Aptitude Questions With Answers for Time & Distance Problems-1

1) A train covers a distance in 50 min ,if it runs at a speed of 48kmph on an average. The speed at which the train must run to reduce the time of journey to 40min will be.

2) Vikas can cover a distance in 1hr 24min by covering 2/3 of the distance at 4 kmph and the rest at 5kmph.the total distance is?

3) Walking at ¾ of his usual speed ,a man is late by 2 ½ hr. the usual time is.

4) A man covers a distance on scooter .had he moved 3kmph faster he would have taken 40 min less. If he had moved 2kmph slower he would have taken 40min more. the distance is.

5) Excluding stoppages, the speed of the bus is 54kmph and including stoppages, it is 45kmph.for how many min does the bus stop per hr.

6) Two boys starting from the same place walk at a rate of 5kmph and 5.5kmph respectively.wht time will they take to be 8.5km apart, if they walk in the same direction

7)2 trains starting at the same time from 2 stations 200km apart and going in opposite direction cross each other ata distance of 110km from one of the stations. what is the ratio of their speeds.

8) Two trains start from A & B and travel towards each other at speed of 50kmph and 60kmph resp. At the time of the meeting the second train has traveled 120km more than the first. the distance between them.

9) A thief steals a ca r at 2.30pm and drives it at 60kmph.the theft is discovered at 3pm and the owner sets off in another car at 75kmph when will he overtake the thief

10) In covering distance, the speed of A & B are in the ratio of 3:4.A takes 30min more than B to reach the destion.The time taken by A to reach the destination is

Answers :

1. Solution::

Time=50/60 hr=5/6hr

Speed=48mph

distance=S*T=48*5/6=40km

time=40/60hr=2/3hr

New speed = 40* 3/2 kmph= 60kmph

2. Solution::

Let total distance be S

total time=1hr24min

A to T :: speed=4kmph

diistance=2/3S

T to S :: speed=5km

distance=1-2/3S=1/3S

21/15 hr=2/3 S/4 + 1/3s /5

84=14/3S*3

S=84*3/14*3

= 6km

3. Solution::

Usual speed = S

Usual time = T

Distance = D

New Speed is ¾ S

New time is 4/3 T

4/3 T – T = 5/2

T=15/2 = 7 ½

 4.Solution::

Let distance = x m

Usual rate = y kmph

x/y – x/y+3 = 40/60 hr

2y(y+3) = 9x ————–1

x/y-2 – x/y = 40/60 hr y(y-2) = 3x —————–2

divide 1 & 2 equations

by solving we get x = 40

5.Solution::

Due to stoppages,it covers 9km less.

time taken to cover 9 km is [9/54 *60] min = 10min

6.Solution::

The relative speed of the boys = 5.5kmph – 5kmph = 0.5 kmph

Distance between them is 8.5 km

Time= 8.5km / 0.5 kmph = 17 hrs

7. Solution::

In same time ,they cover 110km & 90 km respectively

so ratio of their speed =110:90 = 11:9

8. Solution::

Let the distance traveled by the first train be x km

then distance covered by the second train is x + 120km

x/50 = x+120 / 60

x= 600

so the distance between A & B is x + x + 120 = 1320 km

9. Solution::

Let the thief is overtaken x hrs after 2.30pm

distance covered by the thief in x hrs = distance covered by

the owner in x-1/2 hr

60x = 75 ( x- ½)

x= 5/2 hr

thief is overtaken at 2.30 pm + 2 ½ hr = 5 pm

10. Solution::

Ratio of speed = 3:4

Ratio of time = 4:3

let A takes 4x hrs,B takes 3x hrs

then 4x-3x = 30/60 hr

x = ½ hr

Time taken by A to reach the destination is 4x = 4 * ½ = 2 hr

Submitted by Subbaiah