Simplification - Aptitude MCQ Questions and Solutions with Explanations

Answer: Option D

$$\eqalign{ & {\text{According to question,}} \cr & \sqrt {0.00005746} \cr & \Rightarrow \sqrt {\frac{{5746}}{{100000000}}} \cr & \Rightarrow \frac{{75.8}}{{10000}} \cr & \Rightarrow 0.00758 \cr} $$

Answer: Option A

$$\eqalign{ & \frac{{ab}}{{a + b}} = \frac{{\frac{x}{{x + y}} \times \frac{y}{{x - y}}}}{{\frac{x}{{x + y}} + \frac{y}{{x - y}}}} \cr & \frac{{ab}}{{a + b}} = \frac{{\frac{{xy}}{{{x^2} - {y^2}}}}}{{\frac{{x\left( {x - y} \right) + y\left( {x + y} \right)}}{{{x^2} - {y^2}}}}} \cr & \frac{{ab}}{{a + b}} = \frac{{xy}}{{{x^2} + {y^2}}} \cr} $$

Answer: Option C

$$\eqalign{ & \frac{a}{b} = \frac{1}{3}\,\, \Rightarrow b = 3a; \cr & \frac{b}{c} = 2\,\,\, \Rightarrow c = \frac{b}{2} = \frac{{3a}}{2}; \cr & \frac{c}{d} = \frac{1}{2}\,\,\, \Rightarrow d = 2c = 2\left( {\frac{{3a}}{2}} \right) = 3a; \cr & \frac{d}{e} = 3\,\,\, \Rightarrow e = \frac{d}{3} = \left( {\frac{{3a}}{3}} \right) = a; \cr & \frac{e}{f} = \frac{1}{4}\,\,\, \Rightarrow f = 4e = 4a \cr & \therefore \frac{{abc}}{{def}} = \frac{{\left( a \right)\left( {3a} \right)\left( {\frac{{3a}}{2}} \right)}}{{\left( {3a} \right)\left( a \right)\left( {4a} \right)}} \cr & \frac{{abc}}{{def}} = \frac{9}{2}{a^3} \times \frac{1}{{12{a^3}}} \cr & \frac{{abc}}{{def}} = \frac{3}{8} \cr} $$

Answer: Option B

$$\eqalign{ & \frac{m}{n}{\text{ = }}\frac{4}{3}{\text{ and }}\frac{r}{t}{\text{ = }}\frac{9}{{14}} \cr & \Rightarrow \frac{{mr}}{{nt}} = \frac{4}{3} \times \frac{9}{{14}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{6}{7} \cr & \therefore \frac{{3mr - nt}}{{4nt - 7mr}}{\text{ }} \cr & = \frac{{3\frac{{mr}}{{nt}} - 1}}{{4 - 7\frac{{mr}}{{nt}}}} \cr & = \frac{{3 \times \frac{6}{7} - 1}}{{4 - 7 \times \frac{6}{7}}} \cr & = \frac{{\frac{{18}}{7} - 1}}{{4 - 6}} \cr & = \frac{{11}}{7} \times \left( { - \frac{1}{2}} \right) \cr & = - \frac{{11}}{{14}} \cr} $$

Answer: Option B

According to question,
$$\sqrt {{{\left( {0.798} \right)}^2} + 0.404 \times 0.798 + {{\left( {0.202} \right)}^2}} + 1\,$$
$$ \Rightarrow \sqrt {{{\left( {0.798} \right)}^2} + {{\left( {0.202} \right)}^2} + 2 \times 0.202 \times 0.798\,} + 1\,$$
$$ \Rightarrow \sqrt {{{\left( {0.798 + 0.202} \right)}^2}} + 1\,\,\left[ {\because {{\left( {a + b} \right)}^2} = {a^2} + {b^2} + 2ab} \right]$$
$$\eqalign{ & \Rightarrow \sqrt {{{\left( 1 \right)}^2}} + 1 \cr & \Rightarrow 1 + 1 \cr & = 2 \cr} $$

Answer: Option D

$$\eqalign{ & {\text{According to question,}} \cr & \,\,\,\,\,\frac{2}{{2 + \frac{2}{{3 + \frac{2}{{3 + \frac{2}{3}}}}} \times 0.39}} \cr & = \frac{2}{{2 + \frac{2}{{3 + \frac{6}{{11}}}} \times 0.39}} \cr & = \frac{2}{{2 + \frac{{22}}{{39}} \times 0.39}} \cr & = \frac{2}{{2 + \frac{{22}}{{100}}}} \cr & = \frac{{200}}{{222}} \cr & = \frac{{100}}{{111}} \cr} $$

Answer: Option B

According to question,
As we know that a + b + c = 0
Then,
a³ + b³ + c³ - 3abc = 0
$$\therefore {\left[ {{{\left( {0.111} \right)}^3} + {{\left( {0.222} \right)}^3} + {{\left( { - 0.333} \right)}^3} - 3 \times 0.111 \times 0.222 \times \left( { - 0.333} \right)} \right]^3} = 0$$

Answer: Option C

$$\eqalign{ & a + \frac{1}{b} = 1{\text{ }} \cr & \Rightarrow ab + 1 = b \cr & \Rightarrow ab - b = - 1 \cr & \Rightarrow b\left( {a - 1} \right) = - 1 \cr & \Rightarrow b = \frac{1}{{\left( {1 - a} \right)}} \cr & \cr & b + \frac{1}{c} = 1 \cr & \Rightarrow bc + 1 = c \cr & \Rightarrow bc - c = - 1 \cr & \Rightarrow c\left( {b - 1} \right) = - 1 \cr & \Rightarrow c = \frac{1}{{\left( {1 - b} \right)}} \cr & \cr & \therefore c + \frac{1}{a} = \frac{1}{{\left( {1 - b} \right)}} + \frac{1}{a} \cr & c + \frac{1}{a} = \frac{1}{{1 - \left( {\frac{1}{{1 - a}}} \right)}} + \frac{1}{a} \cr & c + \frac{1}{a} = \frac{1}{{\frac{{\left( {1 - a} \right) - 1}}{{\left( {1 - a} \right)}}}} + \frac{1}{a} \cr & c + \frac{1}{a} = \frac{{\left( {1 - a} \right)}}{{ - a}} + \frac{1}{a} \cr & c + \frac{1}{a} = \frac{{\left( {a - 1} \right)}}{a} + \frac{1}{a} \cr & c + \frac{1}{a} = \frac{{a - 1 + 1}}{a} \cr & c + \frac{1}{a} = \frac{a}{a} \cr & c + \frac{1}{a} = 1 \cr} $$

Answer: Option B

$$\eqalign{ & \frac{x}{{\left( {2x + y + z} \right)}} = a \cr & \Rightarrow x = a\left( {2x + y + z} \right)\,....(1) \cr & \frac{y}{{\left( {x + 2y + z} \right)}} = a \cr & \Rightarrow y = a\left( {x + 2y + z} \right)\,....(2) \cr & \frac{z}{{\left( {x + y + 2z} \right)}} = a \cr & \Rightarrow z = a\left( {x + y + 2z} \right)\,....(3) \cr & {\text{Adding (1), (2) and (3) we get:}} \cr & x + y + z \cr & = a\left( {4x + 4y + 4z} \right) \cr & \Rightarrow a = \frac{{\left( {x + y + z} \right)}}{{4\left( {x + y + z} \right)}} \cr & \Rightarrow a = \frac{1}{4} \cr} $$

Answer: Option D

$$\eqalign{ & \frac{a}{{b + c}} = k \cr & \Rightarrow a = k\left( {b + c} \right)\,....(i) \cr & \frac{b}{{c + a}} = k \cr & \Rightarrow b = k\left( {c + a} \right)\,....(ii) \cr & \frac{c}{{a + b}} = k \cr & \Rightarrow c = k\left( {a + b} \right)\,....(iii) \cr & {\text{Adding (i), (ii) and (iii) we get,}} \cr & a + b + c \cr & = k\left( {2a + 2b + 2c} \right) \cr & \Rightarrow k = \frac{{a + b + c}}{{2(a + b + c)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2} \cr} $$