Aptitude - Time and Distance - Discussion

The LCM approach here makes our work easy. This can even be applied in problems on pipes and cisterns, work and time, etc.

So, first, take the LCM of 54 and 45 to be the distance covered i.e. 270 km.

The time taken without stoppage is 270/54 = 5 hrs. The time taken with stoppage is 270/45 = 6 hrs. This implies that out of 6 hrs, 1 hour accounts for the stoppage because 6-5 = 1.
Next, apply the unitary method.

6hrs ---------- 1 hr of stoppage.
1 hr ----------- (1/6)*60 = 10 mins stoppage.

Analogously, in case of pipes and cisterns, the lcm of the time taken by the pipes to fill/empty can be taken as the total capacity of the cistern.

1hr distance without stoppage = 54km,
1hr distance with stoppage = 45km,
=> distance difference = 9.
9/54*60=10.

The bus travelled only 45kms. Hence, it would be correct to divide 45 (and not 50) and multiply 60 with 9 kms (less travelled).

Am I right? please clarify for me.

Bus travels at 54 km/hr.
Due to stoppages say t min, it travels only for (60-t) min.
(60-t) min= (60-t)/60 hr,
Given average speed = 45 km/hr,
Average speed = distance travelled/ total time taken.

45 = (54*(60-t)/60)/1 hr.
45 = 9*(60-t)/10,
450 = 540-9t,
t = 90/9 = 10 min.