Time and Distance

Formulae: 

I)Speed = Distance/Time

II)Time = Distance/speed

III) Distance = speed*time

IV) 1km/hr = 5/18 m/s

V)1 m/s = 18/5 Km/hr

VI)If the ratio of the speed of A and B is a:b,then the ratio of
the time taken by them to cover the same distance is 1/a : 1/b
or b:a

VII) suppose a man covers a distance at x kmph and an equal 
distance at y kmph.then the average speed during the whole 
journey is (2xy/x+y)kmph 

Problems

1)A person covers a certain distance at 7kmph .How many meters 
does he cover in 2 minutes.

Solution::
speed=72kmph=72*5/18 = 20m/s
distance covered in 2min =20*2*60 = 2400m

2)If a man runs at 3m/s. How many km does he run in 1hr 40min

Solution::
speed of the man = 3*18/5 kmph
                 = 54/5kmph
Distance covered in 5/3 hrs=54/5*5/3 = 18km

3)Walking at the rate of 4knph a man covers certain distance 
in 2hr 45 min. Running at a speed of 16.5 kmph the man will 
cover the same distance in.

Solution::
Distance=Speed* time
4*11/4=11km
New speed =16.5kmph
therefore Time=D/S=11/16.5 = 40min
                                               
Complex Problems

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. 

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)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?

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

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)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.

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)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.

Solution::
Due to stoppages,it covers 9km less.
time taken to cover 9 km is [9/54 *60] min = 10min

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

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)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.

Solution::
In same time ,they cover 110km & 90 km respectively
so ratio of their speed =110:90 = 11:9

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.

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)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

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)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 destinstion is.

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

11)A motorist covers a distance of 39km in 45min by moving at a 
speed of xkmph for the first 15min.then moving at double the 
speed for the next 20 min and then again moving at his original 
speed for the rest of the journey .then x=?

Solution::
Total distance = 39 km
Total time = 45 min
D = S*T
x * 15/60 + 2x * 20/60 + x * 10/60 = 39 km
x = 36 kmph

12)A & B are two towns.Mr.Fara covers the distance from A t0 B
on cycle at 17kmph and returns to A by a tonga running at a 
uniform speed of 8kmph.his average speed during the whole 
journey is.

Solution::
When same distance is covered with different speeds,then the 
average speed = 2xy / x+y
=10.88kmph

13)A car covers 4 successive 3km stretches at speed of
10kmph,20kmph,30kmph&:60kmph resp. Its average speed is.

Solution::
Average speed = total distance / total time 
total distance = 4 * 3 = 12 km
total time = 3/10 + 3/20 + 3/30 + 3/60
= 36/60 hr
speed =12/36 * 60 = 20 kmph
                                               
14)A person walks at 5kmph for 6hr and at 4kmph for 12hr.
The average speed is.

Solution::
avg speed = total distance/total time 
= 5*6 + 4*12 / 18
=4 1/3 mph

15)A bullock cart has to cover a distance of 80km in 10hrs.
If it covers half of the journeyin 3/5th time.wht should be its 
speed to cover the remaining distance in the time left.

Solution::
Time left = 10 - 3/5*10
= 4 hr
speed =40 km /4 hr 
=10 kmph 

16)The ratio between the speeds of the A& B is 2:3 an 
therefore A takes 10 min more than the time taken by B to reach 
the destination.If A had walked at double the speed ,he would 
have covered the distance in ?

Solution::
Ratio of speed = 2:3
Ratio of time = 3:2
A takes 10 min more
3x-2x = 10 min 
A's time=30 min 
--->A covers the distance in 30 min ,if its speed is x
-> He will cover the same distance in 15 min,if its speed 
doubles (i.e 2x)

17)A is twice as fast as B and B is thrice as fast as C is.
The journey covered by B in?

Solution::
speed's ratio 
a : b = 2: 1
b : c = 3:1
Time's ratio
b : c = 1:3
b : c = 18:54
(if c covers in 54 min i..e twice to 18 min )

18)A man performed 3/5 of the total journey by ratio 17/20 by
bus and the remaining 65km on foot.wht is his total journey. 

Solution:: 
Let total distance is x
x-(3/5x + 17/20 x) =6.5
x- 19x/20 = 6.5
x=20 * 6.5
=130 km

19)A train M leaves Meerat at 5 am and reaches Delhi at 9am .
Another train N leaves Delhi at 7am and reaches Meerut at 1030am
At what time do the 2 trains cross one another

Solution::
Let the distance between Meerut & Delhi be x
they meet after y hr after 7am
M covers x in 4hr
N covers x in 3 ½ i.e 7/2 hr
speed of M =x/4
speed of N = 2x/7
Distance covered by M in y+2 hr + Distance covered by N in 
y hr is x
x/4 (y+2) +2x/7(y)=x
y=14/15hr or 56 min 

20)A man takes 5hr 45min in walking to certain place and riding 
back. He would have gained 2hrs by riding both ways.The time he 
would take to walk both ways is? 

Solution::
Let x be the speed of walked
Let y be the speed of ride
Let D be the distance

Then D/x + D/y = 23/4 hr -------1
D/y + D/y = 23/4 – 2 hr
D/y = 15/8 --------2
substitute 2 in 1
D/x + 15/8 = 23/4
D/x = 23/4 -15/8 =46-15/8 =31/8
Time taken for walk one way is 31/8 hr
time taken to walk to and fro is 2*31/8 = 31/4 hr
=7 hr 45 min