# Time & Distance Shortcut Tricks & Tips

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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

## Speed Time And Distance Concepts

(Q1)Sriya with her family travelled from Bolpur to Suri by car at a speed of 40 km/hr and returned to Bolpur at a speed of 50 km/hr.The average speed for the whole journey is.
श्रेया ने अपने परिवार के साथ बोलपुर से सूरी तक की यात्रा कार से 40 km/hr की गति से की और बोलपुर की वापसी यात्रा 50 km/hr की गति से की पूरी यात्रा की | पूरी यात्रा की औसत गति कितनी है ?
(a)$44frac { 4 }{ 9 }$ km/h       (b) 45km/h       (c)$45frac { 1 }{ 2 }$ km/h       (d) 44.78 km/h

Solution-: Method-1

[Note : Total Distance means जाने में तय की गई दूरी + आने में तय की गई दूरी इसलिए d +d किया गया है |]

Method-2

Method-3

[Note – L.C.M of 40, 50 = 200. यदि आपलोग 40, and 50 का L.C.M नहीं निकालना चाहते हैं तो आपलोग कोई भी नंबर Assume  कर  ले लेकिन  Assume करते वक्त यह ध्यान रखना है कि वह Number divisible हो 40 और  50 से | Time= Distance/Speed so Time=200/40=5 hr and Time=200/50=4 hr]

[Note: Total Distance means जाने में तय की गई दूरी + आने में तय की गई दूरी इसलिए 2 से गुणा किया गया है | Exam में आपलोग हमेशा Method-3 ही prefer करे|]

(Q2)A train runs from Howrah to Bandel at an average speed of 20km/hr and returns at an average speed of 30 km/hr. The average speed(in km/hr) of the train in the whole journey is.
एक रेलगाड़ी हावड़ा से बंडेल तक 20 km/hr की औसत गति से चलती है और 30 km/hr की औसत गति से वापस आती है तो उस रेलगाड़ी की पूरी यात्रा की औसत गति कितनी है ?
(a)20 km/hr      (b)22.5 km/hr      (c)24 km/hr      (d)25 km/hr

Solution:

(Q3)A man travels a distance of 24 km at 60 km/hr. Another distance of 24 km at 40 km/hr. His average speed for the whole journey is.
एक व्यक्ति 24 km की यात्रा 60 km/hr की चाल से तथा एक अन्य 24 km की यात्रा 40 km/hr की चाल से करता है तो उसकी पूरी यात्रा की औसत चाल कितनी है ?
(a)48 km/hr      (b)60 km/hr      (c)45 km/hr      (d)50 km/hr

Solution: Method-1

Method-2

(Q4)A man travels a distance of 24km at 6 km/hr. Another distance of 24km at 8 km/hr and a third distance of 24km at 12 km/hr. His average speed for the whole journey is.
एक आदमी 24km की दूरी 6 km/hr की चाल से, एक अन्य 24km की दूरी 8km/hr की चाल से तथा एक तीसरी 24km की दूरी है 12km/hr की चाल से तय करता है | पूरी यात्रा के लिए उसकी औसत चाल है |
(a)8 km/hr      (b)9 km/ hr      (c)8.75 km/hr      (d)10 km/hr

Solution:

Fig.-1

[Note: Fig.-1 only for Concept]

(Q5)A person travels four equal distances of 3km each at a speed of 10 km/hr, 20 km/hr, 30 km/hr and 60 km/hr respectively. Find his average speed for the whole journey is.
एक व्यक्ति 3km की चार समान दूरियों को क्रमश: 10 km/hr, 20 km/hr, 30 km/hr और 60 km/hr की गति से तय करता है तो उसकी पूरी यात्रा की औसत चाल कितनी है ?
(a)18 km/hr      (b)25km/hr      (c)21 km/hr      (d)20km/hr

Solution:

Fig.-1

[Note: Fig.-1 only for Concept]

## Time And Distance Concepts Shortcut Tricks For SSC, Bank Railway Exams

(Q1)A boy rides his bicycle 10 km at an average speed of 12 km/hr and again Travels 12 km at an average speed of 10 km/hr. his average speed for the entire trip is approximately.
एक लड़का अपने साइकिल से 10 km की दूरी 12 km/hr की गति से तय करता है तथा फिर 12 km की दूरी 10 km/hr की गति से तय करता है | पूरी यात्रा में उसकी औसत गति लगभग कितनी है ?
(a)10.4 km/hr      (b)10.8 km/hr      (3)11 km/hr      (d)12.2 km/hr

Solution:

(Q2)A man completes 30 km of a journey at the speed of 6 km/hr and the remaining 40 km of the journey in 5 hours. His average speed for the whole journey is.
एक व्यक्ति 30 किलोमीटर की एक यात्रा को 6km/hr की चाल से तथा शेष 40 किलोमीटर की यात्रा को 5 घंटे में पूरा करता है पूरी यात्रा के लिए उसकी औसत चाल क्या है?
(a)7 km/hr      (b)7.25 km/hr      (c)8 km/hr      (d)7.5 km/hr

Solution:

(Q3)One-third of a certain journey is covered at a rate of 40 km/hr one-fourth at the rate of 30 km/hr and the rest at 25 km/hr. The average speed for the whole journey is.
एक निश्चित दूरी का 1/3 भाग 40 km/hr की चाल से, 1/4 भाग 30 km/hr की चाल से और शेष 25 km/hr की चाल से तय करती है तो पूरी यात्रा की औसत चाल क्या है?
(a)35 km/hr      (b)33 km/hr      (c)30 km/hr      (d)34 km.hr

Solution:

[Concept- अगर Mathematics के किसी भी question में fraction आए और आप को assume करना पड़े तो ये हमेशा ध्यान रखिएगा कि denominator का L.C.M, Assume किए हुए Number में जरूर हो | इस question में denominator 3 and 4 है जिसका L.C.M 12 होता है तो अब आप 12 का multiple कुछ भी Assume कर  सकते है | Note:अगर आपको calculation easy करना है तो इस question में speed 2 digit में दिया हुआ है तो आप Distance को  3 digit में  assume कर ले |]

Let distance=120km
Now, 120× ¹/3=40km, 120×¼=30km
Rest=120-70=50 km

(Q4)A person covers 40% distance at a speed of 40 km/hr and 30% distance at a speed of 30 km/hr and the rest distance at a speed of 15 km/hr. Find his average speed in the whole journey.
एक व्यक्ति 40% की दूरी को 40 km/hr की गति से तय करता है और 30% की दूरी को 30 km/hr की गति से तय करता है और शेष दूरी को 15km/hr की गति से तय करता है तो पूरी यात्रा में उसकी औसत गति क्या है ?
(a)22 km/hr      (b)28 km/hr      (c)25 km/hr      (d)30 km/hr

Solution:

Let Distance=100 km
100 का 40%=40 km , 100 का 30%=30 km
Rest=100-70=30 km

(Q1)A boy goes to his school from his house at a speed of 3 km/hr and returns at a speed of 2 km/hr. if he takes 5 hours in going and coming, then the distance between his house and school is?
एक लड़का अपने घर से स्कूल 3 km/hr की चाल से जाता है तथा वापिस की 2km/hr चाल से लौटता है | यदि आने-जाने में वह कुल 5 घंटे का समय लेता है, तो स्कूल और उसके घर के बीच की दूरी है ?
(a)6 km      (b)5.5 km      (c)5 km      (d)6.5 km

Solution: Method-1

[Time = Distance/Speed]

Method-2

(Q2)A man travelled A certain distance by train at the rate of 25 km/hr and walked back at the rate of 4 km/hr. if the whole journey took 5 hours 48 minutes, the distance was.
किसी व्यक्ति ने 25 km/hr की चाल से चलने वाली रेलगाड़ी द्वारा एक दूरी तय की तथा वही दूरी वापसी में 4 km/hr की चाल से पैदल चलकर तय की | यदि आने-जाने में उसे कुल समय 5 घंटे 48 मिनट लगे हो, तो दूरी थी |
(a)25 km      (2)30 km      (3)20 km     (4)15 km

Solution: Method-1

[Time = Distance/Speed]

[Note: Time=5 hr 48 minutes = 5×(48/60)hour = 29/5 hour]

Method-2

## Time And Distance Concepts Shortcut Tricks For SSC, Bank Railway Exams

(Q1)Walking at a speed of 5 km/hr, a man reaches his office 6 minutes late. walking at 6 km/hr he reaches there 2 minutes early. The distance of his office is.
5 km/hr की गति से चलने पर एक व्यक्ति अपने कार्यालय 6 मिनट देरी से पहुंचता है | यदि वह 6 km/hr की गति से चले तो वह 2 मिनट पहले पहुंच जाता है | तो, उसके कार्यालय की दूरी कितनी है ?
(a)3km      (b)4km      (c)3.5 km      (d)2km

Solution: Method-1

Explation:[Let distance = d km.suppose इस d km की दूरी को एक निश्चित गति से तय करने में उस व्यक्ति को T hour लगता है| now, जब उस व्यक्ति की speed 5 km/hr रहता है तो वह 6 minutes late से पहुंचता है तो Time=(T+6/60)hour लगेगा| [Note-6 minutes को hour में change किया गया है इसलिए 60 से divide किया गया है ] Again जब उस व्यक्ति की speed 6 km/hr रहता है तो वह 2 minutes before से पहुंच जाता है तो Time=(T-2/60)hour लगेगा |]

Method-2

[Note: Find Time difference
यदि एक बार lateसे पहुंचता हो तथा दूसरी बार पहले पहुंच जाता है तो Time difference=late(l)+before(b)
यदि दोनों बार late से पहुंचता हो तो Time difference=(l-l)
यदि दोनों बार पहले पहुंचता हो तो Time difference=(b-b)]

Time difference = 6+2=8 minutes =8/60 hours

(Q2)When a person cycled at 10 km/hr he arrived at his office 6 minutes late. He arrived 6 minutes early when he increased his speed by 2 km per hour. The distance of his office from the starting place is.
एक व्यक्ति 10 km/hr की गति से साइकिल चलाकर अपने कार्यालय 6 मिनट देरी से पहुंचा | जब उसने अपनी गति 2 km/hr और बढ़ा दी, तो वह 6 मिनट पहले पहुंच गया | उस व्यक्ति के कार्यालय और उसके आरंभिक स्थान के बीच की दूरी क्या है?
(a)6km      (b)7 km      (c)12 km      (d)16 km

Solution: Method-1

Method-2

Time difference = 6+6=12 minutes =12/60 hours

(Q3)If a train runs at 70 km/hr, it reaches its destination late by 12 minutes. But if it runs at 80 km/hr, it is late by 3 minutes. The correct time to cover the journey is.
यदि एक रेलगाड़ी 70 km/hr की गति से चलती है तो यह अपने गंतव्य तक 12 मिनट देरी से पहुंचती है | किंतु यदि यह 80 km/hr की गति से चलती है तो यह 3 मिनट देरी से पहुंचती है | यात्रा को तय करने का सही समय क्या है ?
(a)58 minutes      (b)2 hours      (c)1 hours      (d)59 minutes

Solution: Method-1

Method-2

दोनों बार late से पहुंचा so, Time difference=(l-l)

Time difference=12-3=9 minutes=9/60 hours

## Speed Time & Distance tricks and concepts

(Q1)Aman when increasing his speed from 24 km/hr to 30 km/hr. He takes 1 hour less than the usual time to cover a certain distance. what is distance usually covered by Aman?
जब अमन अपनी चाल को 24 km/hr से बढ़ाकर 30km/hr कर लेता है तो वह एक निश्चित दूरी को तय करने में वास्तविक समय से एक घंटा कम समय लेता है तो अमन के द्वारा कितनी दूरी तय की गई?
(a)120km      (b)240km      (c)400km      (d)60 km

Solution: Method-1

Method-2

(Q2)Mohan when increasing his speed from 30 km/hr to 40 km/hr. He takes 2 hours less than the usual time to cover a certain distance. what is distance usually covered by Mohan?
जब मोहन अपनी चाल को 30km/hr से बढ़ाकर 40km/hr कर लेता है तो वह एक निश्चित दूरी को तय करने में वास्तविक समय से 2 घंटे कम समय लेता है तो मोहन के द्वारा कितनी दूरी तय की गई?
(a)120 km      (b)160 km      (c)240 km     (d)300km

Solution:

## Time Speed And Distance Shortcut Tricks and Concepts for SSC, Bank, Railway Exams 1

(Q1)A person covers a distance of 840 km with a constant speed when he increases a speed by 10 km/hr. He takes 2 hours less time find the original speed.
एक व्यक्ति 840 किलोमीटर की दूरी को किसी निश्चित गति से तय करता है जब वह अपनी गति को 10 किलोमीटर प्रति घंटा बढ़ा देता है तो वह 2 घंटे कम समय लेता है तो उसकी मूल गति क्या है?
(a)50km/hr      (b)60km/hr      (c)90 km/hr      (d)120 km/hr

Solution:

Take value of X from option
x=60
Then,
=> 4200 = 60x(60+10)
=>4200 = 4200
Ans-60km/hr

(Q2)An aeroplane started its journey half hour late to cover a distance of 1500 km on time it increases its speed by 250 km/hr find its original speed.
कोई हवाई जहाज अपनी यात्रा आधे घंटे देरी से शुरू करता है 1500 किलोमीटर की दूरी को समय पर पूरा करने के लिए इसने अपनी गति को 250km/hr बढ़ा दिया तो उसकी मूल गति क्या है?
(a)650km/hr      (b)750 km/hr      (c)600km/hr      (d)700km/hr

Solution:

Take value of X from option
x=750
Then,
=> 750000 = 750x(750+250)
=>750000 = 750000
Ans-750km/hr

## Time Speed And Distance Shortcut Tricks and Concepts for SSC, Bank, Railway Exams 2

(Q1)A man walks A certain distance in certain time. if he had gone 3 km per hour faster, he would have taken 1 hour less than the scheduled time. if he had gone 2km per hour slower, he would have been 1 hours longer on the road. The distance is.
एक व्यक्ति कुछ समय में कुछ दूरी तक चलता है| यदि वह 3 किलोमीटर प्रति घंटा तेज चला होता तो उसे निर्धारित समय से 1 घंटा कम लगता| यदि वह 2 किलोमीटर प्रति घंटा धीमे चला होता तो उसे सड़क पर एक घंटा अधिक लगता तो दूरी कितनी है?
(a)60km      (b)45km      (c)65km      (d)80km

Solution:

Case-1

[Explation: suppose उस व्यक्ति का speed S km/hr हैं|जब उस व्यक्ति का speed S km/hr हैं तो Point-B तक जाने में suppose T hr लगता है | जब वह अपना speed S km/hr से बढाकर S+3 km/hr कर लेता है तो Point-B तक जाने में Time=T-1 लेगा | Time Diff.=T-(T-1)=1, S1-S2=S-(S+3)=3]

Case-2

Put the value of S in Eqn.-1
12x(12+3)=3D
∴ D=60km Ans.

Method-2

(Q2)A car travels from P to Q at a constant speed. If its speed were increased by 10 km/hr it would have been taken one hour lesser to cover the distance. It would have taken further 45 minutes lesser if the speed was further increased by 10 km/hr. The distance between the two places is?
एक कार स्थान ‘P’ से ‘Q’ तक एक नियत गति से चलती है यदि इसकी गति 10 किलोमीटर प्रति घंटा बढ़ा दी जाए तो उसे यह दूरी तय करने में 1 घंटा कम लगता है यदि अब उसकी गति पुनः 10 किलोमीटर घंटा और बढ़ा दी जाए तो उसे यह दूरी तय करने में अब 45 मिनट का समय और कम लगता है बताइए इन दोनों स्थानों के बीच की दूरी कितनी है?
(a)540km      (b)420 km      (c)600km      (d)620km

Solution:

## Time Speed And Distance Shortcut Tricks and Concepts for SSC, Bank, Railway Exams 3

(Q1)Walking at 3/4 of his usual speed, a man covers a certain distance in 2 hours more than the time he takes to cover the distance at his usual Speed. The time taken by him to cover the distance with his usual speed is.
कोई व्यक्ति अपनी सामान्य चाल की 3/4 चाल से चलने पर, अपनी सामान्य चाल से लगने वाले समय की तुलना में 2 घंटे अधिक समय लेता है उसे सामान्य चाल से चलने पर कुल कितना समय लगेगा?
(a)5 hours      (b)7hours      (c)6hours    (d)5.5 hours

Solution:

Method-2

1unit=2 hours
3unit=6 hours Ans.

[Concept-अगर Distance Constant हो तो जो speed का ratio होता है उसका reciprocal time का ratio होता है क्योंकि speed और time एक-दूसरे का inversely proportional होता है | इसी concept को use करते हुए इस सवाल को solve किया गया है |]

(Q2)By walking at 6/7 of his usual speed, a man reaches his office 25 minutes later than his usual time. The usual time taken by him to reach his office is.
एक आदमी अपनी मूल गति की 6/7 चाल से चलने पर, अपना ऑफिस सामान्य समय से 25 मिनट देरी से पहुंचा तो उसका ऑफिस पहुंचने का सामान्य समय क्या है?
(a)2hurs 30 minutes      (b)2hurs 10 minutes     (c)2hurs 15 minutes      (d)2hurs 25 minutes

Method-2

1unit=25 minutes
6unit=150 minutes = 2 h 30 min. Ans.

## Concept Of Relative Speed

⇒Suppose A and B are two objects and their speeds are respectively S1 and S2 moving in the same direction

⇒suppose A and B are two objects and their speeds are respectively S1 and S2 moving in opposite direction

## Time Speed and Distance relative speed concept for SSC, Bank, Railway Exams 3

(Q1)A train 300 meters long is running at a speed of 54 km/hr. In what time will it cross a telephone pole?
एक 300 मीटर लंबी रेलगाड़ी 54 किलोमीटर प्रति घंटा की गति से चलते हुए एक टेलीफोन के खंभे को पार करने में कितना समय लेगी?
(a)18 seconds      (b)17 seconds      (c)20 seconds      (d)15 seconds

Solution:

(Q2)A train traveling at a speed of 30m/sec crosses a platform 600 meters long in 30 seconds. The length of the train is.
30 मीटर/सेकंड की चाल से चलती हुई एक रेलगाड़ी 600 मीटर लंबे प्लेटफार्म को 30 सेकेंड में पार करती है तो रेलगाड़ी की लंबाई है?
(a)300 m      (b)120 m      (c)200m     (d)150 m

Solution:

(Q3)A train 50 meters long passes a platform of length 100 meters in 10 seconds. The speed of the train in metre/second is.
50 मीटर लंबी एक रेलगाड़ी 100 मीटर लंबे एक प्लेटफार्म को 10 सेकेंड में पार करती है | रेलगाड़ी की गति मीटर/सेकंड में है|
(a)10      (b)50      (c)20      (d)15

Solution:

(Q4)Two trains of lengths 150 m and 180 m respectively are running in opposite directions on parallel tracks. If their speed is 50 km/hour and 58 km/hour respectively. In what time will they cross each other?
दो रेलगाड़ियां जिनकी लंबाई क्रमश: 150 मीटर और 180 मीटर है समांतर ट्रैक पर विपरीत दिशा में चल रही है | यदि उनकी गति क्रमश: 50 किलोमीटर प्रति घंटा और 58 किलोमीटर प्रति घंटा हो, तो कितने समय में वह एक दूसरे को पार करेगी?
(a)15 sec.      (b)22 sec.      (c)11 sec.      (d)30 sec.

Solution:

(Q5)Two trains of equal length are running on parallel lines in the same direction at the rate of 46 km/hour and 36 km/hour. The faster train passes the slower train in 36 seconds the length of each train is.
समान लंबाई की दो रेलगाड़ियां समांतर लाइनों पर एक ही दिशा में क्रमशः 40 किलोमीटर प्रति घंटा और 36 किलोमीटर प्रति घंटा की दर से चल रही है | तेज चलने वाली ट्रेन धीमी ट्रेन से 36 सेकंड में आगे निकल जाती है प्रत्येक ट्रेन की लंबाई कितनी है?
(a)82 m     (b)80 m      (c)72 m      (d)50 m

Solution:

## Time Speed and Distance relative speed concept for SSC, Bank, Railway Exams 4

(Q1)A train passes a platform 110 m long in 40 seconds and a boy standing on the platform in 30 seconds. The length of the train is.
कोई रेलगाड़ी 110 मीटर लंबे किसी प्लेटफार्म को 40 सेकेंड में तथा उस प्लेटफार्म पर खड़े एक लड़के को 30 सेकंड में पार करती है| रेलगाड़ी की लंबाई है|
(a)220 m      (b)110 m      (c)330 m      (d)100 m

Solution:

जब Train platform को पार करती है Then,

(Q2)A train passes two bridges of lengths 500 m and 250 m in 100 seconds and 60 seconds respectively. The length of the train is.
एक रेलगाड़ी 500 मीटर और 250 मीटर लंबे दो पुलों को क्रमशः 100 सेकंड और 60 सेकंड में पार करती है तो रेलगाड़ी की लंबाई है|
(a)250 m      (b)125 m      (c)300 m      (d)175 m

Solution:

जब Train 1st bridge को पार करती है Then,

(Q3)A train traveling at uniform speed crosses two bridges of lengths 300 m and 240 m in 21 seconds and 80 seconds respectively. The speed of the train is.
एक रेलगाड़ी 300 मीटर और 240 मीटर लंबे दो पुलों को क्रमशः 21 सेकंड और 18 सेकंड में पार कर जाती है तो रेलगाड़ी की गति क्या है?
(a)60 km/hr      (b)72 km/hr      (c)68 km/hr      (d)65 km/hr

Solution:

जब Train 1st bridge  को पार करती है Then,

Second Method:

## Time Speed and Distance relative speed concept for SSC, Bank, Railway Exams 5

(Q1)A train passes two persons walking in the same direction at a speed of 3 km/hour and 5 km/hour respectively in 10 seconds and 11 seconds respectively. The speed of the train is.
एक रेलगाड़ी अपनी ही दिशा में क्रमशः 3 किलोमीटर प्रति घंटा तथा 5 किलोमीटर प्रति घंटा की गति से चलने वाले व्यक्तियों को पार करने में क्रमशः 10 सेकंड तथा 11 सेकेंड का समय लेती है| रेलगाड़ी की गति बताएं |
(a)25km/hr      (b)27km/hr      (c)24km/hr      (d)30km/hr

Solution:

Let the speed of train = x km/hr
Relative speed of 1st Person = (x-3)km/hr
Relative speed of 2nd Person = (x-5)km/hr
[Concept-जब कोई Train किसी Pole या व्यक्ति को पार करती है तो वह अपनी लंबाई के बराबर ही distance तय करती है |]
Note: इस Question में एक ही Train दोनों व्यक्तियों को पार करती है so Train का length equal(i.e distance) होगा|
Now,

Method-2

When train crosses 1st person in the same direction then,
Relative speed of 1st Person = (x-3)km/hr

[Note: इस Question में एक ही Train दोनों व्यक्तियों को पार करती है so Train का length equal होगा|]
Now,
(x-3)×10/60 = (x-5)×11/60
10x – 30 = 11x – 55
x = 25 km/hr Ans.

Method-3

(Q2)A train passes two persons walking in opposite direction at a speed of 5 m/second and 10 m/second respectively in 6 seconds and 5 seconds respectively. Find the length of the train is.
एक रेलगाड़ी अपनी विपरीत दिशा में क्रमशः 5 मीटर प्रति सेकंड तथा 10 मीटर प्रति सेकंड की गति से चलने वाले व्यक्तियों को पार करने में क्रमशः 6 सेकंड और 5 सेकंड का समय लेती है| रेलगाड़ी की लंबाई बताएं |
(a)140 metres      (b)150 metres      (c)180 metres      (d)160 metres

Solution:

Let the speed of train = x m/sec
Relative speed of 1st Person = (x+5)m/sec
Relative speed of 2nd Person = (x+10)m/sec
[Concept-जब कोई Train किसी Pole या व्यक्ति को पार करती है तो वह अपनी लंबाई के बराबर ही distance तय करती है |]
[Note: इस Question में एक ही Train दोनों व्यक्तियों को पार करती है so Train का length equal(i.e distance) होगा|]
Now,

Method-2

When train crosses 1st person in opposite direction then,
Relative speed of 1st Person = (x+5)m/sec

[Note: इस Question में एक ही Train दोनों व्यक्तियों को पार करती है so Train का length equal होगा or speed भी equal होगा | इस question में train का length निकलना है तो हमलोग train के speed को equal कर लेंगे |]
Now,

Method-3

Relative speed = (x+5)m/sec=(20+5)=25m/sec.
Length of train=25×6=150 m

(Q3)A train passes two persons walking in the same direction at a speed of 4 km/hour and 5 km/hour respectively in 10 seconds and 12 seconds respectively. Find the length of the train.
एक रेलगाड़ी अपनी ही दिशा में क्रमशः 4 किलोमीटर प्रति घंटा तथा 5 किलोमीटर प्रति घंटा की गति से चलने वाले व्यक्तियों को पार करने में क्रमशः 10 सेकंड और 12 सेकंड का समय लेती है| रेलगाड़ी की लंबाई बताएं|
(a)50/3 meters      (b)60 meters      (c)30 meters      (d)25 meters

Solution:

[Note:Let the speed of train = x km/hr
Relative speed of 1st Person = (x-4)km/hr
Relative speed of 2nd Person = (x-5)km/hr]

Relative speed = (x-4)Km/hr=(10-4)=6km/hr=6×(5/18).=5/3 m/sec.
The length of train=(5/3)×10=50/3 m Ans.

## Time Speed and Distance relative speed concept for SSC, Bank, Railway Exams 1

(Q1)The speed of Two trains in the ratio 3:4 and moving on parallel tracks but in opposite directions.If both crosses a pole in 3 seconds then in how much time they will cross each other.
दो रेलगाड़ियां जिनकी चाल 3:4 के अनुपात में है, समांतर परियों पर विपरीत दिशाओं में जा रही है |यदि प्रत्येक रेलगाड़ी एक खंभे को पार करने में 3 सकेंड ले तो वे कितने समय में एक दूसरे को पार करेगी?
(a)3 sec.      (b)5 sec.      (c)7 sec.      (d)8 sec.

Solution:

Distance traveled by 1st train=3×3=9m(i.e Length of train because जब कोई Train किसी Pole या व्यक्ति को पार करती है तो वह अपनी लंबाई के बराबर ही distance तय करती है |)
Distance traveled by 2nd train=3×4=12m
[Note:जब कोई Train किसी Platform, bridge और Train को पार करती है तो वह अपनी लंबाई के साथ-साथ उस Platform, bridge और Train की लंबाई के बराबर Distance तय करती है|]
Total distance=9 + 12 =21 m
Relative Speed=4 + 3=7 m/sec
Crossing Time=21/7 = 3 sec. Ans

(Q2)The speed of Two trains in the ratio 7:9 and moving on parallel tracks but in same directions.If both crosses a pole in 4 seconds and 6 seconds then in how much time they will cross each other.
दो रेलगाड़ियां जिनकी चाल 7:9 के अनुपात में है ,समांतर परियों पर एक ही दिशा में जा रही है यदि प्रत्येक रेलगाड़ी एक खंबे को पार करने में 4 sec.तथा 6 sec ले तो वे कितने समय में एक दूसरे को पार करेगी?
(a)37 sec.      (b)36 sec.      (c)41 sec.      (d)43 sec.

Solution:

Distance traveled by 1st train=7×4=28m(i.e Length of train because जब कोई Train किसी Pole या व्यक्ति को पार करती है तो वह अपनी लंबाई के बराबर ही distance तय करती है |)
Distance traveled by 2nd train=9×6=54m
[Note:जब कोई Train किसी Platform, bridge और Train को पार करती है तो वह अपनी लंबाई के साथ-साथ उस Platform, bridge और Train की लंबाई के बराबर Distance तय करती है|]
Total distance=28 + 54 =82 m
Relative Speed=9 – 7=2 m/sec
Crossing Time=82/2 = 41 sec. Ans

## Time Speed and Distance relative speed concept for SSC, Bank, Railway Exams 2

(Q1)Two trains of equal length take 10 seconds and 15 seconds respectively to cross a Telephone pole. If the length of each train is 120 m. In what time will they cross Each Other traveling in opposite direction?
समान लंबाई वाली दो रेलगाड़ियां एक टेलीफोन खंबे को क्रमशः 10 सेकंड तथा 15 सेकंड में पार करती है | यदि प्रत्येक रेलगाड़ी की लंबाई 120 मीटर हो, तो विपरीत दिशाओं में चलते हुए वे एक-दूसरे को कितने समय में पार करेगी ?
(a)10 sec.      (b)12 sec.      (c)16 sec.      (d)15 sec.

Solution:

Distance traveled by 1st train=120m(i.e Length of train because जब कोई Train किसी Pole या व्यक्ति को पार करती है तो वह अपनी लंबाई के बराबर ही distance तय करती है |)
Distance traveled by 2nd train=120m
Speed of 1st Train=120/10=12m/sec.
The speed of 2nd Train=120/15=8m/sec.
Relative speed in opposite direction=12+8=20m/sec
Total distance=120+120=240m
Crossing time=240/20 =12 sec. Ans.

(Q2)Two trains running in opposite direction Cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratio of their speeds is.
दो रेलगाड़ियां विपरीत दिशाओं में चलती हुई प्लेटफार्म पर खड़े हुए एक व्यक्ति को क्रमशः27 सेकंड तथा 17 सेकंड में पार करती है जबकि दोनों trains एक दूसरे को 23 सेकंड में पार करती है| तो दोनों रेलगाड़ियों की गति का अनुपात क्या है?
(a)4:3      (b)2:3      (c)3:2      (d)6:4

Solution:

Let the speed of 1st train=x m/sec.
The speed of 2nd train=y m/sec.
Relative speed=(x + y)m/sec.
Distance(i.e Length of 1st Train) traveled by 1st train=27x
Distance(i.e Length of 2nd Train) traveled by 2nd train=17y

By Allegation Method:

## Time Speed and Distance relative speed concept for SSC, Bank, Railway Exams 3

(Q1)The ratio of the length of two trains 4:3 and the ratio of their speeds are 6:5.The ratio of time taken by them to cross a pole is.
दो रेलगाड़ियों की लंबाईयों का अनुपात 4:3 तथा उनकी चालों का अनुपात 6:5 है |तो किसी खंभे को पार करने में उनके द्वारा लिया गया समय का अनुपात है?
(a)11:8      (b)5:6      (c)20:18      (d)27:16

Solution:

Let the length of 1st and 2nd trains are respectively=4x,3x
Speed of 1st and 2nd trains are respectively=6y,5y
Time is taken by 1st train to cross the pole=Total distance/speed
=4x/6y
(Note: takenTotal distance=Length of train because जब कोई Train किसी Pole या व्यक्ति को पार करती है तो वह अपनी लंबाई के बराबर ही distance तय करती है |)
Time is taken by 2nd train to cross the pole=Total distance/speed
=3x/5y
Ratio of their time=(4x/6y):(3x/5y)
=20:18 Ans.

Method-2

The ratio of time to cross a pole=20:18 Ans.

(Q2)The ratio of the length of two trains 5:3 and the ratio of their speed are 6:5. The ratio of time taken by them to cross a pole is.
दो रेलगाड़ियों की लंबाईयों का अनुपात 5:3 तथा उनकी चालों का अनुपात 6:5 है |तो किसी खंभे को पार करने में उनके द्वारा लिया गया समय का अनुपात है?
(a)5:6      (b)27:16      (c)25:18      (d)11:8

Solution:

Let the length of 1st and 2nd trains are respectively=5x,3x
Speed of 1st and 2nd trains are respectively=6y,5y
Time is taken by 1st train to cross the pole=Total distance/speed
=5x/6y
(Note: takenTotal distance=Length of train because जब कोई Train किसी Pole या व्यक्ति को पार करती है तो वह अपनी लंबाई के बराबर ही distance तय करती है |)
Time is taken by 2nd train to cross the pole=Total distance/speed
=3x/5y
Ratio of their time=(5x/6y):(3x/5y)
=25:18 Ans.

Method-2

The ratio of time to cross a pole=25:18 Ans.

Time And Distance: Meeting Point Concept for Competitive Exams Type-6

Suppose point A से एक कार X km/hr के speed से point B की ओर जा रहा है एवं point B से एक कार Y Km/hr की स्पीड से point A की और आ रहा है तो दोनों कार एक- दूसरे से कहीं ना कहीं आपस में मिलेंगे तो माना वह point C है जहां दोनों कार आपस में मिलेंगे |
Note:जहां पर दोनों car आपस में मिलेंगे मैं तो इस स्थिति में दोनों का Time same होगा i.e ऐसा तब होगा जब reference of time दिया होगा| Reference time से तात्पर्य है कि यदि point A से एक कार 9:00 बजे चली है तो point B से भी कार 9:00 बजे ही चली हो|

(Q)Two trains start at the same time from Aligarh and Delhi and proceed towards each other at the rate of 14 km and 21 km per hour respectively. When they meet, it is found that one train has traveled 70 kilometers more than the other. The distance between two stations is.
दो रेलगाड़ियां एक ही समय पर अलीगढ़ और दिल्ली से क्रमशः 14 किलोमीटर प्रति घंटा और 21 किलोमीटर प्रति घंटे की रफ्तार से एक दूसरे की ओर रवाना होती है जब वे एक दूसरे से मिलती है तो यह पता चलता है कि उनमें से एक गाड़ी ने दूसरी गाड़ी के अपेक्षा 70 किलोमीटर अधिक यात्रा की है तो दोनों स्टेशनों के बीच की दूरी क्या है?
(a)300 km      (b)210 km      (c)350 km      (d)140 km

Solution:

[Exp-माना Delhi से जो Train Aligarh की तरफ जा रही है वह meeting point तक पहुंचने में D Km Distance तय की है and जो Train Aligarh से Delhi की ओर जा रही है वह Meeting Point तक पहुंचने में (D+70) Km Distance तय की है|]

Now,

Total distance between two stations=D+(D+70)
=2D+70
=350 km Ans.

Second Method:

Let the trains meet each other after x hr
so, distance traveled by 1st train in x hr=14x
distance traveled by 2nd train in x hr=21x
Distance between Delhi and Aligarh=14x+21x=35x
now,
21x-14x=70
x=10
so, they meet after 10 hr.
Distance between Delhi and Aligarh=35x
=35×10
=350 Km Ans.

(2)A and B are 20 km apart A can walk at an average speed of 4 km/hr and B at 6 km/hr. If they start walking towards each other at 7 a.m. when they will meet?
A और B 20 किलोमीटर दूरी पर है| A 4 किलोमीटर प्रति घंटा और भी 6 किलोमीटर प्रति घंटा की गति से चल सकता है| यदि वह प्रातः 7:00 बजे एक दूसरे की ओर चलना शुरु करते हैं तो वह कब मिलेंगे?
(a)9.00 a.m      (b)8.30 a.m      (c)10.00 a.m      (d)8.00 a.m

Solution:

Second method:

Let they are meeting each other after x hr.
So, distance travelled by A in x hr=4x
distance travelled by B in x hr=6x
Now,
4x+6x=20
x=2 hr
Meeting time=7am+2hr
=9:00 am Ans.

(3)Two places P and Q are 162 km apart. A train leaves P for Q and simultaneously another train leaves Q for P. They meet at the end of 6 hours. If the former train travels 8 km/hr faster than the other, then speed of the train from Q is.
दो स्थान P और Q एक दूसरे से 162 किलोमीटर की दूरी पर है| एक गाड़ी P से Q के लिए प्रस्थान करती है और उसी समय एक अन्य गाड़ी Q से P के लिए प्रस्थान करती है| 6 घंटे के अंत में वे दोनों गाड़ियां मिलती है| यदि पहली वाली गाड़ी दूसरी गाड़ी से 8 किलोमीटर प्रति घंटे तेज चलती है, तो Q से चलने वाली गाड़ी की गति क्या है?

Solution:

According to the question,
The train meet each other after 6 hours
So, distance travelled by 1st train in 6 hours=(x+8)×6 Km
distance travelled by 2nd train in 6 hrs=x×6 Km
Now,
(x+8)×6+6x=162
6x+48+6x=162
x=19/2=9 ½ Ans.

(4)A train A start from Delhi at 4 p.m. and reaches Ghaziabad at 5 p.m. While another train B start from Ghaziabad at 4 p.m. and reaches Delhi at 5:30 P.M. The two trains will cross each other at?
एक ट्रेन A शाम 4:00 बजे दिल्ली से चलती है तथा गाजियाबाद शाम 5:00 बजे पहुंचती है| जबकि दूसरी ट्रेन B शाम 4:00 बजे गाजियाबाद से चलती है एवं दिल्ली 5:30 बजे पहुंचती है दोनों ट्रेन एक-दूसरे को कितने समय में पार करेगी?
(a)4:40 pm      (b) 4:45 pm      (c)4:36 pm      (d)4:30 pm

Solution:

Second Method:

(5)Two trains A and B start from Howrah to Patna and from Patna to Howrah respectively. After passing Each Other they take 4 hours 48 minutes and 3 hours 20 minutes to reach Patna and Howrah respectively if A is moving at 45 km per hour, the speed of B is.
दो रेलगाड़ियां A और B हावड़ा और पटना से क्रमशः पटना और हावड़ा के लिए एक ही समय पर चलना आरंभ करती है| एक-दूसरे के सामने से गुजरने के बाद वे 4 घंटे 48 मिनट और 3 घंटे 20 मिनट क्रमशः पटना और हावड़ा पहुंचने के लिए लेती है| यदि हावड़ा से चलने वाली गाड़ी 45 किलोमीटर प्रति घंटा की गति से चल रही है तो दूसरी गाड़ी की गति क्या है?
(a)64.8 km/hr      (b)45 km/hr      (c)60 km/hr      (d)54 km/hr

Solution:

## Time Speed And Distance: Meeting Point Concept 1

(1)P and Q are 27 km away. Two trains with speed of 24 km/hr and 18 km/hr respectively start simultaneously from P and Q and travel in the same direction. They meet at a point R beyond Q. Distance QR is.
P तथा Q, 27 किलोमीटर की दूरी पर है |दो रेलगाड़ियां क्रमशः 24 किलोमीटर प्रति घंटा तथा 18 किलोमीटर प्रति घंटा की गति से एक साथ P तथा Q से चलती है और एक ही दिशा में यात्रा करती है| तदनुसार, यदि वे Q से आगे एक बिंदु R पर मिलती हो तो QR की दूरी कितनी है?
(a)48 km      (2)81 km      (c)36 km      (d)126 km

Solution:

## Time Speed And Distance: Meeting Point Concept 2

Suppose Train A 10 am में चली हो तथा Train B 11 am में चली हो ऐसी स्थिति में meeting point निकालने के लिए दोनों Train का Reference of Time same करना होगा | Train A 1 hour तक अकेली चलेगी फिर वह Point A’ पर 11:00 बजे पहुंचेगी| इस तरह अब दोनों Train का Reference of time same हो गया|
Now,

(1)The distance between two cities A and B is 330 km. A train starts from A at 8 a.m. and travels towards B at 60 km/hr. Another train starts from B at 9 a.m. and travels towards A at 75 km/hr. At what time do they meet.
दो नगरों A और B के बीच की दूरी 330 किलोमीटर है |एक रेलगाड़ी, A से B की और प्रातः 8:00 बजे 60 किलोमीटर प्रति घंटा की गति से चलती है और एक दूसरी रेलगाड़ी B से A ओर प्रातः 9:00 बजे, 75 किलोमीटर प्रति घंटा की गति से चलती है| वे दोनों कितने बजे मिलेगी|
(a)11:30 am      (b)10 am      (c)10:30 am      (d)11 am

Solution:

(2)A train leaves a station A at 7 a.m. and reaches another station B at 11 a.m. another train leaves B at 8 a.m. and reaches A at 11:30 a.m. The two trains cross one another at.
एक रेलगाड़ी स्टेशन A से प्रातः 7:00 बजे रवाना होती है और दूसरे स्टेशन B पर प्रातः 11:00 बजे पहुंच जाती है| एक दूसरी रेलगाड़ी स्टेशन B से प्रातः 8:00 बजे रवाना होती है और प्रातः 11:30 बजे स्टेशन A पर पहुंच जाती है| दोनों रेलगाड़ी कितने बजे एक-दूसरे को पार करेगी?
(a)9:00 am      (b)9:24 am      (c)8.56 am      (d)8.36 am

Solution: