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

$$\eqalign{ & {\text{Time}} = \frac{{{\text{Distance}}}}{{{\text{Speed}}}} \cr & {\text{Time}} = \frac{{300 + 200}}{{25}} \cr & {\text{Time}} = 20\sec \cr} $$

Distance cover by C in 90 mintue will cover by B $$\frac{90}{3}$$ = 30 minute
Distance cover by B in 30 mintue will cover by A $$\frac{30}{2}$$ = 15 minute

Int e question it is given that, a man who sit in the slower train cross the faster train it means faster train cross the man 18 sec.
⇒ Relative speed of faster train and man in the same direction
= (58 - 30)
= 28 kmph
So, the distance covered by faster train in 18 sec :
= 28 kmph × 18 sec
= 28 × $$\frac{5}{18}$$ × 18
= 140 metres

$$\eqalign{ & {\text{Average speed}} \cr & = \frac{{{\text{2xy }}}}{{x + y}} \cr & = \frac{{2 \times 40 \times 60}}{{40 + 60}} \cr & \therefore {\text{Average speed}} = {\text{4}}8{\text{ km/hr}} \cr} $$

= 6 - 5 = 1 unit → $$\frac{16}{60}$$
15 units → $$\frac{16}{60}$$ × 15 = 4
∴ Required distance = 4 km

First speed = 8 km/hr
Second speed = 12 km/hr
Starting time will be same in both conditions.
Distance travelled by first speed in 2 hours
Time take by second speed to travel distance 16 km
$$\eqalign{ & {\text{Time}} = \frac{{{\text{Distance }}}}{{{2^{{\text{nd}}}}{\text{Speed }} - {\text{ }}{{\text{1}}^{{\text{st}}}}{\text{Speed}}}} \cr & {\text{Time}} = \frac{{16}}{{12 - 8}} \cr & {\text{Time}} = 4{\text{ hrs}} \cr} $$
Total time taken by second speed = 4 hrs
Total distance = 4 × 12 = 48 km
Starting time to travel at second speed :
= 9 am - 4 hrs = 5 am
Total time taken by third speed to reach the office :
= 10 am - 5 am = 5 hrs
$$\eqalign{ & {{\text{3}}^{{\text{rd}}}}{\text{ Speed}} = \frac{{{\text{Total Distance}}}}{{{\text{Time}}}} \cr & {{\text{3}}^{{\text{rd}}}}{\text{ Speed}} = \frac{{48}}{5} \cr & {{\text{3}}^{{\text{rd}}}}{\text{ Speed}} = 9.6{\text{ km/hrs}} \cr} $$

Total distance covered by man in 3 hours 36 minutes is :
$$\eqalign{ & = 5 \times 3\frac{{36}}{{60}} = 5 \times 3\frac{3}{5} \cr & = 5 \times \frac{{18}}{5} \cr & = 18{\text{ km}} \cr} $$
18 km is covered at an speed of 24 km/hr
∴ Time taken is :
$$\eqalign{ & = \frac{{18}}{{24}} \cr & = \frac{3}{4} \times 60 \cr & = 45{\text{ minutes}} \cr} $$

Here, speed of the running train is 90 km/hr
And length of the train is = 120 metres
We know that,
When a train crosses through the platform, it cover the distance equal to the length of platform + length of the train
So, the time will taken by the train :

$$\frac{{{\text{Length of train}} + {\text{Length of platform}}}}{{{\text{Speed}}}}$$
$$\eqalign{ & = \frac{{\left( {120 + 230} \right){\text{ metres}}}}{{90{\text{ km/hr}}}} \cr & = \frac{{350 \times 18}}{{90 \times 5}} \cr & = 14\sec \cr} $$

Time taken by trains to cross each other in opposite direction
$$\frac{{{l_1} + {l_2}}}{{{\text{Relative speed in opposite direction}}}}$$
$$\eqalign{ & {\text{ = }}\frac{{\left( {180 + 120} \right)}}{{\left( {65 + 55} \right)}} \cr & {\text{ = }}\frac{{300}}{{120 \times \frac{5}{{18}}}} \cr & {\text{ = 9 seconds}} \cr} $$

$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Cyclist}}\,:{\text{Jogger}} \cr & {\text{Ratio of distance}} \to \,\,\,\,\,2\,\,\,\,\,\,\,:\,\,\,\,\,\,\,1 \cr & {\text{Ratio of time}}\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,1\,\,\,\,\,\,\,:\,\,\,\,\,\,\,2 \cr & {\text{Ratio of their speed }}\left( {{\text{Jogger}}:{\text{Cyclist}}} \right) \cr} $$
$$\eqalign{ & = \frac{1}{2}:\frac{2}{1} \cr & = 1:4 \cr} $$