1. 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?
[A] 9
[B] 10
[C] 12
[D] 20
Explanation:
Due to stoppages, it covers 9 km less.
Time taken to cover 9 km = [(9/54)*60 min = 10 min.
2. The average speed of a train in the onward journey is 25% more than that in the return journey. The train halts for one hour on reaching the destination. The total time taken for the complete to and from journey is 17 hours, covering a distance of 800 km. The speed of the train in the onward journey is:
[A] 45 km/hr
[B] 47.5 km/hr
[C] 52 km/hr
[D] 56.25 km/hr
Explanation:
Let the speed in return journey be x km/hr.
Then, speed in onward journey =(125/100)x = (5/4)x km/hr
So. Speed in onward journey = [(5/4)*45] km/hr = 56.25 km/hr
3. Walking (6/7) th of his usual speed, a man is 12 minutes too late. The usual time taken by him to cover that distance is:
[A] 1 hour
[B] 1 hr 12 min
[C] 1 hr 15 min
[D] 1 hr 20 min
Explanation:
New speed = (6/7)of usual speed.
New time = (7/6) of usual time.
Therefore (7/6 of usual time)- (usual time) = (1/5) hr.
=> (1/6 of usual time)= (1/5) hr => usual time = (6/5) hr = 1 hr 12 min.
[B] 5 : 3 : 4
[C] 15 : 9 : 20
[D] 15 : 20 : 12
Explanation:
Ratio of speeds = 4 : 3 : 5
Therefore Ratio of times taken = (1/4) : (1/3) : (1/5) = 15 : 20 : 12
5. A motor car starts with the speed of 70 km/hr with its speed increasing every two hours by 10 kmph. In how many hours will it cover 345 kms?
[A] 2 (1/4) hrs
[E] None of these
[C] 4(1/2) hrs
[B] 4 hrs 5 min
[D] Can not be determined
Explanation:
Distance covered in first 2 hours = (70 x 2) km = 140 km
Distance covered in next 2 hours = (80 x 2) km = 160 km
Remaining distance = 345 – (140 + 160) = 45 km.
Speed in the fifth hour = 90 km/hr
Time taken to cover 45 km =(45/90) hr = (1/2) hr
Therefore Total time taken = 2 + 2 + (1/2) = 4 (1/2)hrs
[A] 9
[B] 10
[C] 12
[D] 20
Explanation:
Due to stoppages, it covers 9 km less.
Time taken to cover 9 km = [(9/54)*60 min = 10 min.
2. The average speed of a train in the onward journey is 25% more than that in the return journey. The train halts for one hour on reaching the destination. The total time taken for the complete to and from journey is 17 hours, covering a distance of 800 km. The speed of the train in the onward journey is:
[A] 45 km/hr
[B] 47.5 km/hr
[C] 52 km/hr
[D] 56.25 km/hr
Explanation:
Let the speed in return journey be x km/hr.
Then, speed in onward journey =(125/100)x = (5/4)x km/hr
So. Speed in onward journey = [(5/4)*45] km/hr = 56.25 km/hr
3. Walking (6/7) th of his usual speed, a man is 12 minutes too late. The usual time taken by him to cover that distance is:
[A] 1 hour
[B] 1 hr 12 min
[C] 1 hr 15 min
[D] 1 hr 20 min
Explanation:
New speed = (6/7)of usual speed.
New time = (7/6) of usual time.
Therefore (7/6 of usual time)- (usual time) = (1/5) hr.
=> (1/6 of usual time)= (1/5) hr => usual time = (6/5) hr = 1 hr 12 min.
4. Three persons are walking from a place A to another place B. Their speeds are in the ratio of 4 : 3 : 5. The time ratio to reach B by these persons will be :
[A] 4 : 3 : 5[B] 5 : 3 : 4
[C] 15 : 9 : 20
[D] 15 : 20 : 12
Explanation:
Ratio of speeds = 4 : 3 : 5
Therefore Ratio of times taken = (1/4) : (1/3) : (1/5) = 15 : 20 : 12
5. A motor car starts with the speed of 70 km/hr with its speed increasing every two hours by 10 kmph. In how many hours will it cover 345 kms?
[A] 2 (1/4) hrs
[E] None of these
[C] 4(1/2) hrs
[B] 4 hrs 5 min
[D] Can not be determined
Explanation:
Distance covered in first 2 hours = (70 x 2) km = 140 km
Distance covered in next 2 hours = (80 x 2) km = 160 km
Remaining distance = 345 – (140 + 160) = 45 km.
Speed in the fifth hour = 90 km/hr
Time taken to cover 45 km =(45/90) hr = (1/2) hr
Therefore Total time taken = 2 + 2 + (1/2) = 4 (1/2)hrs