1. A person covers a certain distance at 72 km/h .How many meters does he cover in 2 minutes.
1. 2400 m
2. 2500 m
3. 2450 m
4. 2350 m
2. If a man runs at 3 m/s. How many km does he run in 1 hr 40 minutes?
1. 16 km
2. 18 km
3. 17.5 km
4. 15.5 km
3. Walking at the rate of 4 kmph a man covers certain distance in 2hr 45 min. running at a speed of 16.5 km/h the man will cover the same distance in.
1. 45 min
2. 42 min
3. 40 min
4. none of these
4. A train covers a distance in 50 min, if it runs at a speed of 48 km/h on an average. The speed at which the train must run to reduce the time of journey to 40 min will be.
1. 62 km/h
2. 64 km/h
3. 65 km/h
4. none of these
5. Vikas can cover a distance in 1 hr 24 min by covering 2/3 of the distance at 4 km/h and the rest at 5 km/h. The total distance is?
1. 6 km
2. 8 km
3. 5 km
4. 6 km
6. Walking at ¾ of his usual speed, a man is late by 2 ½ hr. The usual time is?
1. 7.5
2. 7.8
3. 6.5
4. 5.5
7. A man covers a distance on scooter. Had he moved 3km/h faster he would have taken 40 min less. If he had moved 2 km/h slower he would have taken 40 min more. The distance is?
1. 36
2. 38
3. 40
4. 45
8. Excluding stoppages, the speed of the bus is 54 km/h and including stoppages, it is 45 km/h. For how many min does the bus stop per hr?
1. 16 min
2. 10 min
3. 7.5 min
4. 12 min
9. Two boys starting from the same place walk at a rate of 5 km/h and 5.5 km/h respectively. What time will they take to be 8.5 km apart, if they walk in the same direction?
1. 17 hrs
2. 18 hrs
3. 17.5 hrs
4. 15.5 hrs
10. Two trains starting at the same time from 2 stations 200km apart and going in opposite direction cross each other at a distance of 110km from one of the stations. What is the ratio of their speeds?
1. 12 : 7
2. 11 : 8
3. 12 : 9
4. 11 : 9
Answers and Solution:
1.(1) Solution:
Speed=72 km/h=72 x 5/18 = 20 m/s
Distance covered in 2 min =20 x 2 x 60 = 2400 m
2.(2) Solution:
Speed of the man = 3 x 18/5 km/h = 54/5 km/h
Distance covered in 5/3 hrs=54/5 x 5/3 = 18 km
3. (3) Solution:
Distance=Speed x time
4 x 11/4=11 km
New speed =16.5 km/h
Therefore Time=D/S=11/16.5 = 40 min
4. (4) Solution:
Time=50/60 hr=5/6 hr
Speed=48 mph
Distance=S x T=48 x 5 /6=40 km
Time=40/60 hr=2/3 hr
New speed = 40 x 3/2 km/h= 60 km/h
5. (4) Solution:
Let total distance be S
Total time=1 hr 24 min
A to T:
Speed = 4 km/h Distance =2/3 S
T to S:
Speed=5 km/h Distance=1-2/3 S=1/3 S
=> 21/15 hr=2/3 S/4 + 1/3S/5
=> 84=14/3S x 3
S=84 x 3/14 x 3 = 6 km
6. (1)Solution:
Usual speed = S Usual time = T Distance = D
New Speed is 3/4 S
New time is 4/3 T
4/3 T – T = 5/2
T=15/2 = 7 ½
7. (3) Solution:
Let distance = x m Usual rate = y km/h
x/y – x/y+3 = 40/60 hr
2y (y+3) = 9x [Eqn 1]
x/y-2 – x/y = 40/60 hr y(y-2) = 3x [Eqn 2]
Divide 1 & 2 and solving we get
x = 40
8.(2) Solution:
Due to stoppages, it covers 9 km less.
Time taken to cover 9 km is [9/54 x 60] min = 10 min
9. (1) Solution:
The relative speed of the boys = 5.5 km/h – 5 km/h = 0.5 km/h
Distance between them is 8.5 km
Time= 8.5 / 0.5 = 17 hrs
10. (4) Solution:
In same time, they cover 110 km & 90 km respectively
So ratio of their speed =110:90 = 11:9
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