Aptitude - Time and Distance - Discussion

Thanks for the answer @Jaya.

Let's suppose,

Excluding stoppages speed (s1) and including stoppages speed (s2)
then;

The stoppage minutes per hour = (difference b/w the speed(s1~s2))/(Excluding stoppages *60 minute speed (s1)).

(Without stoppage-with stoppage)/Without stoppage.

(54-45) /45.
(9/60) *60.
10 min.

Can't we consider like taking the ratio of speed and time since distant is constant, time is inversely proportional to speed?

From which we can get the ratio of stoppage to total travel time. Please explain to me.

There's a formula for this type of ques:

Speed of bus (excluding stoppage) - Speed of bus (including stoppage) * time/ speed of bus (excluding stoppage).

If you put values in this formula:
Speed of bus without stoppages= 54-45 = 9km/hr,
Time = 1 hr.
So, distance without stoppages = 9 *1= 9 km.

[(54-45) / 54] * 60 = 10.

We have taken the difference of speeds of excl and incl stoppages because we want distance without stoppages. We have taken 54 as denominator because the bus has stopped. We have multiplied it by 60 because the options are in minutes and we were getting answer in hours. If we had options in hours, so we won't have needed to multiply 60.

(Relative speed/maximum speed) * 60min.

For 54 km it is 60 min.
For 45 km how much?

54 = 60
45 = x
X = 50, the difference in time 10 min gives us stoppage time.

Thanks @Prithvi.

Very simple, thanks @Prithvi.

54-45 = 9kmph. i.e., the bus would lose 9km in 1 hr.

Let the time bus stops at a time be t. Then the distance bus would lose will be s. Then s will be,

s = v*t = 54*t.

Let a total number of stoppages be x.

Then the total distance lost by the bus in the second case is,

x*s = x*54*t.

x*54*t = 9.

x*t = 1/6.

So in 1 km, t = 1/6 hour = 10 mins.