Simplification - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & {\text{According to question,}} \cr & \sqrt {\frac{{0.25}}{{0.0009}}} \times \sqrt {\frac{{0.09}}{{0.36}}\,\,} \cr & \Rightarrow \sqrt {\frac{{25}}{9} \times 100} \times \sqrt {\frac{9}{{36}}\,\,} \cr & \Rightarrow \frac{{5 \times 10}}{3} \times \frac{3}{6} \cr & \Rightarrow \frac{{25}}{3} \cr & \Rightarrow 8\frac{1}{3} \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & \sqrt {32} {\text{ }} - \sqrt {128} {\text{ + }}\sqrt {50} \cr & \Rightarrow \sqrt {16 \times 2} - \sqrt {64 \times 2} + \sqrt {25 \times 2} \cr & \Rightarrow 4\sqrt 2 - 8\sqrt 2 + 5\sqrt 2 \cr & \Rightarrow \sqrt 2 \cr & \Rightarrow 1.414 \cr} $$

$$\eqalign{ & a + b + c = 0 \cr & \Rightarrow a = - \left( {b - c} \right) \cr & \Rightarrow {a^2} = {\left( {b + c} \right)^2} \cr & \therefore \frac{{{a^2}}}{{\left( {{a^2} - bc} \right)}} + \frac{{{b^2}}}{{\left( {{b^2} - ca} \right)}} + \frac{{{c^2}}}{{\left( {{c^2} - ab} \right)}} \cr} $$
$$ = \frac{{{{\left( {b + c} \right)}^2}}}{{{{(b + c)}^2} - bc}} + \frac{{{b^2}}}{{{b^2} + c\left( {b + c} \right)}} + \frac{{{c^2}}}{{{c^2} + b\left( {b + c} \right)}}$$
$$ = \frac{{{{\left( {b + c} \right)}^2}}}{{{b^2} + {c^2} + bc}} + \frac{{{b^2}}}{{{b^2} + {c^2} + bc}} + \frac{{{c^2}}}{{{b^2} + {c^2} + bc}}$$
$$\eqalign{ & = \frac{{{b^2} + {c^2} + 2bc + {b^2} + {c^2}}}{{{b^2} + {c^2} + bc}} \cr & = \frac{{2\left( {{b^2} + {c^2} + bc} \right)}}{{{b^2} + {c^2} + bc}} \cr & = 2 \cr} $$

$$\eqalign{ & \frac{{{a^2} - {b^2} - 2bc - {c^2}}}{{{a^2} + {b^2} + 2ab - {c^2}}}\, \cr & = \frac{{{a^2} - ({b^2} + 2bc + {c^2})}}{{({a^2} + {b^2} + 2ab) - {c^2}}} \cr & = \frac{{{a^2} - {{(b + c)}^2}}}{{{{(a + b)}^2} - {c^2}}} \cr & = \frac{{\left( {a + b + c} \right)\left( {a - b - c} \right)}}{{\left( {a + b + c} \right)\left( {a + b - c} \right)}} \cr & = \frac{{a - b - c}}{{a + b - c}} \cr} $$

$$\eqalign{ & {\left( {s - a} \right)^2} + {\left( {s - b} \right)^2} + {\left( {s - c} \right)^2} + {s^2} \cr & = \left( {{s^2} + {a^2} - 2sa} \right) + \left( {{s^2} + {b^2} - 2sb} \right) + \cr & \,\,\,\,\,\,\left( {{s^2} + {c^2} - 2sc} \right) + {s^2} \cr & = 4{s^2} + ({a^2} + {b^2} + {c^2}) - 2s(a + b + c) \cr & = {(2s)^2} + ({a^2} + {b^2} + {c^2}) - \cr & \,\,\,\,\,\,(a + b + c)(a + b + c) \cr & = {(a + b + c)^2} + \left( {{a^2} + {b^2} + {c^2}} \right) - \cr & \,\,\,\,\,\,{(a + b + c)^2} \cr & = {a^2} + {b^2} + {c^2} \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & \sqrt {400} + \sqrt {0.0400} + \sqrt {0.000004} \cr & \Rightarrow 20 + 0.2 + 0.002 \cr & \Rightarrow 20.202 \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & \sqrt {192} - \frac{1}{2}\sqrt {48} - \sqrt {75} \cr & \Rightarrow 8\sqrt 3 - \frac{4}{2}\sqrt 3 - 5\sqrt 3 \cr & \Rightarrow 8\sqrt 3 - 2\sqrt 3 - 5\sqrt 3 \cr & \Rightarrow \sqrt 3 \cr & \Rightarrow 1.732 \cr} $$

$$\eqalign{ & {\text{Let the number be = }}x \cr & {\text{According to question,}} \cr & {x^2} = {\left( {75.15} \right)^2} - {\left( {60.12} \right)^2} \cr & \Rightarrow {x^2} = {\left( {75.15 + 60.12} \right)}\,{\left( {75.15 - 60.12} \right)} \cr & \Rightarrow {x^2} = 135.27 \times 15.03 \cr & \Rightarrow {x^2} = 2033.1081 \cr & \Rightarrow x = 45.09 \cr} $$

$$\eqalign{ & \frac{{{{\left( {a + b} \right)}^2}}}{{\left( {{a^2} - {b^2}} \right)}} \cr & = \frac{{\left( {a + b} \right)\left( {a + b} \right)}}{{\left( {a + b} \right)\left( {a - b} \right)}} \cr & = \frac{{a + b}}{{a - b}}\, \cr} $$

$$\eqalign{ & {\text{Length of reel}} \cr & {\text{ = (32}} \times {\text{120 + 80)}}\,{\text{cm}} \cr & {\text{ = 3920}}\,{\text{cm}} \cr & {\text{ = 39}}{\text{.20}}\,{\text{m}} \cr} $$