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Re: A train with 120 wagons crosses Ram who is going in the same direction [#permalink]
Let the speed of train be s ; Speed of Ram = a ; Speed of Shyam = b and length of train=l mts

36=l/(s-a)---(1)

24=l/(s+b)---(2)

ON DIVIDING 1 BY 2

s= 3a + 2b--(3)
In 30 mins , train covers a distance of 1800s ; Ram covers a distance of 1800a.
Therefore distance b/w Ram and Shyam, when the train has just crossed Shyam is

1800(s-a)-24(a+b)/(a+b)
on substituting value of s from ---(3) , we get answer as 3576
So answer is B
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Re: A train with 120 wagons crosses Ram who is going in the same direction [#permalink]
speed of train ; t
speed of Ram ; r
and speed of Shyam ; s
let train length ; L
so from given info
L/t-r = 36
L/t+s=24
solve we get t=3r+2s
in 30 mins ; 1800 sec
distance of train ; 1800t and Ram 1800r
so distance b/w ram and shyam when train crosses shyam
1800(t-r)- 24(r+s)
time = 1800(t-r)-24(r+s)/ (r+s)
solve we get 3576 sec
IMO B



A train with 120 wagons crosses Ram who is going in the same direction, in 36 seconds. It travels for half an hour from the time it starts overtaking the Ram(he is riding on the horse) before it starts overtaking the Shyam( who is also riding on horse) coming from the opposite direction in 24 seconds. In how much time (in seconds) after the train has crossed the Shyam do the Ram meets to Shyam?

A) 3560
B) 3576
C) 3600
D) 3765
E) 3850
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Re: A train with 120 wagons crosses Ram who is going in the same direction [#permalink]
Ans:B
Let the speed of the train be v kmph & that of the horse be u kmph.
Let L= length of the train.
Hence,
[L/(v+u)]/[L/(v-u)]= 36/24
v= 5u
The distance between Ram and Shyam when train started crossing shyam= 30*60*(v-u)= 7200u
Time required to meet=7200u/2u= 3600s
Final time required= 3600-24= 3576s (time reqred for train to cross shyam =24)
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Re: A train with 120 wagons crosses Ram who is going in the same direction [#permalink]
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The answer is B: 3576 seconds.

EXPLANATION:

First, we have to assume that the horse moves with the same speed. Otherwise, this question can't be solved, as there would be an extra variable to consider with less equations.

Now, let the speed of horse be y, and the speed of train be x.

Now, using relative speed formula:
\( (X-Y)*36= (X+Y)*24 \)

This is because, train is moving towards shyam, who is moving towards train. Hence, their combined speed is X + Y. Similarly, train is moving away from Ram, who is moving in the same direction as the train. So combined speed is X - Y.

Since the distance covered by Shyam would be the same that would be covered by Ram, we calculate the distance covered by Ram in 36 seconds as \( speed*time \) which is \( (X-Y)*36 \), and the distance covered by Shyam in 24 seconds as \( (X+Y)*24 \). Equating both, we get the equation.

Solving them, we get:
\( X = 5Y \)

This means, speed of the train is 1/5th of speed of horse. This relationship gives us following idea:
For L distance covered by Train, Horse would cover L/5 (as Distance = Speed*Time, assuming time as constant, distance would vary by speed directly).
Also, if train covers something in 30 minutes, the horse would take 150 minutes (as Distance = Speed*Time, assuming distance constant, time would vary inversely)

So if the train takes 30 minutes of the time that it starts overtaking Ram before it starts overtaking the Shyam coming from the opposite direction, each horse would take 150 minutes.

Now, let's assume the length of the train as L. This L, is the same distance, that the train has to cover WHEN IT STARTS overtaking Ram, and BEFORE IT STARTS overtaking shyam.

Understand this, by the time train covers this L distance, 30 minutes would have passed (given in the question). This means, in these 30 minutes, the horse in which Ram is riding would also have moved towards Shyam. By how much? We go to our earlier relationship of X = 5Y.
Since we know the speed is 5 times, so Ram WOULD HAVE MOVED L/5 distance, BEFORE TRAIN STARTS OVERTAKING SHYAM.

This means, the remaining distance is 4L/5.

Now, to cover L distance, horse uses 150 minutes (as train uses 30 minutes).
So, to cover 4L/5 distance, horse would use 120 minutes.

This is given for both horses. So for each individual horse (or to get midpoint of their meeting), we would get 120/2 = 60 minutes.

Now, we are not done yet. By the time train was overtaking shyam, shyam moved for 24 seconds. This means we have to subtract 24 seconds from 60*60 = 3600 seconds.

Hence, the final answer would be 3600 - 24 = 3576.

To sum up, following line of thought is followed:
A) First, we get relationship between speed of train (x) and speed of horse (y), which is x = 5y. This is done by equating their relative distance covered (which is same).
B) Then, once we know their relationship between speed, we get the relationship between time and distance.
C) Now, we assume L is the distance train moves in 30 minutes, which is when it STARTS overtaking Ram, and BEFORE it starts overtaking Shyam.
D) For this L distance that is covered by the train in 30 minutes, L/5 distance is covered by the horse in 150 minutes. So we calculate the time taken by the horse for the leftover distance (4L/5), which comes to 120 minutes.
E) Since we need midpoint for each horse, we get 60 minutes the distance to be covered by EACH horse.
F) We also subtract the additional 24 seconds covered by Shyam when the train was overtaking Shyam.
G) So we get final value as 3600 - 24 = 3576!
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Re: A train with 120 wagons crosses Ram who is going in the same direction [#permalink]
Let the length of the train be L metres and speeds of the train, Ram and Shyam be T, R and S respectively, then

L/(T−R)=36 ...(i)

and L/(T+S)=24 ...(ii)



From eq.(i) and (ii)

3(T - R) = 2 (T + S)
T = 3R + 2S



In 30 minutes (i.e 1800 seconds), the train covers 1800T (distance) but Ram also covers 1800R (distance) in the same time. Therefore distance between Ram and Shyam, when the train has just crossed Shyam = 1800 (T - R) - 24(R + S)

Time required = [1800(T−R)−24(R+S)]/(R+S)= [1800(3R+2S−R)-24(R+S)]/(R+S) = (3600 - 24) = 3576 s
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Re: A train with 120 wagons crosses Ram who is going in the same direction [#permalink]
A train with 120 wagons crosses Ram who is going in the same direction, in 36 seconds. It travels for half an hour from the time it starts overtaking the Ram(he is riding on the horse) before it starts overtaking the Shyam( who is also riding on horse) coming from the opposite direction in 24 seconds. In how much time (in seconds) after the train has crossed the Shyam do the Ram meets to Shyam?

A) 3560
B) 3576
C) 3600
D) 3765
E) 3850

Length of the train, L = 120
Let's assume that the speed of the train is a mph & speed of the horse= b mph.
[L/(a+b)]/[L/(a-b)]= 36/24
[L/(a+b)]/[L/(a-b)] = 3/2
[L/(a+b)] * [(a-b)/L] = 3/2
a=5b
The distance between the Ram and shyam when the train started crossing the shyam = 30*60*(a-b)= 7200b
Time required to meet them= 7200b/2b= 3600 seconds where, t=0, when the train started crossing the shyam.
The train takes 24s to cross the shyam.
So, time required= 3600-24= 3576 seconds

Imo. B
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Re: A train with 120 wagons crosses Ram who is going in the same direction [#permalink]
Let the speed of the train, Ram and Shyam be T, R and S respectively.
Let the length of the train be L metres.

\(\frac{L}{(T−R)}\)=36 ...(i)

\(\frac{L}{(T+S)}\)=24 ...(ii)



From (i) and (ii)

\(3(T - R) = 2(T + S)\)
⇒ \(T = 3R + 2S\)



In 30 min. or 1800 seconds, the train covers 1800T distance.
Also, Ram covers 1800R distance in the same time.
Therefore distance between Ram and Shyam, when the train has just crossed Shyam = 1800(T-R) - 24(R+S)

Time required = \(\frac{1800(T−R)−24(R+S)}{(R+S)}\)= \(\frac{1800(3R+2S−R)-24(R+S)}{(R+S)}\) = \((3600 - 24) = 3576 s\)

B is the answer.
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Re: A train with 120 wagons crosses Ram who is going in the same direction [#permalink]
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Re: A train with 120 wagons crosses Ram who is going in the same direction [#permalink]
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