Aptitude - Problems on Trains - Discussion

Discussion Forum : Problems on Trains - General Questions (Q.No. 27)
27.
A train overtakes two persons who are walking in the same direction in which the train is going, at the rate of 2 kmph and 4 kmph and passes them completely in 9 and 10 seconds respectively. The length of the train is:
45 m
50 m
54 m
72 m
Answer: Option
Explanation:

2 kmph = ( 2 x 5 ( m/sec = 5 m/sec.
18 9

4 kmph = ( 4 x 5 ( m/sec = 10 m/sec.
18 9

Let the length of the train be x metres and its speed by y m/sec.

Then, ( x ( = 9 and ( x ( = 10.
y - 5
9
y - 10
9

Therefore 9y - 5 = x and 10(9y - 10) = 9x

=> 9y - x = 5 and 90y - 9x = 100.

On solving, we get: x = 50.

Therefore Length of the train is 50 m.

Discussion:
106 comments Page 10 of 11.

Jayasree said:   4 years ago
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.
(8)

Priyanshu said:   4 years ago
I didn't understand the solution, anyone explain it please.
(2)

Anand K said:   4 years ago
Treat both people separately.

1st compare person A(2 kmph) and train.

Speed = dist/time.
hence time = dist/speed.
Time(for A) = 9s.
Dist = length of train
Speed = relative speed btw A and train=(S-5/9)
S = speed of the train.
9 = LOT/(S-5/9) //LOT=Length of train.

Apply same for person B.
10=LOT/(S-10/9).
solve both equations and get;
LOT= 50mtrs.

Tej said:   4 years ago
5/9 * 10 * 9 = 50.
(6)

Arsalan said:   4 years ago
Why we have not converted the speed of the train in m/s initially? Anyone explain it, please.
(1)

Talha said:   3 years ago
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.
(9)

Jagan Reddy said:   3 years ago
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.
(16)

Anushree Kulkarni said:   3 years ago
Distance = length of the train.
D = S * T.
D = (4-2)*5/18*90,
Therefore D = 50.
(41)

GURU CHARAN MUNDA said:   3 years ago
Good, Thanks @Anushree.
(1)

Sidd said:   3 years ago
How? please anyone explain me.
(3)


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