{ & a - \frac{1}{{a - 3}} = 5 \cr & \Rightarrow a - 3 - \frac{1}{{a - 3}} = 5 - 3 \cr & \Rightarrow \left( {a - 3} \right) - \frac{1}{{\left( {a - 3} \right)}} = 2 \cr & {\text{Cubing both sides}} \cr & \Rightarrow {\left[ {\left( {a - 3} \right) - \frac{1}{{\left( {a - 3} \right)}}} \right]^3} = {\left( 2 \right)^3} \cr} $$
  $$ \Rightarrow {\left( {a - 3} \right)^3} - \frac{1}{{{{\left( {a - 3} \right)}^3}}} - 3 \times {\left( {a - 3} \right)} \times $$       $$\frac{1}{{{{\left( {a - 3} \right)}}}}$$ $$\left[ {\left( {a - 3} \right) - \frac{1}{{\left( {a - 3} \right)}}} \right]$$    $$ = 8$$
$$\eqalign{ & \Rightarrow {\left( {a - 3} \right)^3} - \frac{1}{{{{\left( {a - 3} \right)}^3}}} - 3\left( 2 \right) = 8 \cr & \Rightarrow {\left( {a - 3} \right)^3} - \frac{1}{{{{\left( {a - 3} \right)}^3}}} = 8 + 6 \cr & \Rightarrow {\left( {a - 3} \right)^3} - \frac{1}{{{{\left( {a - 3} \right)}^3}}} = 14 \cr} $$

$$\eqalign{ & \frac{{\left( {3x - 2y} \right)}}{{\left( {2x + 3y} \right)}} = \frac{5}{6} \cr & \Rightarrow 18x - 12y = 10x + 15y \cr & \Rightarrow 8x = 27y \cr & \Rightarrow \frac{x}{y} = \frac{{27}}{8} \cr & \therefore {\left( {\frac{{\root 3 \of x + \root 3 \of y }}{{\root 3 \of x - \root 3 \of y }}} \right)^2} \cr & \Rightarrow {\left( {\frac{{\root 3 \of {27} + \root 3 \of 8 }}{{\root 3 \of {27} - \root 3 \of 8 }}} \right)^2} \cr & \Rightarrow {\left( {\frac{{3 + 2}}{{3 - 2}}} \right)^2} \cr & \Rightarrow {\left( 5 \right)^2} \cr & \Rightarrow 25 \cr} $$

$$\eqalign{ & {\text{According to the question,}} \cr & x = \sqrt 3 + \sqrt 2 \cr & y = \sqrt 3 - \sqrt 2 \cr & \left( {{x^3} - 20\sqrt 2 } \right) - \left( {{y^3} + 2\sqrt 2 } \right) \cr & = \left[ {{{\left( {\sqrt 3 + \sqrt 2 } \right)}^3} - 20\sqrt 2 - {{\left( {\sqrt 3 - \sqrt 2 } \right)}^3} - 2\sqrt 2 } \right] \cr} $$
  $$ = 3\sqrt 3 + 2\sqrt 2 + 9\sqrt 2 + 6\sqrt 3 \, - $$       $$20\sqrt 2 \, - $$  $$3\sqrt 3 \,\, + $$  $$2\sqrt 2\,\, + $$  $$9\sqrt 2 \, - $$  $$6\sqrt 3\, - $$  $$2\sqrt 2 $$
$$\eqalign{ & = 9\sqrt 3 - 9\sqrt 2 - 9\sqrt 3 + 9\sqrt 2 \cr & = 0 \cr} $$

Always do these types of question with the help of,
$$\eqalign{ & {\text{Put }}a = b = c = 1 \cr & {\text{3}}\left( {{a^2} + {b^2} + {c^2}} \right) = {\left( {a + b + c} \right)^2} \cr & 3 = 3{\text{ Satisfied}} \cr & {\text{So, this is answer}} \to a = b = c \cr} $$

$$\eqalign{ & {\text{Let }}m = \sqrt {5 + \sqrt {5 + \sqrt 5 } } \cr & {\text{Factor}} = \left( a \right) \times \left( {a + 1} \right) \cr & {\text{Here }}m = a + 1 \cr & \Rightarrow m - 1 = a\,.........(i) \cr & {\text{Let }}n = \sqrt {5 - \sqrt {5 - \sqrt 5 } } \cr & {\text{Factor}} = \left( a \right) \times \left( {a + 1} \right) \cr & {\text{Here }}n = a\,.........(ii) \cr & {\text{From (i) & (ii)}} \cr & \Leftrightarrow m - 1 = n \cr & \Leftrightarrow m - n - 1 = 0 \cr} $$

$$\eqalign{ & {\text{According to the question,}} \cr & {\text{If ,}}2s = a + b + c \cr & \Leftrightarrow s = \frac{{a + b + c}}{2} \cr & {\text{Let, }} \cr & a = 10 \cr & b = 10 \cr & c = 10 \cr & \because s = \frac{{a + b + c}}{2} \cr & \Rightarrow s = \frac{{10 + 10 + 10}}{2} \cr & \Rightarrow s = \frac{{30}}{2} \cr & \Rightarrow s = 15 \cr & \therefore s\left( {s - c} \right) + \left( {s - a} \right)\left( {s - b} \right) \cr & = 15\left( {15 - 10} \right) + \left( {15 - 10} \right)\left( {15 - 10} \right) \cr & = 75 + 25 \cr & = 100 \cr & {\text{Now check from option, }} \cr & {\text{Option 'A' }}ab = 10 \times 10 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 100\left( {{\text{Satisfied}}} \right) \cr} $$

$$\eqalign{ & {x^m} \times {x^n} = 1 \cr & {x^{m + n}} = {x^0}\left[ {{x^0} = 1} \right] \cr & m + n = 0 \cr & m = - n \cr} $$