Aptitude - Problems on Trains - Discussion

Distance = length of the train.
D = S * T.
D = (4-2)*5/18*90,
Therefore D = 50.

Can solve this way: Same direction a person walk so the subtraction of speed i.e 4-2 = 2kmph.
And (2*5/18)m/s =5/9.
Now the time is 9 and 10-sec addition of two is 90.
So, 90 * 5/9 = 50.

Why we multiply 9 and 10 here? can anyone explain?

S = D/T
S-2 = D/9
9(S-2) = D ------> (1)
S-4 = D/10
10(S-4) = D ------> (2)

Distances are equal as the same cross both, equating 1,2.
10S - 40 = 9S - 18.
1S = 22(Speed of train).

Taking any one person's reference;
22-2 = D/9.
20 * 9 = D,
D = 180 * 5/18,
D = 50mtrs.

Hello everyone. Please have a look at my method of solution. It is giving a different answer.

Let,

the Speed of the train = x.
Then,

With respect to the first person;
The relative speed of the train = (x-2)*5/18 m/s.
Time is taken to cross the first person = 9s,
Length of the train = Speed * Time = (x-2)*(5/18)*9.

With respect to the second person;
The relative speed of the train = (x-4)*5/18 m/s.
Time is taken to cross the first person = 10s.
Length of the train = Speed * Time = (x-4)*(5/18)*10.

As both lengths are of the same body (train), so;
(x-2)*(5/18)*9 = (x-4)*(5/18)*10.
(x-2)*(5/2) = (x-4)*(25/9).

Upon solving, we get:
x = 22m.

Let me explain if anyone wants to know how to solve the value of x,

So we got both the equations ( 1,2 ) by solving (x/ y-5/9) = 9 and (x/ y-10/9) = 10,
9y - 5 = x ->(1) , 10(9y - 10) = 9x ->(2)

By substituting equation no (1) in equation no (2), we will get,
10(9y - 10) = 9(9y - 5),
90y - 100 = 81y - 45.
9y = 55.
| y = 55/9.

Now substitute the value of (y) in equation no (1) to get the value of (x)
9(55/9) - 5 = x.
55-5 = x.
x = 50.
Thank you.

5/9 * 10 * 9 = 50.

Speed = 2 - 4 = 2kmps.
S = 2 * 5/18 = 5/9m/s.
L = s * t
L= 5/9 * 9 * 10
L= 50.

How? please anyone explain me.

Use this formulae:

L = ((T1*T2)/T2-T1) * ( S2 - S1),
= ((9*10)/10-9) * (4-2),
= 90 * 2(5/18),
= 50.4 or 50.