### APTITUDE – problems on 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

1. Vaishali says:

A worker earns a 5% raise.A year later,the worker receives a 2.5% cut in pay & now his salary is Rs 22702.68?What was his salary to begin with??

• Vishy says:

let us consider as x
5% raise means 105x/100
2.5% cut means 97.5/100
so,((105x/100)*(97.5/100))=22702.68
that is: 22176rs

if any queries mail me:
huda_cool_9@yahoo.co.in

• arman says:

Are kya vaishali kya problem bata rahi hai itni simple shaat

2. Vaishali says:

A persons travel equal distance with speed of 3 km/hr , 4 km/hr , 56 km/hr & takes a total time of 47 minute . The total distance is

• Saumya Roy says:

Let,us consider the distance be X km.
So, in three times it will be 3X km.
then time taken in each: X/3+X/4+X/5
then,X/3+X/4+X/5=47/60 hour
or,47X/60=47/60
or,X =47/60*60/47
or,X = 1
then, total distance is 3X = 3*1=3 km.

(in your problem the term 56km/hr must be wrong it have to 5km/hr)

• soniya says:

let the same distance be x.
total time = x/3(kmph)+x/4(kmph)+x/56(kmph)

47/60 = 101x/168
x=1.303 km

3. hemant says:

in a race horse A takes 7 leaps for every 5 leap made by horse B where as, 6 leaps of horse B is equal 9 leaps of horse A. Find the ratio of the time horse A and horse B take to cover a certain distance ?

• \m/-_-\m/ says:

For the same amount of time : A takes 7 leaps whereas B takes 5 leaps. Hence the ratio of their speed is equal to their distance covered.
Also, we know that 6 leaps of B = 9 leaps of A or
1 ALeap = (2/3) BLeap
Therefore, Time A : Time B = Speed B : Speed A
= 5 BLeaps : 7 ALeaps
= 5 Bleaps : 7*(2/3)*BLeap
= 5*3:7*2
= 15:14
Time of A : Time of B :: 15:14

• hiral says:

ratio of A:B=7:5
ratio of B:C=6:9
There fore A:B:B:C=7:5:6:9
HENCE A:B:C=42:30:45

SO RATIO OF A:B=42:30

4. vasuki says:

the number of signals that can be sent by 6 flags of different colours taking one or more at a time is given by ?Ans : 1956.how?

• \m/-_-\m/ says:

No. of signals you can send using 1 flag = 1×6 = 6
No. of signals you can send using 2 flags = 6P2 = 6×5
= 30
No. of signals you can send using 3 flags = 6P3 = 6x5x4
= 120
No. of signals you can send using 4 flags = 6P4 = 6x5x4x3
= 360
No. of signals you can send using 5 flags = 6P2 =
6x5x4x3x2
= 720
No. of signals you can send using 6 flags = 6P1 = 6!
= 720
Hence, the no. of signals that can be sent using 6 flags of different colours taking one or more at a time =
6 + 30 + 120 + 360 + 720 + 720 =
1440 + 480 + 36 =
1956.

5. muralidhara says:

A takes 4 hours more than B to walk 30 km. If he doubles his speed, he takes 1 hour less time than B. Find their walking rate

• Saumya Roy says:

Just observe that,
If speed of A is double.
Then,he will take half time to cover the same distance with that double speed.
So, when A takes the 1/2 time then their time gap is 5 hours(from 4 hours MORE to 1 hour LESS)
So, when A’s time is normal (1) then their time gap is 2*5 = 10 hours.
Now, as per question A takes 4 hours more than B to walk total distance.
So,A’s speed is = 30/10 = 3km/hr
B’s speed is = 30/6 = 5km/hr

• \m/-_-\m/ says:

Cool!

• Ashutosh says:

your a value is when the speed of a is not get twiced…..

• Ashutosh says:

Hi,
ya here is the way u can do it,
let speed of b be x, then b takes x hrs to complete the journey.
Now the same distance a will cover in x+4 hrs.
As the speed of a become twice then the time take to cover the distance is (x+4)/2.
And a will take 1 hr less than b.

So the simle liner equ. relating the term is,

A’s time = (x+4)/2
B’s time = x

so, (x+4)/2 = x-1

we get x=6;

so speed of b is 30/6 = 5 kmph

and speed of a is 30/(6-1)= 6kmph;

• hiral says:

TIME TAKEN BY A & B = X+4 AND X
SO SPEED OF A AND B = X & X+4
NOW IF A DOUBLES SPEED….2X
EQATION WILL BE…2X-X+4=1(HOUR)

SO X=3

6. ramesh says:

1.walking at 6/7th of the usual speed a man is 25 minutes late his usual time is?
2.A train 360m long is running at 40km/h in how much time will it pass a platform 140m long?
3.A is twice as faster as B and B is trice as faster as C is the journey covered by c in 54 min will be covered by A in?

7. srikanth says:

if a man travels 3kmph more than his regular speed he will reach his destiny 1hr earlier,and if he travels 3kmph less than regular speed he will 2hrs late. what is the distance between source and destiny?

8. Jaiswal says:

A man is going to a wedding party. He travels for 2 hours when his vehicle gets a flat tyre. Changing tyres takes 10mins. The rest of the journey he travels at 30 miles/hr. He reaches 30mins behind schedule. He thinks to himself that if that if the flat tyre had occurred 30 miles later, he would have been only 15 mins late.
Find the total distance traveled by the man.

9. narenderreddy says:

9. A car travels from B at a speed of 20 km/hr. The bus travel starts from A at a time of 6 A.M. There is a bus for every half an hour interval. The car starts at 12 noon. Each bus travels at a speed of 25 km/hr. Distance between A and B is 100 km. During its journey , The number of buses that the car encounter is ?

10. Priya says:

Ajay walks from A to B at 10 km/hr and back to A at 8 km/hr.The total time taken for the onward and return journey is 4 1/2 hours.what is the total distance covered by him during the entire journey?

11. dhana says:

there are 45 students in a class, abi is taking 14th rank from last, then wat is the rank of abi from first

12. anurag says:

in a 500m race,A beats B by 40m.In a 1000m race race,B beats C by 40 m.If A beats C by 14.6 seconds in a 500m race,find the time taken by B to run 2.4 km.

13. anurag says:

A had covered 2/3rd of a certain distance when his car had a breakdown.he parked it and covered the remaining distance on foot.his time of travel on foot was 18 times his time of travel on car.how many times was his walking speed with respect to his car’s speed?

14. akash luthra says:

Buses start from a bus terminal
with a speed of 20 km/hr at
intervals of 10 minutes. What is
the speed of a man coming from
the opposite direction towards the
bus terminal if he meets the buses
at intervals of 8 minutes?
give ne explanation…

15. janu says:

A train runs first half of the distance at 40 km/hr and the remaining half at 60 km/hr. What is the
average speed for the entire journey?

16. twinkle says:

DOUBT-
a man starts from B to K,another man from K to B at the same time.after passing each other they complete their journey in 3 1/3 and 4 4/5 hours respectively.find the speed of second man if the speed of the first is 12 km/hr ??