Speed Time And Distance - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{speed}}\,{\text{of}}\,{\text{the}}\,{\text{train}}\,{\text{be}}\,x\,{\text{km/hr}} \cr & {\text{and}}\,{\text{that}}\,{\text{of}}\,{\text{the}}\,{\text{car}}\,{\text{be}}\,y\,{\text{km/hr}} \cr & {\text{Then}},\,\frac{{120}}{x} + \frac{{480}}{y} = 8 \cr & \Rightarrow \frac{1}{x} + \frac{4}{y} = \frac{1}{{15}}\,.........\left( i \right) \cr & {\text{and}},\,\frac{{200}}{x} + \frac{{400}}{y} = \frac{{25}}{3} \cr & \Rightarrow \frac{1}{x} + \frac{2}{y} = \frac{1}{{24}}\,.........\left( {ii} \right) \cr & {\text{Solving}}\,\left( {\text{i}} \right)\,{\text{and}}\,\left( {{\text{ii}}} \right){\text{,}}\, \cr & {\text{we}}\,{\text{get}},\,x = 60\,{\text{and}}\,y = 80 \cr & \therefore {\text{Ratio}}\,{\text{of}}\,{\text{speeds}} \cr & = 60:80 \cr & = 3:4 \cr} $$

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{distance}}\,{\text{travelled}}\,{\text{on}}\,{\text{foot}}\,{\text{be}}\,x\,km \cr & {\text{Then,}}\,{\text{distance}}\,{\text{travelled}}\,{\text{on}}\,{\text{bicycle}} = \left( {61 - x} \right)km \cr & {\text{So}},\,\frac{x}{4} + \frac{{ {61 - x} }}{9} = 9 \cr & \Rightarrow 9x + 4\left( {61 - x} \right) = 9 \times 36 \cr & \Rightarrow 5x = 80 \cr & \Rightarrow x = 16\,km \cr} $$

$$\eqalign{ & {\text{Let}}\,{\text{distance}} = x\,km\,{\text{and}}\,{\text{usual}}\,{\text{rate}} = y\,kmph \cr & {\text{Then}},\,\frac{x}{y} - \frac{x}{{y + 3}} = \frac{{40}}{{60}} \cr & \Rightarrow 2y\left( {y + 3} \right) = 9x\,..........\left( i \right) \cr & {\text{and}},\,\frac{x}{{y - 2}} - \frac{x}{y} = \frac{{40}}{{60}} \cr & \Rightarrow y(y - 2) = 3x\,.............\left( {ii} \right) \cr & {\text{On}}\,{\text{dividing}}\,\left( {\text{i}} \right)\,{\text{by}}\,\left( {{\text{ii}}} \right){\text{,}} \cr & {\text{We}}\,{\text{get}}\,x = 40\,\text{km} \cr} $$

A's speed :
= 72 km/hr
= $$\left( {72 \times \frac{5}{{18}}} \right)$$  m/sec
= 20 m/sec
B's speed = 25 m/sec
Difference :
= (25 - 20) m/sec
= 5 m/sec

$$\eqalign{ & {\text{Speed : }} \cr & = {\text{ }}30{\text{ }}km/hr \cr & = {\text{ }}30 \times \frac{5}{{18}}{\text{ }}m/\sec \cr & = \frac{{25}}{3}m/\sec \cr & {\text{So, time = }}\frac{D}{S} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{100}}{{\frac{{25}}{3}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 12 \,\,\text{seconds} \cr} $$

Let side of the square playground is $$x$$
Ans Area :
$$\eqalign{ & {x^2} = 1127.6164 \cr & x = \sqrt {1127.6164} \cr & x = 33.58 \cr} $$
Perimeter of playground :
$$\eqalign{ & = 4x \cr & = 4 \times 33.58 \cr & = 134.32 \cr} $$
Time taken to complete 1 round :
$$\eqalign{ & = \frac{{134.32}}{{2\frac{9}{{20}}}} \cr & = \frac{{134.32}}{{49}} \times 20 \cr & = 54.82\,\,\text{minutes} \cr} $$

Average speed :
$$\eqalign{ & = \frac{{2 \times 60 \times 45}}{{60 + 45}} \cr & = \frac{{1080}}{{21}}{\text{ km/hr}} \cr} $$
Time :
$$\eqalign{ & = 5 + \frac{1}{4}{\text{ hr}} \cr & {\text{ = }}\frac{{21}}{4}{\text{ hr}} \cr} $$
Distance travelled :
$$\eqalign{ & {\text{ = }}\frac{{1080}}{{21}} \times \frac{{21}}{4} \cr & = 270{\text{ km}} \cr} $$

$$\eqalign{ & {\text{Speed : }} \cr & = {\text{ }}\frac{D}{T} \cr & = \frac{{125}}{{30}}{\text{m /sec}} \cr & {\text{ = }}\frac{{125}}{{30}} \times \frac{{18}}{5}{\text{ km /hr}} \cr & {\text{ = 15 km /hr}} \cr} $$

Let A finish the race of $$x$$ metres
B finish the race of = $$x$$ - 12
C finish the race of = $$x$$ - 18 ..... (i)
In another race of B & C
B finish race of = $$x$$
C finish race of = $$x$$ - 8 ..... (ii)
Ratio of speeds of B & C
$$\eqalign{ & \Rightarrow \frac{{x - 12}}{{x - 18}} = \frac{x}{{x - 8}} \cr & \Rightarrow \left( {x - 12} \right)\left( {x - 8} \right) = x\left( {x - 18} \right) \cr & \Rightarrow {x^2} - 20x + 96 = {x^2} - 18x \cr & \Rightarrow 96 = 2x \cr & \Rightarrow x = 48\,\,\text{metres} \cr} $$

Required distance :
$$\eqalign{ & = 72 \times \frac{5}{{18}} \times 5 \cr & = 100{\text{ metres}} \cr} $$