Aptitude - Time and Distance - Discussion

Good explanation, thanks @Jaya.

Here, the confusion is how come 45/54?

However, the main thing to be noted is that the bus travels at 54kmph in both cases. However, due to stopping for some time, the average speed becomes 45 in case 2. Whereas in the first case, the 54kmph average is the same as its speed because it never stops.

So considering one hour of time, the bus itself covers 45km instead of 54km. We have a difference of nine kilometers in the distance traveled. But we cannot used the average speed of 45 since it travelled at 54 irrespective of the case.
So 9/54 becomes 1/6 hours or 10 minutes.

Without stops, Bus travels - 54 Km in 1 hr.
With stops, the bus travels- 45 Km in 1 hr.

So the extra distance the bus could have travelled if there weren't any stops= 54-45=9 Km.
So the time is taken for travelling that 9 Km is the same as the time taken for the stops,
V = d/t.
54 = 9/t.
t = 9/54 hrs=(9/54)*60=10 min.

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.

54 km in 60 minutes.

9km in x.
Cross multiply.
60 * 9 =540/54.
Is equal to 10 minutes.
So, 9km is taking 10 minutes to stop.

Let me try it in my way
A Particular Distance is travelled by the Bus at 54kmph

When stoppages are included, 54 kmph turns to 45kmph in order to compensate the increase in time.

Let Distance = D, time = t, increase in time = x

D = 54 * t ---> (1)
D = 45 * (t + x) ---> (2)

Equating (1) & (2)

54t = 45 (t+x).
54t = 45t + 45x,
9t = 45x,
9t/45 = x,
(1/5) t = x,
t = 5x.


We need to find x for 1 hr (60 mins)
i.e t+x = 60.
5x+x = 60.
6x = 60.
X = 10.

(Some friends are getting 12mins instead of 10 mins because they are taking (t = 60 min) but (t +x = 60) is the correct one.

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

Simple:

The Speed of a bus without a stoppage is 54kmph.
The speed of bus with a stoppage is 45kmph.
So from this, it is clear that the speed of the bus is 54kmph only;
Now think like this;

54 mango costs 60 rs (1hour = 60minutes);
So 1 mango costs (60/54)rs.
Here the bus travels 9 km less in one hour due to stoppings,
Therefore 9 mango costs = (60/54)*9.
i.e. = 10 rs or 10 minutes.

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.