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 + (p−0.5s)(r+s)
B. (p−0.5s)(r+s)
C. 0.5 + (p−0.5r)r
D. (p−0.5r)(r+s)
E. 0.5 + (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+12 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:
dA=r(t+12)dB=stAnd we know that
dA+dB=p, so we can add both equations to eachother:
r(t+12)+st=pNow, let's isolate
t.
rt+0.5r+st=prt+st=p−0.5rt(r+s)=p−0.5rt=(p−0.5r)(r+s)And we've found
tBut 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=(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=(p−0.5r)(r+s)+0.5Giving E as the answer.