## Friday, December 14, 2012

### MTS / POSTMAN / GDS TO PA / PA RECRUITMENT MATERIAL

This material prepared and compiled by Akula. Praveen Kumar, SPM, Papannapet Sub Office-502 303, Medak Division, Andhra Pradesh (9849636361, 8019549939)

Disclaimer:- All questions/Information provided in this post are Compiled by A. Praveen Kumar for in good faith of Departmental Employees. The types of questions, number of questions and standard of questions may be vary in actual examination. This is my predictions only. Author of blog does not accepts any responsibility in relation to the accuracy, completeness, usefulness or otherwise, of the contents.
TRAIN PROBLEMS
Train problems broadly center on the following three types.
1.      Time taken by a train to cross a stationary point
2.      Time taken by a train to cross a stationary length
3.      Time taken by a train to cross a moving length
* km/hr – m/s conversion:
X km/hr = [X * (5/18)] m/s.
* m/s – km/hr conversion:
X m/s = [X * (18/5)] km/hr.
§  Time taken by a train to cross a stationary point = train length/train speed
§  Time taken by a train to cross a stationary length = (train length + stationary length)/train speed
§  Time taken by a train to cross another train moving in the same direction = sum of length of the two trains/difference in their speeds
§  Time taken by a train to cross another train moving towards it = sum of length of the two trains/sum of their speeds
§   Time taken by a train of length 1 meter to pass a pole or a standing man or a signal post is equal to the time taken by the train to cover 1 meter.

Time taken by a train of length 1 meters to pass a stationary object of length b meters is the time taken by the train to cover (1 + b) meters.

Suppose two trains or two bodies are moving in the same direction at u m / s and v m/s, where u > v, then their relatives speed = (u - v) m / s.

Suppose two trains or two bodies are moving in opposite directions at u m / s and v m/s, then their relative speed is = (u + v) m/s.

If two trains of length a meters and b meters are moving in opposite directions at u m / s and v m/s, then time taken by the trains to cross each other = (a + b)/(u+v)sec.

If two trains of length a meters and b meters are moving in the same direction at u m / s and v m / s, then the time taken by the faster train to cross the slower train = (a+b)/(u-v) sec.

If two trains (or bodies) start at the same time from points A and B towards each other and after crossing they take a and b sec in reaching B and A respectively, then
(A's speet) : (B’s speed) = (b1/2: a1/2).

Example:
1.If a train going at 90 kmph takes 28 seconds to go past a lamp post, but 80seconds to cross a platform, what is the length of (a) the train and (b) the platform?
90 kmph= (90 × 5/18) = 25 m/s.
Solution for (a):
Given the time to cross the lamp post as 28 seconds,
Formula:
(train length/25 )= 28 or train length = 700 m.
Solution for (b):
Given the time to cross the platform as 80 seconds,
Formula:
{(train length + platform length)/25 } = 80
or (train length + platform length) = 2000 m.
Since train length is 700 m, platform length = 1300 m or 1.3 km.
2. How long does it take for a train of length 800 m moving at 80 kmph to cross a train of length 1200 m coming at a speed of 100 kmph from the opposite direction?
The relative speed = 80 + 100 (Moving opposite direction)
= 180 kmph = 50 m/s.
The distance to be covered = 800 + 1200 ( Sum of the train length)
= 2000 m.
Formula:
Time taken for crossing = 2000/50 (Time = Sum of the train length / Relative Speed)         = 40 seconds.
Above two example problems are basic type. Here i will give three more examples with little bit different.
Example 3:
Two trains of length 100m and 200m are 100m apart. They start moving towards each other on parallel tracks, at speed 54 kmph and 72 kmph. After how much time will the trains meet?
In this problem, we need to find the time taken by trains to meet each other. So no need to consider the train length.
Distance between the trains = 100m
Relative speed = 54+72 kmph (Trains are in opposite direction)
= 126 kmph = 35mps
Time taken = Distance/ Relative speed
= 100/35 = 20/7 seconds
Example 4:
Two trains of length 100m and 200m are 100m apart. They start moving towards each other on parallel tracks, at speed 54 kmph and 72 kmph. After how much time will the trains cross each other?
In this problem, we need to find the time taken by trains to cross each other. So we need to consider the train length and also distance between the trains.
Distance to be covered = 100+200 +100 (Train length 1 + Train length 2 +distance between them)
Relative Speed = 35 mps
Time taken = Distance/ Relative speed
= 400/35 = 80/7 seconds
Example 5:
Two trains of length 100m and 200m are 100m apart. They start moving towards each other on parallel tracks, at speed 54 kmph and 72 kmph. After how much time will the trains be 100m apart again ?
In this problem, Trains have to cross each other and then be at 100m apart.
For distance , we need to consider initial distance , train length and their final distance between them.
Relative Speed = 35 mps
Distance to be covered = 100+100+200+100 = 500m
Time taken = 500/35 = 100/7 seconds.
The Questions will involve one train or two trains; therefore read the question whether the question involves one train or two trains, because you can decide what formula to use and solve quickly. The entire aptitude question on trains will contain any of these 5 questions. Note that sometimes trains are replaced by some other thing like person, man, car … etc
Single Train Questions (2 Types of questions)
Question 1:

What is the time taken by a train of length ‘L’ metres and travelling at a speed of ‘S’ to pass a pole or a standing man or a signal post or some stationary object?
Explanation
·         The Question will contain length of the train ‘L’
·         Train’s speed ‘S’ will be given in terms of either km/hr or m/sec.
If speed is given in terms of Km/hr, always remember to convert  it to m/sec which can be done               by this formula: =S x (5/18) m/sec
·         You have to find the time taken by the train to pass the stationary object. Use this formula
Time taken to pass the object= L/S sec

Question 2

What is the time taken by a train of length ‘L’ metres and travelling at speed of ‘S’ to pass a pole or a standing man or a signal post or some stationary object of Length ‘M’?

Explanation
·         The Question will contain length of the train ‘L’
·         Length of the object ‘M’
·         Train’s speed ‘S’ will be given in terms of either km/hr or m/sec.
If the speed is given in terms of Km/hr, always remember to convert it to m/sec which is done                  by this formula: =S x (5/18) m/sec
·         You have to find the time taken by the train to pass the object of length ‘M’. Follow this procedure
1. You know this: Time taken to pass the object= L/S sec
2. Make changes to the above formula in this way…
Time taken to pass the object of lengthM’ = (L+M)/S sec

Two Train Questions (3 Types)
Question 3:

What is the taken by the two trains that are of length ‘a’ and ‘b’ respectively and travelling opposite to each other at a speed of ‘u’ and ‘v’ respectively to cross each other?
Explanation
·         The Question will contain two trains and its lengths as; ‘a’ of the first train and ‘b’ of the second train.
·         The speed of the trains will also be included  as ‘u’ as speed of the first train and ‘v’ as the speed of the second train
If the speed is given in terms of Km/hr, always remember to convert it to m/sec which is done by             this formula
For train 1:      =u x (5/18) m/sec an
For train 2:     = v x (5/18) m/sec
Or
= (u+v) x (5/18) m/sec
·         Now you have find out the time taken by the trains to cross each other; use this formula
= (a+b)/ (u+v)
u+v is called as relative speed of trains travelling in opposite direction

Question 4

What is the taken by the two trains that are of length ‘a’ and ‘b’ respectively and travelling in the same direction or parallel to each other at a speed of ‘u’ and ‘v’ respectively to cross each other?
Explanation
·         The Question will contain two trains and its lengths as: ‘a’ of the first train and ‘b’ of the second train.
·         The speed of the trains will also be included  as ‘u’ as speed of the first train and ‘v’ as the speed of the second train
If the speed is given in terms of Km/hr, always remember to convert it to m/sec which is done by             this formula
For train 1: =u x (5/18) m/sec and
For train 2: = v x (5/18) m/sec
Or
= (u+v) x (5/18) m/sec
·         Now you have find out the time taken by the trains to cross each other;
·         use this formula
= (a+b)/ (u-v)
u-v is called as relative speed of trains travelling in opposite direction

Question 5

What is the time taken by the two trains that start at their points: A for the first train and B for the second train that travels at a speed of ‘u’ and ‘v’ respectively to reach their destination after crossing each other?
Explanation
·         The Question will contain two trains and its’ Speed as: ‘u’ of the first train and ‘v’ of the second train.
·         You have to find out the time taken by the trains to reach their destination after crossing each other.
·         Use this formula             (√v: √u)

EXAMPLES
1.      A train 280 m long, running with a speed of 63 km/hr will pass a tree in:
Sol:Speed =

63 x
 5 18

m/sec =
 35 2
m/sec.

Time taken =

280 x
 2 35

sec = 16 sec.
2.A train moves with a speed of 108 kmph. Its speed in meters per second is:
108 kmph =

108 x
 5 18

m/sec = 30 m/sec.
3. A train 132 m long passes a telegraph pole in 6 seconds. Find the speed of the train.
Speed =

 132 6

m/sec =

22 x
 18 5

km/hr = 79.2 km/hr.

3. How long does a train 110 meters long running at the speed of 72 km/hr take to cross a bridge 132 meters in length?
Sol:
Speed =

72 x
 5 18

m/sec = 20 m/sec.

Total distance covered = (110 + 132) m = 242 m.

Required time =

 242 20

sec = 12.1 sec.

4. The length of the bridge, which a train 130 meters long and travelling at 45 km/hr can cross in 30 seconds, is:
Sol:
Speed =

45 x
 5 18

m/sec =

 25 2

m /sec; Time = 30 sec.

Let the length of bridge be x meters.

Then,
 130 + x 30
=
 25 2
2(130 + x) = 750
x = 245 m.

5. A train speeds past a pole in 15 seconds and a platform 100 m long in 25 seconds. Its length is:
Sol:
Let the length of the train be x meters and its speed be y m/sec.

They,
 x y
= 15
y =
 x 15

 x + 100 25
=
 x 15
x = 150 m.

6.A train crosses a platform 100 m long in 60 seconds at a speed of 45 km/hr. The time taken by the train to cross an electric pole is :

Sol:
Speed =

45 x
 5 18

m/sec =

 25 2

m/sec.

Let the length of the train be x metres.

 Then,
x + 100

 25 2

 = 60   or   x = 650 m

Time taken by the train to cross an electric pole =

650 x
 2 25

sec = 52 sec.

7.A train 240 m long passed a pole in 24 seconds. How long will it take to pass a platform 650 m long?
Sol:
Speed =

 240 24

m/sec = 10 m/sec.

Required time =

 240 + 650 10

sec = 89 sec.

8.A train takes 18 seconds to pass completely through a station 162 m long and 15 seconds through another station 120 m long. The length of the train is:
Sol:
Let the length of the train be x meters.

 x + 162 18
=
 x + 120 15
15 (x + 162) = 18 (x + 120)
x = 90 m.

9.A goods train runes at the speed of 72 kmph and crosses a 250 m long platform in 26 seconds. What is the length of the goods train?

Sol:
Speed =

72 x
 5 18

m/sec = 20 m/sec; Time = 26 sec.

Let the length of the train be x meters.

Then,
 x + 250 26
= 20
x + 250 = 520
x = 270.

10.Two trains are running in opposite directions with the same speed. If the length of each train is 120 meters and they cross each in 12 seconds, then the speed of each train ( in km/hr) is
Sol:
Let the speed of each train be x m/sec.

Then, relative speed of the two trains = 2x m/sec.

So, 2x =
 (120 + 120) 12
2x = 20
x = 10.

Speed of each train = 10 m/sec =

10 x
 18 5

km/hr = 36 km/hr.

11.Two trains travel in opposite directions at 36 kmph and 45 kmph and a man sitting in slower train passes the faster train in 8 seconds. The length of the faster train is:
Sol:
Relative speed = (36 + 45) km/hr =

81 x
 5 18

m/sec =

 45 2

m/sec.

Length of train =

 45 2
x 8

m = 180 m.

12.How many seconds will a 500 meter long train take to cross a man walking with a speed of 3 km/hr in the direction of the moving train if the speed of the train is 63 lm/hr?
Sol: Speed of train relative to man = (63 – 3) km/hr = 60 km/hr

=

60 x
 5 18

m/sec =
 50 3
m/sec.

Time taken to pass the man =

500 x
 3 50

sec = 30 sec.

13.A train of length 150 meters takes 40.5 seconds to cross a tunnel of length 300 meters. What is the speed of the train in km/hr?
Sol:
Speed =

 150 + 300 40.5

m/sec =

 450 40.5
x
 18 5

km/hr = 40 km/hr.
14.A train covers a distance of 12 km in 10 minutes. If it takes 6 seconds to pass a telegraph post, then the length of the train is:
Sol:
Speed =

 12 10
x 60

km/hr =

72 x
 5 18

m/sec = 20 m/sec.

Length of the train = (Speed x Time) = (20 x 6) m = 120 m.

15.A train 360m long is running at a speed of 45 km/hr. In what time will it pass a bridge 140 m long?
Sol:
Speed =

45 x
 5 18

m/sec =
 25 2
m/sec.

Total distance covered = (360 + 140) m = 500 m.

Required time =

500 x
 2 25

sec = 40 sec.

16.Two trains 140 m and 160 m long run at the speed of 60 km/hr and 40 km/hr respectively in opposite directions on parallel tracks. The time (in seconds) which they take to cross each other, is:
Sol:

Relative speed = (60 + 40 ) km/hr =

100 x
 5 18

m/sec =

 250 9

m/sec.

Distance covered in crossing each other = (140 + 160) m = 300 m

Required time =

300 x
 9 250

sec =
 54 5
sec = 10.8 sec.
17.A train 800 meters long is running at a speed of 78 km / hr. If it crosses a tunnel in 1 minute, then the length of the tunnel (in meters) is:
Sol:
Speed =

78 x
 5 18

m/sec =

 65 3

m /sec;

Time = 1 minute = 60 sec.

Let the length of the tunnel be x meters.

Then,
 800 + x 60
=
 65 3
3(800 + x) = 3900
x = 500.
18.A 270 meters long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds. What is the length of the other train?
Sol:
Relative speed = (120 + 80) km/hr =

200 x
 5 18

m/sec =

 500 9

m/sec.

Let the length of the other train be x metres.

Then,

 x + 270 9
=
 500 9
x + 270 = 500
x = 230.

19. Two trains, one from Warangal to Hyderabad and the other from Hyderabad to Warangal, start simultaneously. After they meet, the trains reach their destinations after 9 hours and 16 hours respectively. The ratio of their speeds is:
Sol:
Let us name the trains as A and B. Then,

 (A's speed) : (B's speed) = b : a = 16 : 9 = 4 : 3

20.In what time will a train 100 meters long cross an electric pole, if tis speed be 144 km/hr?
Sol:
Speed =

144 x
 5 18

m/sec = 40 m/sec.

Time taken =

 100 40

sec = 2.5 sec.
21.A train moves past a telegraph post and a bridge 264 m long in 8 seconds and 20 seconds respectively. What is the speed of the train?
Sol:
Let the length of the train be x meters and its speed by y m/sec.

They,
 x y
= 8
x = 8y

Now,
 x + 264 20
y
8y + 264 = 20y
y = 22.

Speed = 22 m/sec =

22 x
 18 5

km/hr = 79.2 km/hr.
22. Two trains, each 100 m long, moving in opposite directions, cross each other in 8 seconds. If one is moving twice as fast the other, then the speed of the faster train is:
Sol:
Let the speed of the slower train be x m/sec.

Then, speed of the faster train = 2x m/sec.

Relative speed = (x + 2x) m/sec = 3m/sec.

 (100 + 100) 8
= 3x
24x = 200
x =
 25 3

So, speed of the faster train =
 50 3
m/sec =

 50 3
x
 18 5

km/hr = 60 km/hr.

23. A train passes a station platform in 36 seconds and a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, what is the length of the platform?
Sol:
Speed =

54 x
 5 18

m/sec = 15 m/sec.

Length of the train = (15 x 20) m = 300 m.

Let the length of the platform be x meters.

Then,
 x + 300 36
= 15
x + 300 = 540
x = 240 m.

24.A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is the length of the train?
Sol:
Speed =

60 x
 5 18

m/sec =

 50 3

m/sec.

Length of the train = (Speed x Time) =

 50 3
x 9

m = 150 m.

Top of Form
Bottom of Form
25.A train 110 m long passes a man, running at 6 kmph in the direction opposite to that of the train, in 6 seconds. The speed of the train is:
Sol:
Speed of the train relative to man =

 110 6

m/sec =

 110 6
x
 18 5

km/hr = 66 km/hr.

Let the speed of the train be x kmph. Then, relative speed = (x + 6) = kmph.

x + 6 = 66  or  x = 60 kmph.

26. A 300 meter long train crosses a platform in 39 seconds while it crosses a signal pole in 18 seconds. What is the length of the platform?
Sol:
Speed =

 300 18

m/sec =
 50 3
m/sec.

Let the length of the platform be x meters

Then,
 x + 300 39
=
 50 3
3 (x + 300) = 1950
x = 350 m.
27. A train 125 m long passes a man, running at 5 kmph in the same direction in which the train is going, in 10 seconds. The speed of the train is:
Speed of the train relative to man =

 125 10

m/sec =

 25 2

m/sec.

 25 2
x
 18 5

km/hr = 45 km/hr.

Let the speed of the train be x kmph. Then, relative speed = (x – 5) kmph.

x - 5 = 45  or   x = 50 kmph.

28. Two stations A and B are 110 km apart on a straight line. One train starts from A at 7 a.m. and travels towards B at 20 kmph. Another train starts from B at 8 a.m. and travels towards A at a speed of 25 kmph. At what time will they meet?
Sol : Suppose they meet x hours after 7 a.m.

Distance covered by A in x hours = 20x km.

Distance covered by B in (x -1) hours = 25 (x – 1) km.

20x + 25 (x - 1) = 110

45x = 135

x = 3.

So, they meet at 10 a.m.

29. Two goods train each 500 m long, are running in opposite directions on parallel tracks. Their speeds are 45 km/hr and 30 km/hr respectively. Find the time taken by the slower train to pass the driver of the faster one.
Relative speed = (45 + 30) km/hr =

75 x
 5 18

m/sec =

 125 6

m/sec.

Distance covered = (500 + 500) m = 1000 m.

Required time =

1000 x
 6 125

sec = 48 sec.
30.A train 108 m long moving at a speed of 50 km/hr crosses a train 112 m long coming from opposite direction in 6 seconds. The speed of the second train is
Sol:
Let the speed of the second train be x km/hr.

Relative speed = (x + 50) km/hr =

(x + 50) x
 5 18

m/sec =

 250 + 5x 18

m/sec.

Distance covered = (108 + 112) = 220 m.

220

 250 + 5x 18

=  6
250 + 5x = 660
x = 82 km/hr.

31. The length of a train and that of a platform are equal. If with a speed of 90 km/hr, the train crosses the platform in one minute, then the length of the train (in metres) is:
Sol:
Speed =

90 x
 5 18

m/sec = 25 m/sec; Time = 1 min. = 60 sec.

Let the length of the train and that of the platform be x metres.

Then,
 2x 60
= 25
x =
 25 x 60 2
= 750.
32.Two trains 140 m and 160 m long run at the speed of 60 km/hr and 40 km/hr respectively in opposite directions on parallel tracks. The time (in seconds) which they take to cross each other, is:
Sol:
Relative speed = (60 + 40 ) km/hr =

100 x
 5 18

m/sec =

 250 9

m/sec.

Distance covered in crossing each other = (140 + 160) m = 300 m

Time taken

300 x
 9 250

sec =
 54 5
sec = 10.8 sec.
33.A train 150 m long passes a km stone in 15 seconds and another train of the same length travelling in opposite direction in 8 seconds. The speed of the second train is:
Sol:
Speed of first train =

 150 15

m/sec = 10 m/sec.

Let the speed of second train be x m/sec.

Relative speed = (10 + x) m/sec.

 300 10 + x
= 8
300 = 80 + 8x
x =
 220 8
=
 55 2
m/sec.

So, speed of second train =

 55 2
x
 18 5

kmph = 99 kmph.

34. Two trains are running at 40 km/hr and 20 km/hr respectively in the same direction. First train completely passes a man sitting in the slower train in 5 seconds. What is the length of the fast train?
Sol:
Relative speed = (40 – 20) km/hr =

20 x
 5 18

m/sec =

 50 9

m/sec.

Length of faster train =

 50 9
x 5

m =
 250 9
m = 27
 7 9
m.
35.Two trains 200 m and 150 m long are running on parallel rails at the rate of 40 kmph and 45 kmph respectively. In how much time will they cross each other, if they are sunning the same direction?
Sol:

Relative speed = (45 – 40 ) kmph = 5 kmph =

5 x
 5 18

m/sec =

 25 18

m/sec.

Total distance covered = Sum of lengths of trains = 350 m.

Time taken

350 x
 18 25

sec = 252 sec.