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:
205 comments Page 17 of 21.
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.
Kaushal said:
1 decade ago
Why it use 54 km/hr with 9 km not use the 45 km/hr?
Mukesh Kumar Jena said:
1 decade ago
Let for a hrs, bus runs at 54 kmph.
And b hrs, bus runs at 0 kmph.
So avg speed => 45=(54*a)/(a+b).
=> a=5b.
Since we need for 1 hour;
a+b=1 hr.
We got ; b=1/6 hr stoppage time = 10 mins.
And b hrs, bus runs at 0 kmph.
So avg speed => 45=(54*a)/(a+b).
=> a=5b.
Since we need for 1 hour;
a+b=1 hr.
We got ; b=1/6 hr stoppage time = 10 mins.
Akriti said:
1 decade ago
Here speed is given - 54 (excluding stoppages), 45(including ).
Time taken at stoppages -d/t.
So 9 kmph, distance - 9*60.
540/54 because (distance /time), we got 10min.
Time taken at stoppages -d/t.
So 9 kmph, distance - 9*60.
540/54 because (distance /time), we got 10min.
Murali said:
1 decade ago
Excluding stoppages= (difference of speeds/excluding Stoppage) *60mins.
= (54-45/54) *60mins.
= (9/54) *60mins.
= (1/6) *60mins.
=10mins.
= (54-45/54) *60mins.
= (9/54) *60mins.
= (1/6) *60mins.
=10mins.
Suresh khichee said:
1 decade ago
D = S*T.
D = 54/1 * T ....(1) T is time taken by bus without stoppages.
D = 45/1 *( T + S ) ...(2) where S is time taken by stoppages.
By (1)& (2)
54*T = 45*T + 45*S
54*T - 45*T = 45*S
9*T = 45*S or 9/T = 45/S so ratio of T/S = 45/9
T/S = 5/1 {T=time ,S= time of stoppage }
T:S ::5:1 shows that
In 6 minute(5+1) bus takes 1 minutes for stoppage.
So In 60 minutes bus will take 10 minutes stoppage.
D = 54/1 * T ....(1) T is time taken by bus without stoppages.
D = 45/1 *( T + S ) ...(2) where S is time taken by stoppages.
By (1)& (2)
54*T = 45*T + 45*S
54*T - 45*T = 45*S
9*T = 45*S or 9/T = 45/S so ratio of T/S = 45/9
T/S = 5/1 {T=time ,S= time of stoppage }
T:S ::5:1 shows that
In 6 minute(5+1) bus takes 1 minutes for stoppage.
So In 60 minutes bus will take 10 minutes stoppage.
Chinmayee said:
1 decade ago
Why we take 54 instead of 45 ? please solve my question.
Pratik said:
1 decade ago
Let us suppose 10 min is the ans..
Let the distance covered be 54km.
Time 1 hr without stoppage.
Then 54/1 = 54 kph..
With stoppage of 10 min= 1/6 hr= 0.1667 hr
54/(1+0.1667)= 46.28 kph
---------------------------------
Let 12 min be the ans
With stoppage 0f 12 min=1/5hr=0.2hr
54/(1+0.2)= 45 kph.
Let the distance covered be 54km.
Time 1 hr without stoppage.
Then 54/1 = 54 kph..
With stoppage of 10 min= 1/6 hr= 0.1667 hr
54/(1+0.1667)= 46.28 kph
---------------------------------
Let 12 min be the ans
With stoppage 0f 12 min=1/5hr=0.2hr
54/(1+0.2)= 45 kph.
Harshita said:
1 decade ago
Hi. Please tell what is wrong in the following approach? because I'm not getting the right answer by this.
54 km/hr means in 1hr it will cover 54km.
Now if it stops of x hr the the new speed.
45=54/ (1+x).
So x=9/45 hr= (9/45) 60 min. =12 min?
54 km/hr means in 1hr it will cover 54km.
Now if it stops of x hr the the new speed.
45=54/ (1+x).
So x=9/45 hr= (9/45) 60 min. =12 min?
Afsar said:
1 decade ago
How it take 9km. ?
It is because there is no time limit given so.
We take reference time 1hr. And each having same time.
So distance=speed*time.
It is because there is no time limit given so.
We take reference time 1hr. And each having same time.
So distance=speed*time.
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