# Speed, Time & Distance Tricks & Tips

Time, Speed & Distance is one of the most important topics in quantitative aptitude section in various SSC and Banking exams.

### INTRODUCTION:

Time is referred as time taken to complete a certain distance and its unit is hr/min/sec, Speed is referred as the rate at which a certain distance is covered and its unit is km/hr,m/sec , Distance is referred as the total distance covered its unit is km/m.

Time and Distance Formulas

Distance=Speed*Time(km/m)

Speed=Distance/Time(km/hr (or) m/sec)

Time=Distance/Speed(hr/sec)

 SPEED TIME DISTANCE Km/hr Hr km m/sec Sec m

### AVERAGE SPEED WHEN DISTANCE ARE SAME :

Average Speed = Total Distance/Total Time

Average Speed (when distance is same and there is 2 speed) =2xy/(x+y)

Average Speed (when distance is same and there is 3 speed) =3xyz/(xy+yz+zx)

### TYPE CONVERSIONS:

If the unit is in km/hr and it is converted into m/sec by multiplying it with 5/18,

If the unit is in m/sec and it is converted into km/hr by multiplying it with 18/5,

1 km/hr  is converted into 1m/sec by *(5/18)

1 m/sec is converted into 1 Km/hr by *(18/5)

### PROPORTIONAL:

If Distance is Constant then, Time(T1 &T2) and Speed(S1 and S2) is inversely proportional to each other S1/S2=T2/T1

If Time is Constant then, the ratio between Distance and Speed is directly proportional to each other.

(i.e. If Speed increases then Distance is also increases,If Speed decreases then Distance also decreases)

## #1 TYPE 1: BASED ON TIME ,DISTANCE AND SPEED:

If a man rides a distance of 540km at the rate of 60Km/hr .Then at what speed he should ride in order to reach the destination less than 3 hrs than his usual speed express in m/sec?

Time=Distance/Speed

=540/60

=9hr.

Speed=Distance/Time(time =(9-3hr)=6hr)

=540/6

=90km/hr.

Type Conversion:

1 Km/hr ⇒ 1m/sec *(5/18)

90km/hr=90*(5/18)=25m/sec

### Try & Practice Questions:

1. A increases his speed by 5km/hr then time taken reduced by 20%.What is the initial speed of journey?

a) 10km/hr b) 15km/hr c) 20km/hr d) 25km/hr

2.A Car travelling with 5/7 of the actual speed covers 42km in 1hr 40min 48sec.Find the actual speed of the car?

a) 17km/hr b) 32km/hr c) 31km/hr d) 35km/hr

3. A man travels on a Scooter from A to B at a speed of 30km/hr and returns back from B to A at 20km/hr. The total journey was performed by him in 10hr.Find the distance from A to B?

a) 100km b) 110km c) 120km d) 125km

## #2 TYPE 2 : BASED ON REACHING THE DESTINATION EARLY AND LATE:

If a man walks from home to his office at the rate of 40km/hr and he is late by 20 minutes and if man goes to his office from home by riding a bike at the rate of 60km/hr then he reaches the office 10 minutes early. Then what is the distance between his home and his office?

Solution:

Let us consider a man starts from his home by 9.00A.M.

By walking he would reach at 9.20 A.M.

By Riding he would reach at 8.50 A.M.

Here we don’t know the distance between them so lets take L.C.M of speed and consider it as the total distance

L.C.M. of 40,60=60km

By walking a man takes 60/40=1hr 30 mins

By Riding a man takes 60/60=1hr.

The difference between the time taken here is 30 mins(1hr 30 mins and 1hr is 30 mins)

The difference between 8.50 AM -9.20 AM=30 minutes.

Here the difference between the time taken here is 30 mins(1hr 30 mins and 1hr is 30 mins) and The difference between 8.50 AM – 9.20 AM=30 minutes is same.

Therefore the distance between his home and office is 60Km.

Shortcut Method:

LCM of 40,60=60 KM

By Walking he takes =60/40=1hr 30 min

By Riding he takes=60/60=1 hr

1hr30Min-1hr=30 Min

8.50 A.M. – 9.20 A.M.=30 Mins

Here they are same (30 Mins=30 Mins) So the distance is 60 Km

TRY & Practice Questions:

1)Starting from his house one day ,a student walks at a speed of 2(1/2)kmph and reaches his school 6minutes late .Next day increases his speed by 1kmph and reaches the school 6 minutes early. How far the is the school from his house?

a)1km b)15km c)14km d)1(3/4)km

2) If I walk at 4km/hr, I miss the bus by 10 min. If I walk at 5km/hr, I reach 5 min before the arrival of the bus. How far I walk to reach the bus stand?

a)5km b)5.5km c)6km d)7.5km

3.If a man goes to his office by bicycle from home at the rate of 60km/hr he reaches 10 minutes early and If a man goes to his office by bike from home at the rate of 80km/hr then he reaches the office 25 minutes early so what is the distance between the his home and his office?

A) 60km B) 200km C) 300km D) 400km

## #3 TYPE 3: BASED ON AVERAGES:

### For Different Distance:

Average Speed=(Total distance/Total Time Taken)

For Same Distance :

 Number Of Speed given Average Speed 2 Speed namely x and y 2xy/(x+y) 3 Speed namely x,y and z 3xyz/(xy+yz+zx)

1.A bus moves along the road which is in the shape of square.It’s speed along each side of the square are recorded as 50km/hr,25km/hr,60km/hr and 90km/hr .Find the approximate average speed of the bus in km/hr)?

Solution:

Here we don’t know the distance  bus moves so, take

LCM(50,25,60,90=900km)

But here they say the Bus moves in a square path so the distance covered by the bus=900*4=3600.

Average Speed=Total Distance /Total Time

Time taken in 1st path=900/50=18hr

Time taken in 2nd path=900/25=36hr

Time taken in 3rd path=900/60=15hr

Time taken in 4th path=900/90=10hr

Average Speed=3600/(18+36+15+10)=3600/79=45km/hr

2 .A man walks 7.5 km at a speed of 3km/hr. At what speed would the man need walk during the next 2hr to have an average of 4km/hr for the entire session?

Solution

Time=Distance/Speed

=7.5/3  ⇒ 2hr30min.

 Speed Time Distance 3 km/hr 2hr 30min 7.5km X km/hr 2hr 2X Average Speed=4km/hr

Average Speed =Total Distance /Total time taken

Therefore,

Total Distance=7.5+2X

Average Speed=4km/hr

Total time taken=4hr30min

Therefore 4=(7.5+2X)/4hr 30min

18=7.5+2X

10.5=2X

5.25=X

### Try & Practice Questions

1. When a man travels a distance with different speed separating the distance in to 3 equal halves and it reaches the destination after travelling with velocity of 12km/hr, 15km/hr and 20km/hr. Find the average speed of the man?

a)12km/hr b)13km/hr c)14km/hr d)15km/hr

2.A Car Driver covers a distance between two cities at a speed of 60km/hr and on the return his speed is 40km/hr.He goes again from the 1st to the 2nd city at twice the original speed and returns at the half the original return speed.Find the average speed for entire journey?

a)55km/hr b)50km/hr c)48km/hr d)40km/hr

3. Mac travels from A to B a distance of 250 miles in 5hrs30 minutes. He returns to A in 4hrs 30mins.his average speed is?

a)42m/h b)48m/h c)49m/h d)50m/h

## #4 TYPE 4:BASED ON DIRECTIONS:

 Direction Speed Relative Speed Same Direction S1,S2 S1-S2/S2-S1 Opposite Direction S1,S2 S1+S2

1 .Two men starting from the same place walk at the rate of 4km/hr and 4.6km/hr respectively. What time will they take to be 3 km apart ,if they walk in the same direction?

 Relative Speed(Same Direction) 1st men 2nd men 0.6km/hr 4km/hr 4.6km/hr Same Direction=4.6km/hr – 4km/hr=.6km/hr

Time =Distance/Speed

=3/0.6hr

Time =5hours

2. Two cars A and B are running towards each other from two different place which is 88km apart. If the ratio of the speeds of the cars A and B is 5:6 and the speed of the car B is 90km/hr, after what time they meet each other express in mins?

 Relative Speed(opposite direction) Car A Car B 11x 5x 6x 75+90=165 75 90

Time=Total Distance/Relative Speed

=88/165hr

=88*60/165

Time =32mins

TRY:

1.Ram and Sita are walking towards each other with the speed of 10km/hr and 20km/hr ,respectively over a road 900km long. How long will they meet?

a)30hr b)40hr c)50hr d)60hr

2 .The thief escaped from the jail at 12am on Monday and escapes at the rate of 40km/hr. Police notice at 6am on the same day and searches him at the velocity of 50km/hr at what time and on which day they catch the thief?

a)6pm on Monday b)6 am on Tuesday c)6pm on Tuesday d)6am on Wednesday

## #5 TYPE 5: DISTANCE COVERED WITH DIFFERENT SPEED:

1.I started on my bicycle on 7 AM to reach a mall. But after going for a certain distance my bicycle went out of order. Then I rested there for 35mins then returned home at 1PM.What is the distance covered by me if my cycling speed be 10km/hr and walking speed be 1km/hr?

Let my Distance be X,

From 7AM to 1PM=6Hours.

Time=Distance /Speed

 Time Taken for Cycling(hr) Rest Time(hr) Time Taken for walking(hr) X/10 35/60 X/1

(X/10) + (35/60)+(X/1)=6

66X+35=360

66X=325

X=4(61/66)km

Q. Two persons A and B started walking from two places P and Q respectively towards Q and P at 8.20 A.M. Their speeds of walking were in the ratio 4:5. They met at a place between P and Q ,spent some time together for coffee and then both started towards their respective destinations at 9.27 A.M. If A reached Q at 10.32 A.M., how much time they spent together?

Options:

a) 17 Minutes b) 15 Minutes C) 10 Minutes d) 22 Minutes e) 25 Minutes

As the problem is silent on the distance covered either by A or by B, the ratio of the time taken by them will be helpful in quickly solving it. As speed and time are inversely proportional, the ratio of time taken by A and B is 5:4.

How much time did A take to cover the distance between P and Q as per the Question? Let the point at which they met be R. A covers the distance between P and R in (9.27 Hours – 8.20 Hours) i.e. in 67 minutes including the time spent together by them at R. A specific point to remember here is that the difference in time should not be computed by simply subtracting the fractions, but by converting the time elements into minutes.( one Hour= 60 Minutes). The actual time taken by him to cover the distance between R and Q is (10.32 Hours-9.27 Hours) i.e. 65 Minutes. The total time taken by A to cover the distance between P and Q = Time taken to cover the distance between P and R + The time spent with B for coffee + Time taken to cover the distance between R and Q. Let the time (in minutes) they spent together be t. For computing the speed at which A traveled  the resting time (the  time taken for coffee) should not be included. Thus in relation to speed, the actual time taken by A to cover the entire distance between P and Q is 67 Minutes-t Minutes + 65 Minutes. While A takes ( 67-t) and 65 Minutes as actual traveling time, B takes  only (67 -t) X 5/4 Minutes (from Q to R) plus  65X 5/4 Minutes(from R to Q) to complete his journey from Q to P.  While A takes 65 Minutes to cover the distance between R and P, Q takes 65 X 4/5 Minutes to complete the remaining distance between R and Q.

It is therefore clear that :

67-t (Minutes) = 65 X 4/5 (Minutes) = 52 Minutes.

Solving the simple equation, t = 15 Minutes.

The solution is that of finding the time spent together by them which is t.

Illustration:

A

67-t Minutes              (Time taken)                     65 Minutes

P –———————––————- R –——-–———————— Q

52 Minutes (Time taken) 41 Minutes +36 Seconds B

Ratio of their time (A:B) = 52+ 65 Minutes : 52 +(41 Minutes + 36 Seconds).

i.e 117 Minutes : 93 Minutes plus 36 Seconds.

7020 Second: 5616 Seconds.

5 : 4

Which is given.

1)Akash,Vikash & Vinesh are the three friends left City A and move to City B with equal intervals of time between them starting from Akash.And they reach City B simultaneously and again move to City  C which is 300 km from City B.Vikash reach City C before an hour Akash reach the City C.Vinesh reaches City C & immediately travel  to City B and meets Akash 100km away from City C.Find the Speed of Akash?

Friends don’t Panic by just seeing the length of the Question this Problem is quite big but can be solved by just knowing Relationship between Time,Speed & Distance.Nothing more is needed to solve these kinds of Problem.

### Finding The Value Of Akash:

Friends this Problem is quite bigger and these type of Problem can be solved if we know to find the Relationship between them.

As per given in the question

When Akash Travels 200 Km Vinesh Travels 400 Km From this we can Say Vinesh is 2(Akash)

Then it is Given that they start with equal intervals of time,

Ta-Tv=Tv-Tvi

From the Above Step we will be Substitute the Value of Vinesh in terms of Akash

We have Solved this And Found that Sv=4Sa/3(only this Part is Calculative)

After the next Part is Quite easy ,

To find the Value  of Sa,

Distance given in the question/Sa=Difference in Time between 2 Persons

Then we will find the Sa Value from which we can find Sv & Svi

### RACES:

There are two types of Races are there they are Linear Races and Circular Races

### Type  1: LINEAR  RACES:

1.In a Race of 100 metre ,A beats B by 20 metre and beats C by 30m.Then in a race of 100 m by how many metre  will B beats C?

Explanation:

If   A runs 100 metre then B will run 80 metre ,If A ru 100 metre then C will run 70 metre.

 A B C 100 80 70

B beats C by

(10/80)*100=12.5metres

2.In a Race of 100 m A beats B by 20 m and in other Race B beats C by 30 m .then fow many metres A beats C?

Explanation:
 A    B      A B     C     C
 100    100                        100 80       70                          56

A beats C by 44m(100-56=44m)

### CIRCULAR  RACE:

1.A,B and C participated in a Circular Race and A completes the race in 20 sec B in 15 sec and C in 10 secs if all the three starts at the same time and the same point then find when all will meet at the first time?

Explanation:

A  completes the circular field at 20,40,60..sec

B completes the circular field at 15,30,45,60…sec

C completes the circular field at 10,20,30,40,50,60.sec.

So they will meet in 60th sec

Hint :

For first meet at starting  point just take LCM if they move in same direction as well as in opposite direction

LCM(20,15,10=60sec)

2.A and B moves around a circular field if A completes the circular field in 25 sec and B completes the circular field in 20 sec then at what time they will meet the first time?

Explanation:
 Opposite Direction: Same Direction Speed of A=ASpeed of B=B Relative Speed=A+B Speed of A=ASpeed of B=B Relative Speed=A-B

 Speed Time Distance Speed of A=D/25 25sec D Speed of B=D/20 20sec D

=D/((D/25)+(D/20))=500D/45D=100/9sec

1.Speed, Time and Distance:

Speed = Distance / time

Time = distance /speed

Distance =speed*distance

2. km/hr to m/sec conversion:
Y km/hr = [Y × 5/18] m/sec

3. m/sec to km/hr conversion:
Y m/sec = [Y × 18/5] km/h

4. If the ratio of the speeds of A and B is a : b , then the ratio of the
times taken by then to cover the same distance is 1/a:1/b or b : a.

5. Suppose a man covers a certain distance at x km/hr and an equal distance at y km/hr. Then, the average speed during the whole journey is[xy/(x+y)] km/hr.

6Average speed: If both the time taken are equal i.e t1 = t2 = t ,then, t1 + t2 / 2

7. The average of odd numbers from 1 to n is = [Last odd no. + 1] / 2.

8. The average of even numbers from 1 to n is = [Last even no. + 2] / 2.

TRICKS:

Average speed should not be calculated as average of different speeds, i.e., Ave. speed ≠ =Sum of speed / No. of different Speed

There are two different cases when average speed is required.

1. Case I

When time remains constant and speed varies :

If a man travels at the rate of x km/h for t hours and again at the rate of y km/h for another t hours, then for the whole journey, his average speed is given by

Average speed = Total Distance÷Total Time

2. Case II

When the distance covered remains same and the speeds vary :

When a man covers a certain distance with a speed of x km/h and another equal distance at the rate of y km/h. then for the whole journey, the average speed is given by

Average speed =2xy/(x+y)km/h.

3. Velocity

The speed of a moving body is called as its velocity if the direction of motion is also taken into consideration

Velocity = Net Displacement Of The Body ÷ Time Taken

4. Relative speed

4(a) Bodies moving in same direction

When two bodies move in the same direction, then the difference of their speeds is called the relative speed of one with respect to the other.

When two bodies move in the same direction, the distance between them increases (or decreases) at the rate of difference of their speeds

4(b) Bodies moving in opposite direction

The distance between two bodies moving towards each other will get reduced at the rate of their relative speed (i.e., sum of their speeds). =Initial distance between two bodies/ Some of their Speed

Relative speed of one body with respect to other body is sum of their speeds.

Increase or decrease in distance between them is the product of their relative speed and time.

Important tricks to solve the problems:-

• When a moving body covers a certain distance at x km/h and another same distance at the speed of y km/h, then average speed of moving body during its entire journey will be
[2xy/(x+y)]km/h
• A man covers a certain distance at x km/h by car and the same distance at y km/h by bicycle. If the time taken by him for the whole journey by t hours, then Total distance covered by him = 2txy/(x+y) km.
• A boy walks from his house at x km/h and reaches the school ‘ t 1 ‘ minutes late. If he walks at y km/h he reaches ‘ t 2 ‘ minutes earlier. Then, distance between the school and the house. =[xy / (y-x) ] [( t1+ t2 ) / 60 ]km
• If a man walks with (x/y) of his usual speed he takes t hours more to cover a certain distance.Then the time to cover the same distance when he walks with his usual speed, x t / (y-x) hours.
• If two persons A and B start at the same time in opposite directions from the points and after passing each other they complete the journeys in ‘ x ‘ and ‘ y’ hrs. respectively, then A’s speed : B’s speed = √y :√x
• If the speed is (a/b) of the original speed, then the change in time taken to cover the same distance is given by Change in time = [(b/a)-1] × original time

Key notes to solve problems on Trains

• The time taken by a train in passing a signal post or a telegraph pole or a man standing near a railway line = Length of the train / Speed of the train.
• The time taken by a train passing a railway bridge or a platform or a tunnel or a train at rest = (x + y ) / speed where, x = length of the train y = length of the bridge or platform or standing train or tunnel.
• Time taken by faster train to pass the slower train in the same direction = x +y / u-v
• where, x = length of the first trainy = length of the second train
u = speed of the first trainv = speed of the second train and u > v
• Time taken by the trains in passing each other while moving in opposite direction = x +y / u-v.
• Time taken by the train to cross a man = x / (u-v) where, both are moving in the same direction andx = length of the trainu = speed of the train andv = speed of the manTime taken by the train to across a man running in the opposite direction = x / ( u+ v)
• If two trains start at the same time from two points A and B towards each other and after crossing, they take a and b hours in reaching B and A respectively. Then, A’s speed : B’s speed = ( √b :√a)
• A train starts from a place at u km/h and another fast train starts from the same place after t hours at v km/h in the same direction. Find at what distance from the starting place both the trains will meet and also find the time of their meeting.Distance = uvt / (v-u) kmTime = ut / v-u hours
• The distance between two places A and B is x km. A train starts from A to B at u km/h. One another train after t hours starts from B to A at v km/h. At what distance from A will both the train meet and also find the time of their meetingTime=[ ( x – u t / u + v ) + t ] hours.Distance from A = {u [ (x – u t ) / ( u + v ) ] + t } km
• Two trains starts simultaneously from the stations A and B towards each other at the rates of u and v km/h respectively. When they meet it is found that the second train had traveled x km more than the first. Then the distance between the two stations = [ x ( u + v ) / (v – u ) ] km. (i.e., between A and B)

### Relationship Between Speed, Time & Distance

Speed = Distance / Time

We can deduce the following from this formula:-

– When time is constant, distance covered is directly proportional to the speed.
– When distance is the same, speed is inversely proportional to time.

### Average Speed

Average Speed = Total Distance Traveled / Total Time Taken
– Remember that average speed is NOT the arithmetic mean of the speeds.
– Also, average speed can never be double or more than double of any of the original speeds.

### Relative Speed

When two objects with speed S1 and S2 respectively and they are traveling in:
– Same direction, the relative speed (S’) is the difference of the individual speeds
S’ = S1 – S2
– Opposite direction, the relative speed (S’) is the sum of the individual speeds
S’ = S1 + S2

### Trains Crossing

If L1 and L2 are the lengths of two trains moving at speeds V1 and V2 respectively, then the time taken by them to cross each other given by,
Time to Cross = ( L1 + L2 ) / (Relative Speed)

### Boats & Streams

If a boat traveling at the speed (B) is in a stream, the speed of which is denoted by S and it is traveling:
– Upstream (against the direction in which the stream is flowing)
Upstream Speed = B – S
– Downstream (in the same direction as that of the flow of the stream)
Downstream Speed = B+ S

#### Circular Motion

– When two runners are on the same circular track, the time taken for them to meet for the first time is given by the following expression:-
Length of the track / Relative speed of the runners

– Number of times two runners meet on the circular track = Number of rounds gained by faster

runner over the slower one.

– If ratio of speeds of two runners running in circular track is x : y, they will meet at the starting point again in the time given by the following expressions:-

|x – y| time (if running in the same direction)
(x + y) time (if running in the opposite direction)

### Questions on Races

Some points to remember while solving questions based on these are as follows:-
– the distance covered by the winner = length of the race
– loser’s distance = winner’s distance – (beat distance + start distance)
– winner’s time = loser’s time – (beat time + start time)

– if a race ends in a deadlock, i.e. both reach the winning post together then beat time = 0 and beat distance = 0

# Practice Sets for Speed Time Distance

• Two places are A and B are 200 kms from each other. A train leaves from A for B at the same time another train leaves B for A. The two trains meet at the end of 8 hours. If the train travelling from A to B travels 8km/hr faster than the other. Find the speed of the faster train?
A.15 km/hr
B.17.5 km/hr
C.16.5 km/hr
D18 km/hr
E.None of these

Explanation :
Speed of trains  = s + 8 and s km/hr.
At some P distance from A both train will meet. So,
8 = P/(s +8) ; 8 = (200 – P)/s
Solve both equation, we get s = 8.5 so faster train speed 16.5 km/hr
SHORTCUT:
200/8 = 25
A+B = 25
A- B = 8      (given )
A= 16.5 km/hr
• A truck driving on a highway passed a man walking at the rate of 15km/hr in the same direction. He could see the truck for 2 minutes and up to 500 meters. Find the speed of the truck?
A.27km/hr
B.30km/hr
C.32km/hr
D.42km/hr
E. None of these

Explanation :
x-15=(500/120) *18/5
x=30
• A train leaves Chennai for Bangalore at 2:15 p.m. and travels at the rate of 50 kmph. Another train leaves Bangalore for Chennai at 1:35 p.m. and travels at the rate of 60 kmph. If the distance between Bangalore and Chennai is 590 km at what distance from Chennai will the two trains meet?
A.280 km
B.320 km
C.250 km
D.225 km
E.None of these

Explanation :
Total Distance = 590 km
Distance in 40 min = 60*40/60=40km
Remaining=550 km
Time = 550/110=5hr
Reqd Distance = 50*5=250 km
• A bus was travelling from Mumbai to Pune was delayed by 16 minutes and made up for the delay on a section of 80 km travelling with a speed 10 km per hour higher than its normal speed. Find the original speed of the bus?
A.60 km/h
B.66.66 km/h
C.50 km/h
D.40 km/h
E.None of these

Explanation :
80/x – 80/(x+10) = 16/60= 4/15
300/x -300/x+10 =1
X2 + 10x – 3000 = 0
X = -60, +50
So 50km/hr
• Two cities A and B are at a distance of 120 km from each other. Two persons P and Q start from First city at a speed of 20km/hr and 10km/hr respectively. P reached the second city B and returns back and meets Q at Y. Find the distance between A and Y.
A.50 km
B.66 km
C.55 km
D.40 km
E.None of these

Explanation :
P = 120/20 = 6hrs
In 6hrs Q travels 60km
120-60 = 60km
60/(20+10) = 2hrs
Point Y → 60 + 10*2hrs = 80
• Vijay takes 4 hr in walking at certain place and return back. While it takes 3 hrs in walking at certain place and riding back. Find the time Vijay will take to ride both sides
A.1.5hr
B.2 hr
C.2.5 hr
D. 3.5 hr
E. None of these

Explanation :
W + W = 4. W =2
W+ R = 3. R =1
So to ride both direction, it will take 1+1 = 2 hrs.
• Rakesh travelled 2000 kilometre by air which formed 3/5 of the total journey. He travels 1/4 of the trip by car and the remaining trip by train. Find the distance travelled by train.
A.2800
B.4500
C.2500
D.3800
E.None of these

Explanation :
3=2000
5=10000/3
Distance by train=3×10000/3×4=2500
Another method :
Total 20
12====2000
20====2000*20/12
5=====2000*5/12
3=====2000*5*3/12=2500
• Rahul has to travel from one point to another point in a certain time. Travelling at a speed of 6kmph he reaches 40m late and travelling at a speed of 8kmph he reaches 12 m earlier.What is the distance between this two points ?
A.27km
B.18km
C.15km
D.21km
E.None of these

Explanation :
t +40/60 = d/6
t – 12/60 = d/8
By the solving these two equations we get.
d = 20.8 km ~ 21 km
Another method :
6………………..4.           (40m late)
………24…………
8…………………..3.            (12 m ear)
60 = 52
1 = 52/60 = 13/15
So distance = 24 *13/15
= 8*13/5 = 104/5 = 20.8km
• Vivek travelled a distance of 50km in 7hrs. He travelled the distance partly on foot at 5kmph and partly on bicycle at 8kmph. What is the distance that he travelled on foot ?
A.12km
B.5km
C.10km
D.18km
E.None of these

Explanation :
x/5 + 50-x/8 =7
8x+250-5x = 7*40
3x+250 = 280
3x = 30
X = 30/3 = 10km
• Mani drove at the speed of 45 kmph. From home to a resort. Returning over the same route, he got stuck in traffic and took an hour longer, also he could drive only at the speed of 40 kmph. How many kilometers did he drive each way ?
A.520km
B.240km
C.360km
D.420km
E.None of these

Explanation :
x/40 – x/45 = 1
9x-8x/360 = 1
x/360 = 1
x = 360km
• Rohit starts cycling along the boundaries of the squares. He starts from a point A and after 90 minutes he reached to point C diagonally opposite to A. If he is travelling with 20km/hr, then find the area of square field.
A.150
B.225
C.350
D.455
E.None of these

Explanation :
D = 20*3/2 = 30 km. So side of square is 15km, so area – 225km^2

• The distance of the School and house of Suresh is 80km. One day he was late by 1 hour than the normal time to leave for the college, so he increased his speed by 4km/h and thus he reached to college at the normal time. What is the changed speed of Suresh?
A. 28 kmph
B. 25 kmph
C. 20 kmph
D. 24 kmph

Explanation :
80/x – 80/(x+4) = 1
x(x+20) – 16(x+20) = 0
x = 16kmph
Increased speed = 20 kmph
• Anita goes to College at 20 km/h and reaches college 4 minutes late. Next time she goes at 25 km/h and reaches the college 2 minutes earlier than the scheduled time. What is the distance of her school?
A. 16 km
B. 12 km
C. 15 km
D. 10 km

20*25/(25-20)*6/60=10.
• Two places R and S are 800 km apart from each other. Two persons start from R towards S at an interval of 2 hours. Whereas A leaves R for S before B. The speeds of A and B are 40 kmph and 60 kmph respectively. B overtakes A at M, which is on the way from R to S. What is the ratio of time taken by A and B to meet at M?
A. 1:3
B. 1:2
C. 1:4
D. 3:2
E. None of these

Explanation :
Time taken by B to reach at M = 4h
Time taken by A to reach at M = 6h
Ratio = 6:4 = 3:2
• Two places R and S are 800 km apart from each other. Two persons start from R towards S at an interval of 2 hours. Whereas A leaves R for S before B. The speeds of A and B are 40 kmph and 60 kmph respectively. B overtakes A at M, which is on the way from R to S. What is the extra time taken by A to reach at S?
A. 6hrs 20 minutes
B. 6hrs 40 minutes
C. 6hrs 30 minutes
D. 6hrs 10 minutes

Answer – B. 6hrs 40 minutes
Explanation :
Time taken by A to reach at Q = 800/40 = 20 hours
Time taken by B to reach at Q = 800/60 = 13 hours and 20 min
A takes 6hr 40 minutes extra time to reach at Q.
• Ajay covers certain distance with his own speed but when he reduces his speed by 10kmph his time duration for the journey increases by 40 hours while if he increases his speed by 5 kmph from his original speed he takes 10 hours less than the original time taken. Find the distance covered by him.
A. 1000 km
B. 1200 km
C. 1500 km
D. 1800 km

Explanation :

x/(y – 10)  – x/y = 40
x = 4y(y-10) —(i)
x/y  – x/(y + 5) = 10
x = 2y(y + 5) — (ii) From (i) and (ii) => y = 25; x = 1500
• The driver of an ambulance sees a college bus 40 m ahead of him after 20 seconds, the college bus is 60 meter behind. If the speed of the ambulance is 30 km/h, what is the speed of the college bus?
A. 10 kmph
B. 12 kmph
C. 15 kmph
D. 22 kmph

Explanation :
Relative Speed = (Total distance)/total time
= (60+40) /20 = 5 m/s = (5*18)/5 = 18 kmph
Relative Speed = (speed of ambulance – speed of College bus)
Speed of College bus = speed of ambulance – relative speed.
= 30-18 = 12 kmph.
• Two places R and S are 800 km apart from each other. Two persons start from R towards S at an interval of 2 hours. Whereas A leaves R for S before B. The speeds of A and B are 40 kmph and 60 kmph respectively. B overtakes A at M, which is on the way from R to S. What is the distance from R, where B overtakes A?
A. 260 km
B. 235 km
C. 240 km
D. 300 km

Explanation :
Distance between R and M = 4 * 60 = 240
• Two rabbits start running towards each other, one from A to B and another from B to A. They cross each other after one hour and the first rabbit reaches B, 5/6 hour before the second rabbit reaches A. If the distance between A and B is 50 km. what is the speed of the slower rabbit?
A. 20 kmph
B. 10 kmph
C. 15 kmph
D. 25 kmph

Explanation :
Let second rabbit takes x hr with speed s2
First rabbit takes x-5/6 hr with speed s1
Total distance = 50km
S1 = 50/(x-(5/6))
S2= 50/x
As they cross each other in 1hr…
Total speed = s1 + s2
Now, T = D / S
50/(s1+s2) = 1
x = 5/2, 1/3
Put x= 5/2 in s2 –> 20km/hr
• Pranav walked at 5 kmph for certain part of the journey and then he took an auto for the remaining part of the journey travelling at 25 kmph. If he took 10 hours for the entire journey, what part of journey did he travelled by auto if the average speed of the entire journey be 17 kmph
A. 750 km
B. 100 km
C. 150 km
D. 200 km

Explanation :

Total distance = 17*10=170
Let Journey travelled by auto in x hr
25 * x + (10-x ) 5 = 170
25 x + 50 – 5x = 170
x = 6
Required Distance = 6 * 25 = 150 km
• Aravind started for the station half a km from his home walking at 1 km/h to catch the train in time. After 3 minutes he realised that he had forgotten a document at home and returned with increased, but constant speed to get it succeded in catching the train. Find his latter speed in kmph?
A. 1.25
B. 1.1
C. 11/9
D. 2

Explanation :

Distance covered in 3 minutes = 3*(1000/60) = 50
Now he has to cover (500+50)m in (30-3) minutes
Required speed = (550/1000)/(27/60) = 11/9 km/h

• A train 150 meters of length travels at the rate of 60 km/hr. In what time the train will pass a man who is walking at a speed of 10 km/hr in opposite direction.
a) 53/7 sec
b) 56/7 sec
c) 54/7 sec
d) 57/7 sec
e) None of these

Explanation :
150 = (70)*5/18*t
• Two trains of length 100 meter and 125 meter are travelling at a speed of 45 km/hr and 60km/hr respectively in same direction. In what time they will completely cross each other.
a) 52 sec
b) 54 sec
c) 56 sec
d) 58 sec
e) None of these

Explanation :
225 = (60 – 45)*5/18*T
• Two trains are travelling in same direction with 60 km/hr and 75 km/hr respectively. The faster train crosses a man sitting in the slower train in 30 sec. find the length of faster train.
a) 100 meter
b) 125 meter
c) 140 meter
d) 150 meter
e) None of these

Explanation :
L = 15*5/18*30 = 125 meter
• A train running at 45 km/hr takes 36 sec to pass a platform. Next, the train takes 12 sec to pass a man walking at the speed of 15 km/hr in the same direction. Find the length of the platform.
a) 250m
b) 300m
c) 350m
d) 400m
e) None of these

Explanation :
Let ‘T’ and ‘P’ are the length of train and platform respectively
T = 12*30*5/18 = 100 meter
P + 100 = 45*5/18*36
P= 350
• Two stations P and Q are 400 km apart from each other. One train start from P at a speed of 60km/hr towards Q and after 2 hours another train starts from Q towards P at 45 km/hr. At what distance from P the train will meet.
a) 220 km
b) 240 km
c) 260 km
d) 280 km
e) None of these

Explanation :
First train will travel 120 km before the start of second train, Now the distance between them is 280km.
Now,  x/60 = (280 – x)/45
We get x = 160 km, so distance from P = 120 + 160 = 280 km
• Two stations A and B are 150 km apart from each other. One train starts from A at 6 AM at a speed of 30 km/hr and travels towards B. Another train starts from station B at 7 AM at a speed of 20 km/hr. At what time they will meet.
a) 9:34 AM
b) 10:34 AM
c) 8:34 AM
d) 7:34 AM
e) None of these

Explanation :
Distance travel by first train in one hour = 30, now the distance remains 120 km only.
x/30 = (120 – x)/20, so we get x = 72 km
Now, time = (30 + 72)/30 = 3hrs and 24minutes i.e. 9: 24 am
• Two trains of length 120 meter and 150 meter crosses a stationary man in 10 and 15 seconds respectively. In what time they will cross each other when they are moving in same direction.
a)120 sec
b) 125 sec
c) 135 sec
d) 140 sec
e) None of these

Explanation :
120 = a*10, a = 12 m/sec (speed of first train)
150 = b*15, b = 10 m/sec (speed of second train)
270 = (2)*T, T = 135 seconds
• A train running with 72km/hr takes 20sec to cross a platform 200 m long. How much it takes to cross a stationary train having twice the length of platform.
a) 20 sec
b) 30 sec
c) 40 sec
d) 50 sec
e) None of these

Explanation :
200+L =  72*(5/18)*20. L= 200m.
400 + 200 = 72*(5/18)*t. So, t = 30sec
• A train travelling with 54 km/hr takes 20 second to cross a bridge. Another train 70 meter shorter crosses the same bridge at 36 km/hr. Find the time taken by the second train to cross the bridge.
a) 23 sec
b) 24 sec
c) 25 sec
d) 26 sec
e) None of these

Explanation :
Let L and B are length of train and bridge respectively.
L + B = 54*5/18*20 = 300 meter
L + B – 70 = 36*5/18*t = 230, se we get t  = 23 sec
• Two trains are moving in opposite direction having speed in the ratio 5:7. First train crosses a pole in 12 second and second train crosses the same pole n 15 second. Find the time in which they can cross each other completely.
a) 55/4 sec
b) 53/4 sec
c) 57/4 sec
d) 59/4 sec
e) None of these

Answer – a) 55/4 secExplanation :
Let the length of first train and second train be a and b meter. Then
a = 5x*12 = 60x and b = 7x*15 = 105x
They are moving in opposite direction, 165x = (12x)*T
T = 165/12 = 55/4 sec

• Sofi started travelling from a place A to B and Priya started travelling from a place B to A which are 576 km apart. They meet after 12 hours. After their meeting, Sofi increased her speed by 2 km/hr and Priya reduced her speed by 2 km/hr, they arrived at B and A respectively at the same time. What is their initial speed?
A. 21 kmph, 23 kmph
B. 25 kmph, 27 kmph
C. 25 kmph, 23 kmph
D. 24 kmph, 26 kmph

Answer – C. 25 kmph, 23 kmph
Explanation :
Sum of their speeds = Distance/time = 576/12 = 48 kmph
Respective Speed of Sofi and Priya = (25 + 23) = 48 kmph
• A and B set out at the same time to walk towards each other respectively from a place P and Q 144 km apart. A walks at the constant speed of 8 km/h, while B walks 4 km in the first hour, 5 km in the second hour, 6 km in the third hour and so on. Then the “A” will meet “B” at?
A. 36 km
B. 72 km
C. 56 km
D. 26 km

Explanation :
Distance travelled by them in first hour = 12 km
Distance travelled by them in second hour = 13 km and so on
In 9 hours both will cover exactly 144 km.
In 9 hours each will cover half the total distance.
• Two Vans start from a place with a speed of 50 kmph at an interval of 12 minutes. What is the speed of a car coming from the opposite direction towards the place if the car meets the vans at an interval of 10 minutes?
A. 13 kmph
B. 10 kmph
C. 14 kmph
D. 16 kmph
E. None of these

Explanation :
50*12/60 = 10/60 * (50+x)
600 = 500 + 10x
x = 10 kmph
• A car travels from a place A to B in 7 hour. It covers half the distance at 30 kmph and the remaining distance at 40 kmph, what is the total distance between A and B?
A. 120 Km
B. 250 Km
C. 240 Km
D. 150 Km

Explanation :
Total Distance = x
(x/2*30) +(x/2*40) = 7
x = 240
• Two persons A and B start from the opposite ends of a 450 km straight track and run to and from between the two ends. The speed of the first person is 25 m/s and the speed of other is 35 m/s. They continue their motion for 10 hours. How many times did they pass each other?
A. 1
B. 4
C. 3
D. 2

Explanation :

First person speed = 25 m/s * 18/5 = 90 kmph
Second person speed = 35 m/s * 18/5 = 126 kmph
First person covers 90 * 10 = 900km
900/450 = 2
• A truck travelled to a place Q from P, the first 50 km at 10 kmph faster than the usual speed, but it returned the same distance at 10 kmph slower than usual speed. If the total time taken by the truck is 12 hours, then how many hours will travel at the faster speed?
A. 8 hours
B. 6 hours
C. 2 hours
D. 3 hours

Explanation :
Total time taken,
[50/(x-10)] + 50/(x +10)] = 12 hours.
By solving the equation, we get
x = 15
Time is taken by the truck at faster speed = 50/(15+10) = 2 hours.
•  Mr.Kavin walks at 4/5 of his normal speed and takes 60 minutes more than the usual time. What will be the new time taken by Mr. Kavin?
A. 260 minutes
B. 235 minutes
C. 220 minutes
D. 300 minutes

Explanation :
4/5 of speed = 5/4 of original time
5/4 of original time = original time + 60 minutes;
1/4 of original time = 60 minutes;
Thus, original time = 60*4 = 240 minutes = 240 + 60 = 300 minutes
• A travel bus normally reaches its destination at 60 kmph in 20 hours. Find the speed of that travel bus at which it travels to reduce the time by 5 hours?
A. 80 kmph
B. 60 kmph
C. 50 kmph
D. 40 kmph

Explanation :
60 * 20 = x *15
x = 80 kmph
• A Lion starts chasing a Giraffe. It takes 4 hours to catch the Giraffe. If the speed of the Lion is 40 km/h. What is the speed of Giraffe?
A. 20 km/h
B. 50 km/h
C. 40 km/h
D. 70 km/h

Explanation :
Giraffe Speed = x Kmph
4 = 4*x/(40-x)
x = 20 km/h.
• Anu and Purvi are running on a circular track of length 500m. The Speed of Anu is 40 m/s and that of Purvi is 30 m/s. They start from the same point at the same time in the same direction. When will they meet again for the first time?
A. 25 s
B. 23 s
C. 50 s
D. 48 s

Explanation :
Time = Distance/Relative Speed = 500/10 = 50 s

• Two trains A and B start from two places P and Q towards  Q and P respectively. After passing each other they take 4 hours 48 minutes and 3 hours and 20 minutes to reach Q and P respectively. If the train from P is moving at 45 km/h then find the speed of other train.
A)69 km/h
B)74 km/h
C)54 km/h
D)64 km/h

Explanation :
Speed of 2nd train= Speed of first train ×(√Time taken by 1st train after meeting/√Time taken by 2nd train after meeting)
• The speeds of the two trains are in the ration 3:4. they are moving in opposite tracks on parallel tracks. The time taken by both the trains to cross a telegraph line are 5 and 10 second respectively. Find the time taken by the trains to cross each other completely.
A)4 5/7seconds
B)6 4/7 seconds
C)1 3/7 seconds
D)7 6/7 seconds

Answer – D) 7 6/7 seconds
Explanation :5×3k+10×4k/ 3k+4K
• A train travels a certain distance by taking 3 stops of 20min each. Considering the period of stoppage, the overall speed of the train comes to 40 km/h. While without consideration of stoppage, it is 45 km/h. How much distance must the train have travelled?
A)400km
B)550km
C)360km
D)200km

Explanation :
45×y=40(y+1)
distance travelled= 45×8
• Two trains of length 115 m and 110 m respectively run on parallel rails. When running in the same direction the faster train passes the slower one in 25 seconds, but when they are running in opposite directions with the same speeds as earlier, they pass each other in 5 seconds. Find the speed of each train?
A)27m/s 18 m/s
B)22m/s 15m/s
C)16m/s 30m/s
D)24m/s 18m/s

Explanation :
x-y=115+110/25
x+y=115+110/5
• A train covers a distance of 3584 km in 2 days 8 hours. If it covers 1440 km on the first day and 1608 km on second day. By how much does the average speed of the train for the remaining part of the journey differ from that for the entire journey?
A)3 km/h less
B)4 km/h more
C)3 km/h more
D)4 km/h less

Explanation :
Total time=56 hrs
speed=3584/56
remaining part=536 remaining time=8 hrs
speed of remaining journey=536/8=67 km/hr
67.64=3 km/h more
• A train travelling with a speed of 60 km/hr catches another train travelling in the same direction and then leaves it 120m behind in 18 seconds. The speed of the second train is
A)26 km/hr
B)35 km/hr
C)36 km/hr
D)63 km/hr

Explanation :50/3 – x=120/18 => x=10 * 18/5
• Two stations P and Q are 110 km apart on a straight track. One train starts from P at 7 a.m. and travels towards Q at 20 kmph. Another train starts from Q at 8 a.m. and travels towards P at a speed of 25 kmph. At what time will they meet?
A)11:00 a.m.
B)09:00 a.m.
C)10:00 a.m.
D)09:10 a.m.

Explanation :20x + 25(x-1) = 110

• An express train travelled at an average speed of 100 km/hr, stopping for 3 minutes after every 75 km. How long did it take to reach its destination 600 km from the starting point?
A)5 hrs 30 min
B)6 hrs 23 min
C)6 hrs 21 min
D)5 hrs 46 min

Answer – C)6 hrs 21 min
Explanation :
Time taken to cover 600= 6hrs
No of stoppages=600/75 -1
total time=3*7
• A train covers a distance in 50 minutes, if it runs at a speed of 48 km/hr on an average. The speed at which the train must run to reduce the time of journey to 40 minutes will be
A)50 km/hr
B)60 km/hr
C)65 km/hr
D)70 km/hr

Explanation :
D=48 * 5/6 = 40 km
Time = 40/60 hr = 2/3 hr
New speed = 40*3/2 km/h
• Two friends starting from the same place walk at a rate of 5 km/h and 5.5 km/h respectively. What time will they take to be 8.5km apart, if they walk in the same direction?
A)15 hrs
B)18 hrs
C)17 hrs
D)16 hrs