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

The length of pole is considered as negligible i.e., = 0
i.e., when a train crosses the pole, it covers the distance equal to the length of train
So, the time will be taken by the train = 30 sec
And speed = 60 km/hr
Then the length of the train :
= 60 kmh × 30 sec
= 60 × $$\frac{5}{18}$$ × 30 metres
= 10 × $$\frac{5}{3}$$ × 30
= 500 metres

$$\eqalign{ & {{\text{S}}_{{\text{average}}}} = \frac{{2ab}}{{a + b}} \cr & = \frac{{2 \times 25 \times 4}}{{25 + 4}} \cr & = \frac{{200}}{{29}}{\text{ km/hr}} \cr & {\text{Now, 2D = }}\frac{{200}}{{29}} \times \left( {5 + \frac{4}{5}} \right) \cr & = \frac{{200}}{{29}} \times \frac{{29}}{5} \cr & = 40{\text{ km}} \cr & {\text{ = D = 20 km}} \cr} $$

According to the question,


5 units → 12 hr
1 unit → $$\frac{12}{5}$$ hr and,
4 units → $$\frac{12}{5}$$ × 4
             = $$\frac{48}{5}$$
Distance travelled on foot :
= $$\frac{48}{5}$$ × 5
= 48 km

Total distance :
= 42 × 10
= 420 km
New given time = 7 hr and
Required new speed :
= $$\frac{420}{7}$$ km/hr
= 60 km/hr
∴ Required increase in speed :
= (60 - 42) km/hr
= 18 km/hr

Distance travelled by the 1st car in 15 seconds
$$\eqalign{ & {\text{OA}} = \left( {36 \times \frac{5}{{18}}} \right) \times 15\,{\text{m}} \cr & \,\,\,\,\,\,\,\,\,\,\, = 150\,{\text{m}} \cr} $$
Distance travelled by the 2nd car in 15 seconds
$$\eqalign{ & {\text{OB}} = \left( {48 \times \frac{5}{{18}}} \right) \times 15\,{\text{m}} \cr & \,\,\,\,\,\,\,\,\,\, = 200\,{\text{m}} \cr} $$
∴ Distance between them after 15 seconds
$$\eqalign{ & {\text{AB}} = \sqrt {{\text{O}}{{\text{A}}^2} + {\text{O}}{{\text{B}}^2}} \cr & \,\,\,\,\,\,\,\,\,\, = \sqrt {{{150}^2} + {{200}^2}} \cr & \,\,\,\,\,\,\,\,\,\, = \sqrt {22500 + 40000} \cr & \,\,\,\,\,\,\,\,\,\, = \sqrt {62500} \cr & \,\,\,\,\,\,\,\,\,\, = 250\,{\text{m}} \cr} $$

In the first situation,
⇒ Total distance covered by train :
= 80 × $$4\frac{1}{2}$$
= 360 kms
⇒ Therefore,
The speed of the train to cover the same distance 360 km in 4 hours is :
$$\eqalign{ & = \frac{{360}}{4}\left\{ {{\text{Speed }} = \frac{{{\text{Distance }}}}{{{\text{Time}}}}} \right\} \cr & = 90{\text{ km/h}} \cr} $$

$$\eqalign{ & {{\text{V}}_{{\text{rel}}{\text{.}}}} = \left( {21 - 15} \right){\text{ m/min}} \cr & \,\,\,\,\,\,\,\,\,\,\, = {\text{ 6 m/min}} \cr} $$
Time taken to catch the thief :
= $$\frac{114}{6}$$ min
= 19 min

$$\eqalign{ & {\text{Average speed }} = \frac{{{\text{Total distance }}}}{{{\text{Total time}}}} \cr & {\text{Average speed }} = \frac{{{\text{10 + 12 }}}}{{\frac{{10}}{{12}} + \frac{{12}}{{10}}}} \cr & {\text{Average speed }} = 10.8{\text{ km/hr}} \cr} $$

Difference between his reaching time :
= (14 - 10) hrs
= 4 hrs
= 4 hrs → 6m + 6m
(late + before)
= 4 hrs → 12 minutes
= 1 unit = $$\frac{12}{4 × 60}$$   km
($$\because $$ 1 m = 60 seconds)
1 unit = $$\frac{1}{20}$$ km
Then, 35 units :
= 35 × $$\frac{1}{20}$$ km
= $$\frac{7}{4}$$ km
Then the distance between his house and school is = $$\frac{7}{4}$$ km

$$\eqalign{ & {\text{Speed of train :}} \cr & = \frac{{450 + 150}}{{20}} \cr & = 30{\text{ m/s}} \cr & {\text{Speed (in km/hr):}} \cr & = 30 \times \frac{{18}}{5} \cr & = 108{\text{ km/hr}} \cr} $$