## 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

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??

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

Are kya vaishali kya problem bata rahi hai itni simple shaat

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

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)

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

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 ?

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

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

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?

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.

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

Reply

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

Cool!

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

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;

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

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?

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?

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. 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 ?

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?

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

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.

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?

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…

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?

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 ??