Time and Distance Multiple Choice Questions

Time and Distance

Question 1.

 A car covers its journey at the speed of 80km/hr in 10hours. If the same distance is to be covered in 4 hours, by how much the speed of car will have to increase ?

A.

40km/hr

B.

60km/hr

C.

90km/hr

D.

120km/hr

Explanation :

Initial speed = 80km/hr
Total distance = 80 * 10 = 800km
new speed = 800/4 =200km/hr
Increase in speed = 200 - 80 = 120km/hr

Question 2.

 two cars A and B are running in the same direction. Car A has already covered a distance of 60kms when car B started running. The cars meet each other in 3 hours after car B started running. What was

A.

20km/hr

B.

30km/hr

C.

15km/hr

D.

None of these

Explanation :

Difference between speeds of car A and car B = 60/3 = 20km/hr
But from the result it is impossible to find the speed of car A because here the unkowns are two (Speeds of cars A and B ) but the equation is only one


Question 3.

 Walking 3/4th of his usual rate, a man is 15min late. Find his usual time in minutes

A.

30

B.

35

C.

45

D.

25

Explanation :

Walking at 3/4th of usual rate implies that time taken would be 4/3th of the usual time. In other words, the time taken is 1/3rd more than his usual time
so 1/3rd of the usual time = 15min
or usual time = 3 * 15 = 45min

Question 4.

 A monkey climbs up a greased pole, ascends 20m and slip 4m in alternate minutes. If the pole is 96m high, how many minutes will it take to reach the top?

A.

54/5

B.

12

C.

44/5

D.

9

Explanation :

net height climbed in 2 min = 20m - 4m = 16m
In 10 min the net height climbed = 16 * 10/2 = 80m
Remaining height = 96m - 80m = 16m
In the 11th min , the monkey will be ascending up. 
Time taken to ascend the last 16min
16/20 = 4/5min
Total time taken = 10 + 4/5 = 54/5 min

Question 5.

 A car is running at 7/10 of its own speed reached a place in 22 hours. How much time could be saved if the train would run at its own speed ?

A.

4hours

B.

6hours

C.

7hours

D.

8hours

Explanation :

Since the car runs at 7/11 th of its own speed, the time it take is 11/7th of its usual speed.
let the usual time taken be t hours
then we can write, 11t / 7 = 22
or t  = 22 * 7 / 11 = 14 hours
Time saved  = 22 - 14 = 8hours