Aptitude - Problems on Trains - Discussion

Very useful. Thanks all.

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.

Given data:
t1=27 sec t2=17 sec
S=d/t,d=s * t.

d is consider as l,

l1=s1 * t1 =17 s1,l2 = s2* t2 =27 s2.

In question ,it clearly move in a opposite direction,so we have to use this formula->(l1+l2) /s1+s2.

Total time = (17s1+27s2)/s1 + s2,
23(s1+s2)= 17s1+27s2.

Rearrange it,
23s1-17s1=27s2-23s2,
6s1 = 4s2,
s1/s2 = 6/4=3/2.

So,the answer is 3:2

I do not understand this. Please explain to me.

Very well done. Thanks.

Isn't it like X/Y = 4/6?

How come in the above solution like X/Y 6/4=3:2.
X=4 and Y=6 isn't it?

Can the ratio be 2:3? Anyone explain me, please.

Short method is;

x person 27.
y person 17 add both distances and divide the total distance and multiple the same for both persons.

27+17/23 *2.
37/23 = 1.6 *2.
1.6*2.
Ans : 3.2.

@Poornima.

Thank you for the explanation.

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