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

If train crosses the platform i.e., it covers the distance equal to the length of train and platform.
In the question train crosses the man who stands on the platform in 10 seconds and crosses the man + platform in 25 seconds i.e., train crosses the platform whose length is 300 metres in 25 - 10 = 15 seconds, here train's length is not added.
So, speed of the train = $$\frac{300}{15}$$ = 20 m/sec
Length of the train = 10 × 20 = 200 metres (If train crosses the only man in 10 seconds)
Time taken by the train to cross a platform 200 metre long :
$$\eqalign{ & = \frac{{{\text{ Length of train + platform}}}}{{{\text{Speed}}}} \cr & = \frac{{\left( {200 + 200} \right)}}{{20}} \cr & = \frac{{400}}{{20}} \cr & = 20 \cr} $$
Time taken by train = 20 seconds

Distance covered in 1 min = 50 m
Distance covered in 2 min = 90 m
Similarly, 1^st min, 2^nd min, 3^rd min . . . . . 15^th
Distance → 50 m + 90 m + 130 m + . . . . . . . .
By using A.P.
a = 50 m
d = (90 - 50) = 40 m
Tn = a + (n - 1)d
     = 50 + (15 - 1) × 40
     = 50 + 560
     = 610 minutes

Let the speed of truck is = x km/hr
Their relative speed in same direction = (45 - x) km/hr
[Here (45 - x) has been written because bus crosses the truck which is running 150 metres ahead of it. i.e., Truck speed will be lower than that of bus]
According to the question,
$$\eqalign{ & {\text{Time}} = \frac{{{\text{Distance}}}}{{{\text{Speed}}}} \cr & \Rightarrow \frac{{150}}{{\left( {45 - x} \right) \times \frac{5}{{18}}}} = 30 \cr & \Rightarrow \frac{{150 \times 18}}{{\left( {45 - x} \right) \times 5}} = 30 \cr & \Rightarrow x = 27{\text{ km/hr}} \cr} $$

$$\eqalign{ & {\text{Average speed }} = \frac{{{\text{Total distance }}}}{{{\text{Total time}}}} \cr & {\text{Average speed}} = \frac{{60 \times 3}}{{\frac{{60}}{{10}} + \frac{{60}}{{20}} + \frac{{60}}{{60}}}} \cr & {\text{Average speed}} = \frac{{180{\text{ km}}}}{{\left( {6 + 3 + 1} \right){\text{ hr}}}} \cr & {\text{Average speed}} = \frac{{180{\text{ km}}}}{{{\text{10 hr}}}} \cr & {\text{Average speed}} = 18{\text{ km/hr}} \cr} $$
Short trick formula :
$$\left( {{\text{Average speed}} = \frac{{3xyz}}{{xy + yz + zx}}} \right)$$

Let the speed = x km/hr
And, time = y hr
According to the question,
x × y = (x + 3) (y - 1)
xy = xy + 3y - x - 3
x - 3y = - 3 ..... (i)
Also,
x × y = (x - 2)(y + 1)
xy = xy - 2y + x - 2
x - 2y = 2 ..... (ii)
Solve equation (i) and (ii)
x = 12
y = 5
Distance = Speed × Time
Distance = 12 × 5
               = 60 km

Let the man travels 1 unit distance
So, remaining distance
$$\eqalign{ & = 1 - \left( {\frac{1}{6} + \frac{3}{4}} \right) \cr & = 1 - \frac{{22}}{{24}} \cr & = \frac{1}{{12}} \cr & \because \frac{1}{{12}}{\text{ unit}} = {\text{2 km}} \cr & {\text{So, 1 unit}} = 24{\text{ km}} \cr} $$

$$\because $$ Distance covered by the truck in a minute = 550 metres
Then, the speed of the truck will be :
$$\eqalign{ & \frac{{550 \to {\text{Metres}}}}{{60 \to {\text{Seconds}}}}\left\{ {{\text{Speed = }}\frac{{{\text{Distance}}}}{{{\text{Time}}}}} \right\} \cr & \left( {1{\text{ minute = 60 seconds}}} \right) \cr & = \frac{{550}}{{60}} \Rightarrow \frac{{55}}{6}{\text{ m/s}}.....{\text{(i)}} \cr} $$
Whereas, distance covered by the bus in 45 minutes = 33 km
Then, the speed of the bus will be :
$$\frac{{33{\text{ km}}}}{{45\min }} \Rightarrow \frac{{33 \times 1000}}{{45 \times 60}}$$

\[\left\{ \begin{gathered} 1{\text{ km = 1000 metres}} \hfill \\ {\text{1 min = 60 seconds}} \hfill \\ \end{gathered} \right\}\]

$$\eqalign{ \Rightarrow \frac{{110}}{9}{\text{ m/s}}.....{\text{(ii)}} \cr} $$
So, the ratio of their speeds will be,
$$\eqalign{ & = \frac{{55}}{6}:\frac{{110}}{9} \cr & = \frac{1}{2}:\frac{2}{3} \cr & = {\bf{3}}\,\,\,\,\,{\bf{:}}\,\,\,\,\,\,{\bf{4}} \cr & (\text{Truck : Bus}) \cr }$$

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$$\eqalign{ & {\bf{Alternate:}} \cr & {\text{Speed of truck}} = 550\,{\text{metres/min}} \cr & {\text{Speed of bus}} = \frac{{33}}{{45}}\,{\text{km/min}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{33000}}{{45}}\,{\text{metres/min}} \cr & {\text{Speed of truck}}:{\text{Speed of bus}} \cr & = 550:\frac{{33000}}{{45}} \cr & = 55:\frac{{3300}}{{45}} \cr & = 5:\frac{{300}}{{45}} \cr & = 1:\frac{{60}}{{45}} \cr & = 1:\frac{4}{3} \cr & = 3:4 \cr} $$

Let the train's speed = x km/hr
  $$\because \,$$ When train will cross a man then it covers only its length.
$$\eqalign{ & \therefore \left( {x - 3} \right) \times \frac{{10}}{{60}} = \left( {x - 5} \right) \times \frac{{11}}{{60}} \cr & x = 25{\text{ km/hr}} \cr} $$
Note :
Products of time × speed is always subtracted if both the men are running in the same direction and the products of time × speed is added only if the men are running opposite direction.
⇒ Here train's direction is not considered.
But attention please ⇒ Always divided by the difference of time.

Here length of pole is considered 0 metre
Time will be taken by train to cross the poll :
$$\eqalign{ & = \frac{{{\text{300 m}}}}{{54 \times \frac{5}{{18}}{\text{ m/s}}}} \cr & = \frac{{300}}{{15}} \cr & = 20 \cr} $$
Required time = 20 seconds

Let the total distance = 1200 km


$$\eqalign{ & {\text{Total time taken :}} \cr & = \frac{{400}}{{25}} + \frac{{300}}{{30}} + \frac{{500}}{{50}} \cr & = {\text{ }}16 + 10 + 10 \cr & = {\text{ 36 hours}} \cr & \therefore {\text{Average speed}} \cr & = \frac{{{\text{Total distance }}}}{{{\text{Total time}}}} \cr & = \frac{{1200}}{{36}} \cr & = 33\frac{1}{3}{\text{ km/hr}} \cr} $$