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.
LT =
Length of train
LO =
Length of object
ST =
Speed of train
SO =
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 LT = ST × t
When a train crossing a
stationary object with length then (LT + LO) = ST ×
t
When a train
crossing a moving a moving object without length
1.
In opposite direction: LT =(ST + SO)
× t
2.
In same direction: LT =(ST - SO) ×
t
When a train
crossing a moving object with length
1.
In opposite direction (ST + SO) × t = ( LT
+ LO)
2.
In same direction (ST + SO) × t = ( LT +
LO)
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
To
view quantitative study notes on other topics CLICK HERE
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