Algebra MCQ Questions and Solutions with Explanations | Math Aptitude

$$\eqalign{ & y:x = 4:15 \cr & \therefore \frac{y}{x} = \frac{4}{{15}} \cr & \therefore \frac{{x - y}}{{x + y}} \cr & \Rightarrow \frac{{x\left( {1 - \frac{y}{x}} \right)}}{{x\left( {1 + \frac{y}{x}} \right)}} \cr & {\text{Taking }}x{\text{ common}} \cr & \Rightarrow \frac{{1 - \frac{4}{{15}}}}{{1 + \frac{4}{{15}}}} \cr & \Rightarrow \frac{{11}}{{15}} \times \frac{{15}}{{19}} \cr & \Rightarrow \frac{{11}}{{19}} \cr} $$

$$\eqalign{ & {\left( {x - 3} \right)^2}{\text{ + }}{\left( {y - 5} \right)^2}{\text{ + }}{\left( {z - 4} \right)^2} = 0 \cr & \therefore {\left( {x - 3} \right)^2} = 0{\text{ }}x = 3 \cr & {\left( {y - 5} \right)^2} = 0{\text{ }}y = 5 \cr & {\left( {z - 4} \right)^2} = 0{\text{ }}z = 4{\text{ }} \cr & \therefore \frac{{{x^2}}}{9}{\text{ + }}\frac{{{y^2}}}{{25}}{\text{ + }}\frac{{{z^2}}}{{16}} \cr & \Rightarrow \frac{9}{9} + \frac{{25}}{{25}} + \frac{{16}}{{16}} \cr & \Rightarrow 3 \cr} $$

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

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

$$\eqalign{ & \frac{{4x}}{3} + {\text{2P}} = 12 \cr & \left[ {x = 6\left( {{\text{Given}}} \right)} \right]\,\, \cr & \Rightarrow \frac{{4 \times 6}}{3} + 2{\text{P}} = 12 \cr & \Rightarrow 2{\text{P}} = 12 - 8 \cr & \Rightarrow {\text{P}} = \frac{4}{2} \cr & \Rightarrow {\text{P}} = 2 \cr} $$

$$\frac{{4 + 3\sqrt 3 }}{{7 + 4\sqrt 3 }}$$
(By rationalization of denominator)
$$\eqalign{ & = \frac{{4 + 3\sqrt 3 }}{{7 + 4\sqrt 3 }} \times \frac{{7 - 4\sqrt 3 }}{{7 - 4\sqrt 3 }} \cr & = \frac{{\left( {4 + 3\sqrt 3 } \right)\left( {7 - 4\sqrt 3 } \right)}}{{49 - 48}} \cr} $$

$$\eqalign{ & a = \sqrt 6 - \sqrt 5 \cr & b = \sqrt 5 - 2 \cr & c = 2 - \sqrt 3 \cr & {\text{Rationalize all the terms}} \cr & a = \frac{{\sqrt 6 - \sqrt 5 }}{{\sqrt 6 + \sqrt 5 }} \times \sqrt 6 + \sqrt 5 \cr & \,\,\,\,\, = \frac{1}{{\sqrt 6 + \sqrt 5 }} \cr & b = \frac{{\sqrt 5 - 2}}{{\sqrt 5 + 2}} \times \sqrt 5 + 2 \cr & \,\,\,\,\,\, = \frac{1}{{\sqrt 5 + 2}} \cr & \,\,\,\,\,\, = \frac{1}{{\sqrt 5 + \sqrt 4 }} \cr & c = \frac{{2 - \sqrt 3 }}{{2 + \sqrt 3 }} \times 2 + \sqrt 3 \cr & \,\,\,\,\,\, = \frac{1}{{2 + \sqrt 3 }} \cr & \,\,\,\,\,\, = \frac{1}{{\sqrt 4 + \sqrt 3 }} \cr & {\text{Now, }} \cr & a = \frac{1}{{\sqrt 6 + \sqrt 5 }} \cr & b = \frac{1}{{\sqrt 5 + \sqrt 4 }} \cr & c = \frac{1}{{\sqrt 4 + \sqrt 3 }} \cr} $$
Now the term whose denominator is largest is the smallest term.
So, a < b < c

$$\eqalign{ & x = 5 - \sqrt {21} \cr & 2x = 10 - 2\sqrt {21} \,......(i) \cr & \Rightarrow 2x = {\left( {\sqrt 7 } \right)^2} + {\left( {\sqrt 3 } \right)^2} - 2\left( {\sqrt 7 } \right)\left( {\sqrt 3 } \right) \cr & \Rightarrow 2x = {\left( {\sqrt 7 - \sqrt 3 } \right)^2} \cr & \Rightarrow x = \frac{1}{2}{\left( {\sqrt 7 - \sqrt 3 } \right)^2} \cr & \Rightarrow \sqrt x = \frac{1}{{\sqrt 2 }}\sqrt {{{\left( {\sqrt 7 - \sqrt 3 } \right)}^2}} \cr & \Rightarrow \sqrt x = \frac{1}{{\sqrt 2 }}\left( {\sqrt 7 - \sqrt 3 } \right) \cr & \therefore \frac{{\sqrt x }}{{\sqrt {32 - 2x} - \sqrt {21} }} \cr & = \frac{{\sqrt 7 - \sqrt 3 }}{{\sqrt 2 \left[ {\sqrt {32 - \left( {10 - 2\sqrt {21} } \right)} - \sqrt {21} } \right]}} \cr & = \frac{{\sqrt 7 - \sqrt 3 }}{{\sqrt 2 \left[ {\sqrt {22 + 2\sqrt {21} } - \sqrt {21} } \right]}} \cr & = \frac{{\sqrt 7 - \sqrt 3 }}{{\sqrt 2 \left[ {\sqrt {{{\left( {\sqrt {21} + 1} \right)}^2}} - \sqrt {21} } \right]}} \cr & = \frac{{\sqrt 7 - \sqrt 3 }}{{\sqrt 2 \left[ {\sqrt {21} + 1 - \sqrt {21} } \right]}} \cr & = \frac{{\sqrt 7 - \sqrt 3 }}{{\sqrt 2 }} \cr & = \frac{1}{{\sqrt 2 }}\left( {\sqrt 7 - \sqrt 3 } \right) \cr} $$

$$\eqalign{ & {\left( {{x^{b + c}}} \right)^{b - c}} \times {\left( {{x^{c + a}}} \right)^{c - a}} \times {\left( {{x^{a + b}}} \right)^{a - b}}\left( {x \ne 0} \right) \cr & = {x^{{b^2} - {c^2}}} \times {x^{{c^2} - {a^2}}} \times {x^{{a^2} - {b^2}}} \cr & = {x^{{b^2} - {c^2} + {c^2} - {a^2} + {a^2} - {b^2}}} \cr & = {x^0} \cr & = 1 \cr} $$

$$\eqalign{ & \frac{x}{a} = \frac{1}{a} - \frac{1}{x} \cr & \Rightarrow \frac{x}{a} - \frac{1}{a} = - \frac{1}{x} \cr & \Rightarrow \left( {\frac{{x - 1}}{a}} \right) = - \frac{1}{x} \cr & \Rightarrow \frac{{1 - x}}{a} = \frac{1}{x} \cr & \Rightarrow x\left( {1 - x} \right) = a \cr & \Rightarrow x - {x^2} = a \cr} $$