Aptitude - Problems on Trains - Discussion

Let me explain in detail if anyone still not understood yet.

Please remember the formula for finding speed. Speed = Distance/Time.
Therefore, Distance = Speed x Time.

If a train crosses a man/pole/post/tree in T seconds, and speed of the train is S m/s, then

Then, the Length of the Train = (Speed x Time) = ST metres.

Note: Here, the distance traveled by train in T seconds at the Speed of S m/s is equal to the length of the train.

Now, lets come to the given problem.

Let speed of the first train = X.
Time taken taken by the first train to cross a man = 27 seconds.
Therefore, Length of the first train = Speed x Time = X x 27 = 27X metres.

Let speed of the second train = Y.
Time taken taken by the second train to cross a man = 17 seconds.
Therefore, Length of the second train = 17Y metres.

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

Given that, (a + b) / (u + v) = 23 seconds.

Here, a = 27X, b = 17Y and u = X, v = Y.

Therefore, (27X + 17Y)/(X + Y) = 23.

=> 27X + 17Y = 23X + 23Y

=> 4X = 6Y

=> X/Y = 6/4 = 3/2

=> X : Y = 3 : 2.

Guys, there is an easy way to understand this question.

Let us consider the SPEED of Train'A' and Train'B' as X and Y.

Trains moving in opposite direction which crosses a man standing on the platform signifies the TIME of trains taken to cross the man which are 27s and 17s respectively, not the time taken by trains to cross each other(given separately in the question).

Therefore,

DISTANCE of the trains taken to cross the man = SPEED * TIME.
D (TRAIN A)= X * 27.
D (TRAIN B)= Y * 17.

So, now we know that DISTANCE(LENGTH)covered by the trains and its SPEED(mentioned above).

From this, it is understood that its time to find out the TIME take by the trains.
But the time has already given;

TIME = DISTANCE/SPEED.

23 = 27x + 17y /x + y.
Cross multiply 23 (x + y) = 27x + 17y.

Bring all like terms on same sides; So,

23x + 23y = 27x + 17y.
23x - 27x = 17y - 23y.
- 4x = - 6y.

Since both the sides have a negative sign, it will automatically get deleted)ie, 4x = 6y .

Now its time to form the answer in ratio form.
Ratio forms when one digit is divided by the other digit, always remember NUMERATOR should be greater than DENOMINATOR.

4x = 6y .
(Bring all like terms on same sides).

Then, x/y = 6/4.
= 3/2 ie, 3 : 2.

Answer = 3 : 2.

Given:
1. time taken by the first train to cross a man (t1) = 27 sec
2. time taken by the second train to cross a man (t2) = 17 sec

3. Time taken by the 1st & 2nd train to cross each other (T)= 23 sec

Find: Ratio
1st train's speeds (x) / 2nd train's speed (y) = x/y = ?

Formula:

speed =
distance(or length)
---------------------
time

Let,
Speed of 1st train to cross a man (x) = distance(d1) / time(t1)
Speed of 2nd train to cross a man (y) = distance(d2) / time(t2)

Thus,
x = d1/27 .................I
y = d2/17 .................II

So, Total speed of the train (S) = speed of 1st train (x) + speed of 2nd train (y)
Therefore, S = x + y

Now, we can say that
Total distance (D) = d1 + d2
D = 27x + 17y............from I & II

Total time (T) = 23 seconds

Total speed of the train (S) = Total distance (D) / Total time (T)

x + y =

27x + 17y
-----------
23

23x + 23y = 27x + 17y
23y - 17y = 27x - 23x
6y = 4x
x/y = 6/4 = 3/2
Thus,
1st train's speeds (x) / 2nd train's speed (y) = x/y = 3/2

When u refer the important formula we can come to know that "Time taken by a train of length x metres to pass a pole or a signal post is equal to the time taken by the train to cover x metres"

This theorem indirectly show us that the length of the train is 27 metres and 17 meters respectively ok

Then according to the 8th formula that "If the train of length a metres and b metres are moving in the opposite direction at u m/s and v m/s then the time taken by the trains to cross each other = a+b/u+v sec
we don't know the value of u and v so let take it as x and y
then:
a=27x
b=17y
therefore 27x+17y/x+y = 23

take x+y to the right side though it is in division it will turn to multiplication
like this:
27x+17y = 23(x+y)
27x+17y = 23x+23y
then calculate difference between x and y
you will get 4x = 6y
2x = 3y
and x/y = 3/2

Speed=(distance/time).

It rewritten as (distance/speed)=time.
Same case:
(relative distance/relative speed)=time
We understand the only thing how to calculate relative speed and relative distance,
Consider car A travel at x km/hr and car B travels at y km/hr
Both A and B travel in same direction means
Their relative speed=subtraction of two speed (x+y)km/hr.
Relative distance= (speed1*time1)-(s2+t2).
=27x-17y.

Here 27 and 17 are time1 and time2.
X and y are speed1 and speed2.

Case2:
CarA and carB travels opposite direction.
Relative speed= x+y(addition of speeds)
Relative distance=(speed1*time1)+(s2*t2)
So we get. = 27x+17y.
Finally (relative distance/relative speed)=time,
Therefore (27x+17y)/(x+y)=23,

Maybe my solutions help for you

Step (1)
Speed= (Distance/Time) meters/sec. Distance= Length here.
From this formula, Length = Speed* Time

So,Length of 1st Train= Speed * Time = 27* x where x=speed
Likewise,length of 2nd Train .= 17* y where y=speed

Step (2):
Note:Imagine one train is stationary. Then the distance traveled by each train would be equal to the length of the other train + its own length.Because a train can cross the other train after the last bogie of the 1st train crosses the last bogie of the other train.
So total length of both trains = 27x+17y.
Now the relative speed of the trains = x+y (since both run in opposite directions).

So Time = Length/speed= 27x+17y/x+y =23(given)
Cross multiply,transpose and simplify. Then you get the answer.

Hi friends in smart way....

To find out the speed ratio so ,

formula is Length = speed*time
so,

speed = length /time

here 2 trains so two different length and speed so the formula

s1+s2=L1+L2/time

speed we don't know so we take X,Y for speed for 2 trains
respectively ....
use above formula:

X+Y=L1+L2/23

HERE
L1=speed*time (speed is not given so take it as X,time is given that is 27)

L2= speed*time (speed is not given so take it as Y,time is given that is 17)
SO L1=27X
L2=17Y
apply it,
X+Y=27X+17Y/23

SO SOLVE IT
23(X+Y)=27X+17Y
23X+23Y=27X+17Y

separate the X and Y terms so we get

23X-27X+23Y-17Y=0

WE GET:
-4X+6Y=0
6Y=4X

RATIO (means difference of X and Y)
SO FRIEND,

X/Y=6/4=3/2
SO THE RATIO IS 3:2

Let the speeds of the two trains be x m/sec and y m/sec respectively.

Then, length of the first train = 27x metres,(length=speed*time..so length=x*27=27x metres)

And length of the second train = 17y metres.(length=speed*time..so length=y*17=17y metres)

If two trains of length a metres and b metres are moving in opposite directions at u m/s and v m/s, then:

The time taken by the trains to cross each other = (a + b)/u + v)sec.
By using above formula we get d solution for this.
Here it is given that they cross each other in 23 seconds.
So

23 = (27x+17y) / (x+y).

23x + 23y = 27x + 17y
23y - 17y = 27x - 23x
6y = 4x
x/y = 6/4 = 3/2
Thus,
1st train's speeds (x) / 2nd train's speed (y) = x/y = 3/2.

Let speed of train1: x m/s.
Speed of train2: y m/s.

Then relative speed of train1 = (x+y) m/s............(1).

Train are crosses a man means they cross their length.

Length of train1: 27x m.
Length of train2: 17y m.

According to Qst:

Train crosses Each other in 23sec.
Means train1 crosses his length & also the length of train2.

=> Total length crosses by train1 = Length of train1+Length of train2.
=>Total length covered by train1(D)=27x+17y...........(2).

We knows,

Speed(s) = Distance(D)/Time(t)............Eqn(3).

Put Values OF eqn(1) &(2) in eqn(3).

x+y = (27x+27y)/23.
=> 23x+23y = 27x+17y.
=> 27x-23x = 23y-17y.
=> 4x = 6y.

=> x/y = 6/4 = 3/2..........Answer.

Let Speed of Train A= x m/s
And Speed of Train B= y m/s

Time taken by Train A to cover distance equals to its length = 27 s

Time taken by Train B to cover distance equals to its length = 17 s

Total distance covered
---------------------- = Time taken to Cross each other
Total speed


(Length of train A)+ (Length of train B)
so, --------------------------------------- = 23 seconds
(Speed of train A) + (Speed of train B)

(Speed x time) of Train A + (Speed x time) of Train B
----------------------------------------------------- =23
(Speed of train A) + (Speed of train B)

27x + 17y
--------- = 23
x + y

27x + 17y = 23x + 23y
4x =6y

x/y= 3/2