Aptitude - Time and Distance - Discussion
Discussion Forum : Time and Distance - General Questions (Q.No. 5)
5.
Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour?
Answer: Option
Explanation:
Due to stoppages, it covers 9 km less.
| Time taken to cover 9 km = |  |
9 | x 60 |  min |
= 10 min. |
| 54 |
Discussion:
204 comments Page 5 of 21.
Harsh said:
4 years ago
@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.
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.
(1)
Lohith said:
7 years ago
We can do this simply by ratios.
Given speeds 54kmph,45kmph ratio:54:45,
Speeds ratio- 6:5.
Times ratio will be- 5:6 (since speed and time are inversely proportional).
In times ratio 1 part difference is there hence in total 6 parts 1 part is stoppage we calculate (1/6)*60=10.
Given speeds 54kmph,45kmph ratio:54:45,
Speeds ratio- 6:5.
Times ratio will be- 5:6 (since speed and time are inversely proportional).
In times ratio 1 part difference is there hence in total 6 parts 1 part is stoppage we calculate (1/6)*60=10.
Satwik said:
6 years ago
For 1hr ---> 54km.
For how much time with 45km/hr it takes to travel 54km is the total time taken by the bus to travel with stops.
So t=(54/45)*60=72min;.
Its the total time taken subtracting an hour travelled common and we get 12 min.
then how come 10 min be the answer?
For how much time with 45km/hr it takes to travel 54km is the total time taken by the bus to travel with stops.
So t=(54/45)*60=72min;.
Its the total time taken subtracting an hour travelled common and we get 12 min.
then how come 10 min be the answer?
Umar said:
4 years ago
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.
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.
(8)
Sahil said:
7 years ago
Let 54 km and 45 km is a distance that travelled by bus in *1hour*.
45 km distances travelled by car in 1 hour due to the stoppage.
If it travels without a stoppage, it would cover in a 45km/ 54kmph = 5/6h.
5*60/6 = 50 min,
60-50 = 10 minutes.
45 km distances travelled by car in 1 hour due to the stoppage.
If it travels without a stoppage, it would cover in a 45km/ 54kmph = 5/6h.
5*60/6 = 50 min,
60-50 = 10 minutes.
Hema said:
4 years ago
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.
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.
Ranjeet bind said:
6 years ago
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)).
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)).
SANDEEP said:
1 decade ago
I will suggest one more formula to solve this kind of problems when stoppage time has to be calculated.
Then, stoppage time = 1-Slower speed/Faster speed.
Hence in this case.
= 1-45/54.
= 1- 5/6.
= (6-5)/6.
= 1/6.
Which can be converted to = 1/6*60 = 10 min.
Then, stoppage time = 1-Slower speed/Faster speed.
Hence in this case.
= 1-45/54.
= 1- 5/6.
= (6-5)/6.
= 1/6.
Which can be converted to = 1/6*60 = 10 min.
Bipana said:
1 decade ago
1hr----- 54 km without stoppage.
1hr----- 45 km with stoppage.
Speed = 54 km/hr.
Bus losses (54-45=) 9 km in 1hr.
Time = distance/speed.
= 9 km/54 km/hr.
= 9*60/54.
= 10 min.
Bus needs 10 min. To travel 9 km. So it losses 10 min to travel 9 km per hour.
1hr----- 45 km with stoppage.
Speed = 54 km/hr.
Bus losses (54-45=) 9 km in 1hr.
Time = distance/speed.
= 9 km/54 km/hr.
= 9*60/54.
= 10 min.
Bus needs 10 min. To travel 9 km. So it losses 10 min to travel 9 km per hour.
Suresh said:
1 decade ago
Excluding stoppages, the speed of a bus is 54 kmph.
Including stoppages, the speed of a bus is 45 kmph.
The difference between 54 and 45 is 9 that means the bus stopped.
Overall about 9 kms.
Formula : (Difference/Bigger value)*60.
Therefore (9/54)*60.
Including stoppages, the speed of a bus is 45 kmph.
The difference between 54 and 45 is 9 that means the bus stopped.
Overall about 9 kms.
Formula : (Difference/Bigger value)*60.
Therefore (9/54)*60.
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