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

Let the speed of train = x km/ hr
Length of train = 300 metres
Their relative speed in same direction
= (x - 3) km/hr
According to the question,
$$\frac{{\left( {300 + 0} \right)m}}{{\left( {x - 3} \right) \times \frac{5}{{18}}m/s}} = 33$$
[Here man's length is 0 metre]
$$\eqalign{ & \frac{{100 \times 18}}{{5x - 15}} = 11 \cr & 1800 = 55x - 165 \cr & 55x = 1965 \cr} $$
∴ Speed of the train = $$\frac{{1965}}{{55}}$$  = $$35\frac{8}{{11}}$$  kmph

A →____________________← B
Ratio of the speed of trains is given by;
$$\eqalign{ & \frac{{{\text{Speed}}\,{\text{of}}\,{\text{train}}\,A}}{{{\text{Speed}}\,{\text{train}}\,{\text{of}}\,B}} = \frac{{\sqrt B }}{{\sqrt A }} \cr & = \frac{{\sqrt 9 }}{{\sqrt 4 }} \cr & = \frac{3}{2} \cr & = 3:2 \cr} $$

Let length of each train be x meter.
Then, speed of 1^st train = $$\frac{x}{{18}}$$ m/sec
Speed of 2^nd train = $$\frac{x}{{12}}$$ m/sec
Now,
When both trains cross each other, time taken
$$\eqalign{ & = {\frac{{2x}}{{ { {\frac{x}{{18}}} + {\frac{x}{{12}}} } }}} \cr & = \frac{{2x}}{{ {\frac{{ {2x + 3x} }}{{36}}} }} \cr & = \frac{{2x \times 36}}{{5x}} \cr & = \frac{{72}}{5} \cr & = 14.4\,\,{\text{seconds}} \cr} $$

Let the length of the train traveling at 48 kmph be 2x meters.
And length of the platform is y meters.
$$\eqalign{ & {\text{Relative speed of train}} \cr & = \left( {48 + 42} \right)\,{\text{kmph}} \cr & = {\frac{{90 \times 5}}{{18}}} \cr & = 25\,m/\sec \cr & {\text{And}}\,{\text{48}}\,{\text{kmph}} \cr & = \frac{{48 \times 5}}{{18}} \cr & = \frac{{40}}{3}\,m/\sec \cr & \cr & {\text{According to the question}}, \cr & \frac{{ {2x + x} }}{{25}} = 12; \cr & Or,\,3x = 12 \times 25 = 300 \cr & Or,\,x = \frac{{300}}{3} = 100m \cr & {\text{Then, length of the train}} \cr & = 2x \cr & = 100 \times 2 = 200m \cr & \frac{{200 + y}}{{ {\frac{{40}}{3}} }} = 45 \cr & 600 + 3y = 40 \times 45 \cr & {\text{Or}},\,3y = 1800 - 600 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1200 \cr & {\text{Or}},\,y = \frac{{1200}}{3} = 400m \cr & {\text{Length of the platform}} \cr & = 400m \cr} $$

Length of the 1^st train = 105 m
Length of the 2^nd train = 90 m
Relative speed of the trains,
= 45 + 72 = 117 kmph
= $$\frac{{117 \times 5}}{{18}}$$ = 32.5 m/sec
Time taken to cross each other,
= $$\frac{{{\text{Length}}\,{\text{of}}\,{1^{{\text{st}}}}\,{\text{train}} + {\text{length of}}\,{2^{{\text{nd}}}}\,{\text{train}}}}{{{\text{relative }}\,{\text{speed }}\,{\text{of}}\,{\text{the }}\,{\text{trains}}}}$$
∴ Time taken = $$\frac{{195}}{{32.5}}$$  = 6 secs.

1^st method:
Let the speed of the train be S. And length of the train be x.
When a train crosses a man, its travels its own distance.
$$\eqalign{ & {\text{According to question}}; \cr & \frac{x}{{ {\left( {s - 3} \right) \times {\frac{5}{{18}}} } }} = 10 \cr & {\text{or}},\,18x = 50 \times s - 150.....({\text{i}}) \cr & {\text{and}} \cr & \frac{x}{{ {\left( {x - 5} \right) \times {\frac{5}{{18}}} } }} = 11 \cr & 18x = 55 \times s - 275......({\text{ii}}) \cr & {\text{Equating equation }}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr & 50 \times s - 150 = 55 \times s - 275 \cr & {\text{or}},\,5 \times s = 125 \cr & {\text{or}},\,s = 25\,{\text{kmph}} \cr} $$

1^st Method:
Let length of the train be x m and speed of the train is s kmph.
Speed, s = $$\frac{{x + 800}}{{100}}$$ . . . . . (i)
Speed, s = $$\frac{{x + 400}}{{60}}$$ . . . . . (ii)
Equating equation (i) and (ii), we get,
Or, $$\frac{{x + 800}}{{100}}$$  = $$\frac{{x + 400}}{{60}}$$
Or, 5x + 200 = 3x + 2400
Or, 2x = 400
Or, x = 200m

2^nd Method:
As in both cases, the speed of the train is constant, and then we have;
Time α distance
$$\eqalign{ & \frac{{100}}{{60}} = \frac{{x + 800}}{{x + 400}} \cr & {\text{or,}}\,\,x = 200{\text{m}} \cr} $$

Let speed of the 2^nd train is S m/sec.
And,
60 kmph $$ = \frac{{60 \times 5}}{{18}} = \frac{{50}}{3}$$    m/sec.
As trains are traveling in same distance, Then Relative distance,
$$\eqalign{ & = \frac{{60 \times 5}}{{18}} = \frac{{50}}{3} \cr & \frac{{50}}{3} - S = \frac{{120}}{{18}} \cr & \Rightarrow \frac{{50 - 3S}}{3} = \frac{{20}}{3} \cr & \Rightarrow 50 - 3S = 20 \cr & \Rightarrow 3S = 50 - 20 \cr & \Rightarrow 3S = 30 \cr & \therefore S = 10\,\,{\text{m/sec}} \cr} $$
Or, Speed of the 2^nd train = $${\text{10}} \times \frac{{18}}{5}$$  = 36 kmph.

Let the distance of the destination from the starting point = x km.
Speed downstream = (10 + 4) = 14 kmph
Speed upstream = (10 - 4) = 6 kmph
According to the question,
Total time taken = 5 hours
$$\eqalign{ & {\frac{x}{{14}}} + {\frac{x}{6}} = 5 \cr & \frac{{ {3x + 7x} }}{{42}} = 5 \cr & {\text{or}},\,10x = 42 \times 5 \cr & {\text{or}},\,x = \frac{{42 \times 5}}{{10}} \cr & {\text{So}},\,x = 21\,{\text{km}} \cr} $$

Let the distance covered be x km and speed of stream = y kmph.
Speed downstream = $$\frac{{15}}{2} + y$$   kmph
Speed upstream = $$\frac{{15}}{2} - y$$   kmph

$$\eqalign{ & {\text{According}}\,{\text{to}}\,{\text{question,}} \cr & {\frac{{2x}}{{ { {\frac{{15}}{2}} + y} }}} = {\frac{x}{{ { {\frac{{15}}{2}} - y} }}} \cr & or,\,15 - 2y = {\frac{{15}}{2}} + y \cr & or,3y = 15 - {\frac{{15}}{2}} = \frac{{15}}{2} \cr & or,y = \frac{{15}}{6} = 2.5\,{\text{kmph}} \cr} $$