**Quantitative Aptitude Study Notes**

**BANK AND SSC EXAM**

**TIME DISTANCE AND TRAINS**

You know that quantitative
aptitude section is most important in

**bank exams in PO and****Clerk**and for other competitive exams because if you want good score in bank exam then you have to score good in maths. In competitive exams the most important thing is time management, if you know how to manage your time then you can do well in**Bank Exams.**That’s where maths shortcut tricks and formula are comes into action. So continuously we are providing shortcut tricks on different maths topics.
Today’s topic is

**TIME DISTANCE AND****TRAINS,**it is similar to time and distance topic which we have covered in our recent topic. This is the one of the most important topic in quantitative aptitude section in bank and ssc exam. You should know how to calculate trains questions and answers in very short time for bank exam. From this chapter around 1-2 questions are given in the**SBI and IBPS exams.**For this here we are providing shortcut tricks and**quicker method**to calculate TIME DISTANCE AND**TRAINS**in very short time.
If we want to solve

**time distance and Trains**questions or any other type of questions, then the first thing we need that is Formulas about that topic. So here is the list of formulas that is used in time distance and trains quantitative topic.**L**

_{T }= Length of train**L**

_{O }= Length of object**S**

_{T }= Speed of train**S**

_{O }= Speed of object**t = time**

**NOTE: Length of train as well as length of an object is always added in all cases:**

When a train crossing a
stationary object without length then L

_{T }= S_{T }× t
When a train crossing a
stationary object with length then (L

_{T }+ L_{O}) = S_{T }× t**When a train crossing a moving a moving object without length**

1.
In opposite direction: L

_{T }=(S_{T }+ S_{O}) × t
2.
In same direction: L

_{T }=(S_{T }- S_{O}) × t**When a train crossing a moving object with length**

1.
In opposite direction (S

_{T }+ S_{O}) × t = ( L_{T }+ L_{O})
2.
In same direction (S

_{T }+ S_{O}) × t = ( L_{T }+ L_{O})**TIME DISTANCE AND TRAINS**

This topic is the same as the
previous chapter (Time and Distance). The only difference is that the length of
the moving object (train) is also considered in trains topic.

Some important things to be
noticed in this chapter are:

1)
When two trains are moving in opposite directions their speeds
should be added to find the relative speed.

2)
When they are moving in the same direction the relative speed is
the difference of their speeds.

3)
When a train passes a platform it should travel the length equal
to the sum of the lengths of train and platform both.

Here is the various case on speed
time and distance problems on train.

**CASE 1: Trains passing a telegraph post or a stationary man**

**Example 1:**How many seconds will a train 100 meters long running at the rate of 36km an hour take to pass a certain telegraph post?

**Solution:**In passing the post the train must travel its own length.

Now, 36 km/hr = 36 × 5/18

= 10 m/sec

Required time = 100/10 = 10
seconds

**CASE 2: Trains crossing a bridge or passing a railway station**

**Example 2:**How long does a train 110 meters long running at the rate of 36 km/hr take to cross a bridge 132 meters in length?

**Solution:**In crossing the bridge, the train must travel its own length plus the length of the bridge.

Now, 36 km/hr = 36 × 5/18

= 10 m/sec

Required time = 242/10 = 24.2 sec

**CASE 3: Trains running in opposite direction**

**Example 3:**Two trains, 121 meters and 99 meters in length respectively, are running in opposite directions, one at the rate of 40 and the other at the rate of 32 km an hour. In what time will they be completely clear of each other from the moment they meet?

**Solution:**As the two trains are moving in opposite directions their relative speed = 40 + 32 = 72 km/hr or 20 m/sec.

The required time = total length
/ relative speed

= 121 + 99 / 20

= 11 secs

**CASE 4: Trains running in the same direction**

**Example 4:**Two trains, 121 meters and 99 meters in length respectively, are running in same directions, one at the rate of 40 and the other at the rate of 32 km an hour. In what time will they be completely clear of each other from the moment they meet?

**Solution:**Relative speed

= 40 – 32 = 8 km/hr

=20/9 m/sec

Total length = 121 + 99 = 220 m

Required time = total length /
relative speed

= 220/20 × 9

= 99 sec

**CASE 5: Train passing a man who is walking**

**Example 5:**A train 110 meters in length, travels at 60 km / hr . In what time will it pass a man who is walking at 6 km an hour

i.
Against it

ii.
In the same direction?

**Solution:**This question is to be solved like the above example 3 and 4, the only difference being that the length of the man is zero.

1.
Relative speed = 60 + 6 = 66 km/hr

= 55 / 3 m/sec

Required time = 110/55 × 3 = 6
seconds

2.
Relative speed = 60 – 6 =
54 km/hr

= 15 m/sec

Required
time = 110 / 15 = 22/3 sec

**Example 6:**Two trains are moving in the same direction at 50 km/hr and 30 km/hr The faster train crosses a man in the slower train in 18 seconds. Find the length of the faster train.

**Solution:**Relative speed = (50 – 30) km/hr

= (20 × 5/18) m/sec

= 50/9 m/sec

Distance covered in 18 sec at
this speed = 18 × 50/9

= 100 m

Length of the faster train = 100
m

**Example 7:**A train running at 25 km/hr takes 18 seconds to pass a platform. Next, it takes 12 seconds to pass a man walking at 5 km/hr in the opposite direction. Find the length of the train and that of the platform.

**Solution:**speed of the train relative to man = 24 + 5 = 30 km/hr

= 30 × 5/18

= 25 / 3 m/sec

Distance traveled in 12 seconds
at this speed = 25/3 × 12 = 100 m

Length of the train = 100 m

Speed of train = 25 km//hr

= 25 × 5/18

= 125/18 m/sec

Distance traveled in 18 secs at
this speed = 125/18 × 18 = 125 m

Length of the train + length of
the platform = 125 m

Length of the platform = 125 –
100 = 25 m

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