Train & Distance : 7
Important Shortcuts Explained
Concept: Problems on trains
and ‘Time and Distance’ are almost same. The only difference is we have to
consider the length of the train while solving problems on trains.
A train is said to have
crossed an object (stationary or moving) only when the last coach (end) of the
train crosses the said object completely. It implies that the total length of
the train has crossed the total length of the object.
Hence, the
distance covered by the train = length of train + length of object
Points To Remember
1. Time taken by a train of
length of L meters to pass a stationary pole is equal to the time taken by
train to cover L meters.
2. Time taken by a train of
length of L meters to pass a stationary object of length P meters is equal to
the time taken by train to cover (L + P) meters.
3. Relative speeds :
i. If two trains are moving in
same direction and their speeds are x km/h and y km/h (x > y) then their
relative speed is (x –y) km/h.
ii. If two trains are moving
in opposite direction and their speeds are x km/h and y km/h then their
relative speed is (x + y) km/h.
Unit Conversion:
i. To convert 'X' Km/hr into
m/s
- Multiply
X with 5/18
ii. To convert 'x' m/s into
Km/hr
- Multiply
x with 18/5
Some Shortcut Methods
Trick-1:
A train has
leangth 'L' and its speed is 'S'.Time taken to cross a constant object/man is
= (Leangth/Speed)
Ex: A train has leangth of 180
meters and is going with 54 km/hr. time taken to cross a pole/man ?
a. 12 s b. 10/3 s
c. 18 s d. 20 s
Sol: = 180/(54*5/18) =
20 sec
Trick-2:
A train having
length 'L' and its speed is 'S'. To cross a platform having length 'X' is given
by
= (L + X)/S
Ex: A train's length is 240 m
and its speed is 36 Km/hr.To cross a platform having length is 160 m in how
much time ?
a. 40 s b. 30 s
c. 36 s d. 54 s
Sol: [250+160]/(36*5/18) = 40
sec
Trick-3:
If two trains of p
meters and q meters are moving in same direction at the speed of x m/s and y
m/s (x > y) respectively then time taken by the faster train to overtake
slower train is given by
= (p + q)/( x - y)
Ex: If two trains of 360 and
140 meters are moving in same direction at speed of 54 m/s and 44 m/s.how much
time taken by faster train to overcome slower train ?
a. 50 s b. 54 s
c. 44 s d. 10 s
Sol: = (360+140)/(54-44) = 50
sec
Trick-4:
If two trains of p
meters and q meters are moving in opposite direction at the speed of x m/s and
y m/s respectively then time taken by trains to cross each other is given by
= (p + q)/( x + y)
Ex: If two trains of 360 and
140 meters are moving in same direction at speed of 54 m/s and 46 m/s.how much
time taken by faster train to overcome slower train ?
a. 50 s b. 5 s
c. 63.3 s d. 10 s
Sol: = (360+140)/(54+46) = 5 sec
:
Trick-5:
Two trains of
length 'p' m and 'q' m respectively. When running in the same direction the
faster train passes the slower one in 'a'seconds, but when they are running in
opposite directions with the same speeds as earlier, they pass each other in
'b' seconds.
Then, Speed of the
faster train
= [( p + q)/ 2] x[
( a+b) / (a xb)] Speed of the slower train
= [(p-q) / 2] x[
(a-b) / (a xb)]
Note : The speeds obtained
using the above formula are in mts/ sec, if the speeds are to be expressed in
kmph, they have to be multiplied by 18/5.
Ex: Two trains of length 100 m
and 250 m run on parallel lines. When they run in the same direction it will
take 70 seconds to cross each other and when they run in opposite direction,
they take 10 seconds to cross each other. Find the speeds of the two trains?
Sol: Speed of the faster train
= [(100 + 250) / 2] [ (70 + 10) / ( 70 x10) ]. = 175 x(8 /70) = 20 m/sec.
Speed of the slower train = [
( 100 + 250) / 2] [ ( 70-10) / (70 x10) ] = 175 x( 6/ 70) = 15 m/ sec.
Trick-5:
If a train passes
by a stationary man in 'p' seconds and passes by a platform /bridge, the length
of which is 'm' mts, completely in 'q' sec. Then Length of the train
= (m * p) / (q-p).
Ex: A train crosses by a
stationary man standing on the platform in 7 seconds and passes by the platform
completely in 28 seconds. If the length of the platform is 330 meters, what is
the length of the train?
a. 100m b. 110m c.
200m d. 210m
Sol: Length of the train = (
330 x7) / ( 28-7) = 330x 7 / 21 = 110 mts.
Trick-6:
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.The time taken by two trains is given by
= square root of v
: square root of u
Ex: 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 ‘4’ and ‘9’ respectively to reach their
destination after crossing each other?
Sol: = (9)^0.5 : (4)^0.5 = 3 :
2
Trick-7:
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.
The distance between A and B is 'X' .The time taken by two trains to meet is
given by
From A side : (X *
u)/(u + v)
From B side : (X *
v)/(u + v)
Ex: What is the time taken by
the two trains to meet from first train side, that start at their points: A for
the first train and B for the second train that travels at a speed of ‘4’ km/hr
and ‘9’ km/hr respectively.The distance between A and B is 13 km?
a. 4 km b. 5 km
c. 6 km d. 7 km
Sol: = (13*4)/(4+9) = 4 km
For more Tricks you can buy R.S Agarwal's Arithmetic book . link is given below
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