Time and Distance problems
Posted by
Ravi Kumar at Monday, November 24, 2008
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Time and Distance problems
Time and Distance problems
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=12/3S=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/y2 – x/y = 40/60 hr y(y2) = 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 x1/2 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 4x3x = 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
3x2x = 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 =4615/8 =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