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Problems on Train

Quantitative Aptitude ( CCS) Problems on trains

Problems on Train

Problems on Trains Methods shortcut tricks - Math Shortcut Tricks

Shortcut Tricks are very important things in competitive exam. Time is the main factor in competitive exams. If you know how to manage time then you will surely do great in your exam. Most of us skip that part. We provide examples on Problems on Trains shortcut tricks here in this page below. And we try to provide all types of shortcut tricks on problems on trains here. We request all visitors to read all examples carefully. You can understand shortcut tricks on Problems on Trains Methods by these examples.

Few Important things to Remember

You all know that math portion is very much important in competitive exams. It doesn’t mean that other topics are not so important. You can get a good score only if you get a good score in math section. A good score comes with practice and practice. All you need to do is to do math problems correctly within time, and you can do this only by using shortcut tricks. But it doesn’t mean that you can’t do math problems without using any shortcut tricks. You may have that potential that you may do maths within time without using any shortcut tricks. But so many people can’t do this. Here we prepared problems on trains shortcut tricks for those people. Here in this page we try to put all types of shortcut tricks on Problems on Trains. But if you see any tricks are missing from the list then please inform us. Your help will help others.

Now we will discuss some basic ideas of Problems on Trains Methods. On the basis of these ideas we will learn trick and tips of shortcut problems on trains methods. If you think that how to solve problems on trains methods questions using problems on trains methods shortcut tricks, then further studies will help you to do so.

Remember some important formulas of train problems for instant solutions of problems on trains.

1. x km/hr = m/s.

2. x m/s = km/hr

3. Time taken by a train to pass a pole or a standing man or a single post is equal to the time taken by the train to cover its lenght .

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

5. 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.

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

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

8. If two trains of length a metres and b metres are moving in the same direction at u m /s and v m/ s, then the time take by the faster train to cross the slower train =

9. If two trains (or bodies) starts 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 (As speed): (B’s speed) =

Problem on trains with solutions :

Let’s have some solved examples to understand the topic clearly.

Question1. A train 100 m long is running at the speed of 30 km/hr. find the time taken by in to pass a man standing near the railway line.

Solution:

Speed of the train = 30km/h =30x5/18 m/sec = 25/3 m/sec.

Distance moved in passing the standing man = 100 m.

Required time taken = 100/(25/3)=100 x 3/25 sec = 12 sec.

Question.2. A train is moving at a speed of 132 km/hr. If the length of the train is 110 metres, how long will it take to cross a railway platform 165 metre long?

Solution:

Speed of the train = 132 km/h = 132 x 5/18 m/sec. = 110/3 m/sec.

Distance covered in passing the platform = (110+165) m = 275 m.

Time taken = (275 x 3/110) sec=15/2 sec= $7\frac{1}{2} sec$

Question.3 A man is standing on a railway bridge which is 180 m long. He finds that a train crosses the bridge in 20 seconds but himself in 8 seconds. Find the length of the train and its speed.

Solution:

Let the length of the train be x metres.

Then, the train covers x metres in 8 seconds and (x + 180) metres in 20 seconds.

$\therefore \; \frac{x}{8}=\frac{x+180}{20}$ = ↔ 20 x = 8(x + 180) ↔ x = 120.

Length of the train = 120 m.

Speed of the train = 120/8 m/sec = 15 x 18/5 kmph = 54 kmph.

OR
when the train crosses a bridge the it will cover a total distance of { lenght of train + lenght of bridge }
and , when train crosses person standing then it will cover a distance of equal to lenght of train
i.e in 20 seconds { lenght of train + lenght of bridge }
and in 8 seconds { lenght of train }

20 seconds = 8 seconds ( lenghjt of train) + 12 seconds ( lenght of bridge )
now , from above we can conclude that
train will cover 180 m in 12 seconds

d = speed x time

180 = speed x 12
speed = 180 / 12 = 15 m/s

or 15 x (18/5)

54 kmph

Question.4 A train 150 m long is running with a speed of 68 kmph. In what time will it pass a man who is running at 8 kmph in the same direction in which the train is going ?

Solution:

Speed of the train relative to man = (68 – 8) kmph = 60 x 5/18 m/sec= 50/3 m/sec

Time taken by the train to cross the man = Time taken by it to cover 150 m at (50/3) m/sec = 150x3/50sec = 9 sec.

Question.5. A train 220 m long is running with a speed of 59 kmph. In what time will it pass a man who is running at 7 kmph in the direction opposite to that in which the train is going ?

Speed of the train relative to man = (59+7) kmph = 66 x 5/18 m/sec = 55/3 m/sec.

Time taken by the train to cross the man = Time taken by it to cover 220 m at 55/3 m/sec. = 220x3/55 sec = 12 sec.

Question.6. Two trains 100 meters and 120 meters long are running in the same direction with speed of 72 km/hr and 54 km/hr. In how much time will the first train cross the second?

Solution

Relative speed of the trains =(72-54) km/h = 18 km/h
=(18 x 5/2) m/sec =5 m/sec

Time taken by the trains to cross each other

= Time taken to cover (100+120) m at 5 m/sec

=(220/5)sec=44 sec

Question.7. A train 100 metres long take 6 second to cross a man walking at 5 kmph in a direction opposite to that of the train. Find the speed of the train.

Solution

Let the speed of the train be x kmph.

Speed of the train relative to man = (x+5) kmph

(x +5) x 5/18 m/sec

$\frac{100}{(x+5)\times \frac{5}{18}}=6$

=> 30(x+5)=1800

=> X=55

So Speed of the train is 55 kmph.

Question.8. Two train 137 metres and 163 metres in length are running towards each other on parallel lines, one at the rate of 42 kmph and another at 48 kmph. In what time will they be clear of each from the moment they meet ?

Solution:
Relative speed of trains=(42+48)=90 km/h= 90x5/18 m/sec = 25m/sec
To cross each other total distance will be equal to the total length of both the train=(137+163)=300 m
Time taken by the trains to pass each other
=Time taken to cross each other=distance/speed=300/25sec= 12 sec

Question.9. A train running at 54 kmph takes 20 seconds to pass a platform. Next it takes 12 seconds to pass a man walking at 6 kmph in the same direction in which the train is going. Find the length of the train and the length of the platform ?

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

Speed of the train relative to man = (54 – 6) kmph = 48 kmph.=48 x 5/18 =40/3 m/sec
In passing a man, the train covers its own length with relative speed.
Length of the train is =(Relative speed x time to pass the man)
=40/3 x 12=160m

Also, speed of the train=54 km/h= 54 x 5/18= 15 m/sec
So to pass the plateform to distance will be =length of train + length of train=speed x time to cross the plateform
x+y=20 x 15=300
160+y=300
y=140
Question.10. A man sitting in a train which is travelling at 50 kmph observes that a goods trains, travelling in opposite direction, takes 9 seconds to pass him. If the goods trains is 280 m long, find its speed.

Solution: Relative Speed= distance/time=280/9m/sec=280/9 x18/5=112 km/h
Speed of goods train= (112-50)km/h=62 km/h

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