Aptitude - Time and Distance - Discussion

Why we take 54 instead of 45? Please explain.

Min per hour= (difference of two speeds÷ without stoppage)*60.
Or
Min per hour=(diff of two speeds÷ large number)*60.

Let's we take that a bus reach an 54 km distance in 1 hour without any stoppages while the bus was covered 45 distance with some stop.

So we can understand that the remaining 9 kms were not reached the bus.
So, 9 km travelling time will be a stop time. We all know that, time=9/54*60=10 mins.

@Sumit Dubey.

Use Arithmetic Progression (A.P.)
Sum to n-terms = n/2(2a + (n-1)d).
n/2 (2*40 + (n-1)5) = 385
40n + 5n^2/2 - 5n/2 = 385
5n^2 + 75n - 650 = 0
(5n - 35)(n + 21) = 0.

So,
n = 7 or n = -21.
As n (time) cannot be negative we will discard -21.
Hence, the answer is 7 hours.

Non stoppages speed= 54kmph,
with stoppages speed= 45kmph.

Reduce speed due to stoppages = 54-45 = 9 kmph.
we need to find the time taken by bus at a stop in each hour i.e t
t = s/v,
t = 9/54 ph ( we take 54 because we are talking about non-stoppages).
= 1/6 *60 min.
=10 min.

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.

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.

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.

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

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.