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Re: At 1:00 PM, Train X departed from Station A on the road to S [#permalink]
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I do think the explanation above is well perfect.

Simple and straight.

:?:
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Re: At 1:00 PM, Train X departed from Station A on the road to S [#permalink]
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Carcass wrote:
At 1:00 PM, Train X departed from Station A on the road to Station B. At 1:30 PM, Train Y departed Station B on the same road for Station A. If Station A and Station B are p miles apart, Train X’s speed is r miles per hour, and Train Y’s speed is s miles per hour, how many hours after 1:00 PM, in terms of p, r, and s, do the two trains pass each other?

A. \(0.5\) + \(\frac{(p - 0.5s)}{(r + s)}\)

B. \(\frac{(p - 0.5s)}{(r + s)}\)

C. \(0.5\) + \(\frac{(p - 0.5r)}{r}\)

D. \(\frac{(p - 0.5r)}{(r + s)}\)

E. \(0.5\) + \(\frac{(p - 0.5r)}{(r + s)}\)



This is very tricky!

First let's denote \(t\) as the time Train Y travelled to get to Station B.

Remember this, as we'll be using this at the end.

When the two trains pass eachother what we're actually asking is: when do they meet? They meet when the sum of the distances of Train X and Train Y have covered the entire distance \(p\).

We're also told that Train X departs at 1:00 and Train Y departs at 1:30. So Train X has been travelling for a half hour longer than Train Y. Since we let \(t\) be the time it takes Train Y to get to Station B, \(t + \frac{1}{2}\) must be the time Train X travels to get to Station A.

Given that the rate of Train X is \(r\) and the rate of Train Y is \(s\), putting all these pieces together:

\(d_A = r(t + \frac{1}{2})\)

\(d_B = st\)

And we know that \(d_A + d_B = p\), so we can add both equations to eachother:

\(r(t + \frac{1}{2}) + st = p\)

Now, let's isolate \(t\).

\(rt + 0.5r + st = p\)

\(rt + st = p - 0.5r\)

\(t(r+s) = p - 0.5r\)

\(t = \frac{(p - 0.5r)}{(r+s)}\)

And we've found \(t\)


But we're not done


Recall (first line above in blue) that we let \(t\) be the time it takes train Y to get to Station B, which left at 1:30. We're being asked how long it took them to meet after 1:00.

So if it took \(t = \frac{(p - 0.5r)}{(r+s)}\) for them to meet when we started counting at 1:30, then we need to factor in the extra half an hour of travel. So:

\(t = \frac{(p - 0.5r)}{(r+s)} + 0.5\)

Giving E as the answer.
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Re: At 1:00 PM, Train X departed from Station A on the road to S [#permalink]
1
easy solution is,
x-r m/hr
y-s m/hr
y starts 1/2 hr later
means x has travelled 1/2 more
ie r/2 miles
now total distance is p-r/2
when two things travel towards each other we add up the rates
r+s m/hr=(P-r/2)/t .ie is t=(P-0.5r)/r+s.
final answer from A it has travelled 0.5 hr more i.e why 0.5+(P-0.5r)/r+s.
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Re: At 1:00 PM, Train X departed from Station A on the road to S [#permalink]
1
By taking real numbers this problem is pretty approachable:
let p be the distance = 90
let x be the speed of the train A = 10
let y be the speed of the train B = 20

:: Note the time taken to pass each other = total_distance / total_speed
Now X has a lead time of 30 mins
so distance covered in 30 mins= 0.5 * 10 = 5 #####; (30/60) =0.5 (since everything is in hrs)
Now the given X traveled 5 miles remaining dist = 90-5 =85
and 85 miles is the distance to be covered by 2 trains travelling at 10+20 rate so
p- (5)/ r+s

now given all options we can do a simple trick
0.5(bob) = 5 {since we're trying to put back the formula}
notice bob = 5/0.5 = 10 which equals x
so eliminate option A, B
we know denominator should be r+s so eliminate C and then finally eliminate D as the function accounts for the time taken to travel 85 miles; Remember we need to add 30 mins or 0.5 at the beginning
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Re: At 1:00 PM, Train X departed from Station A on the road to S [#permalink]
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