Aptitude - Time and Distance - Discussion

I am not getting solution clearly. Anyone please help me.

v1 = 54km/hr.
v2 = 45km/hr.
Relative speed = v2 - v1 = 9km/hr.
x min = 9km/hr.
60min = 54 km/hr.
x = (9 * 60)/54 = 10min.

Ok, thanks for your explanation, it's helpful for me.

Well-said. Thanks all.

See, the problem is simple, but we just require a way of understanding it.

Speed is 54kmph. That is clear. Now it says that due to the inclusion of 'stoppage' it is 45kmph. We need to understand that this just means that it appears to be 45kmph due to stoppage, which further means that where it had the potential of covering 54km in 1hr, it is now covering 45km in 1hr (but the speed is going to remain, just that the bus stops, hence time is consumed there, otherwise the speed is same only).

So (actual speed)*(1hr - time wasted due to stopping) = 45

Actual speed = 54kmph, and the time wasted has to be found out.

1hr - wasted time shows the effective moving time, due to which, while travelling at 54kmph, the bus covered a 45km distance only.

I hope it clears the doubt.

@All.

Very easy approach:

S1 = 54.
S2 = 45.

By ratio method:
The speed diff is 9kmph.
distance = 9 km.
T= d/s.
t = 9/54 = 1/6 hr=(1/6) * 60min.
t = 10 min.

@Shravani.

Here is my explanation for your doubt.

It doesn't represent the actual speed, it includes the stoppages to cover less speed. Using 45kmph does not give us the correct answer that's why we use actual speed to find the time taken.

12 km is the right answer.

Why have we taken the speed of the train without stoppages (that is 54kmph) for calculating the time taken by the bus to cover 9 km?

Anyone, please clarify.

Speed1 = 54kmp
Speed2 = 45kmph
By ratio method:
Speed diff is 9kmph.
It means the distance is 9 km.
Time = distance/speed
t = 9/54 = 1/6 hr.
Time = 10 min.