SOLUTION: A train traveling at 40 miles per hour leaves for a certain town. One hour later, a bus traveling at 50 miles per hours leaves for the same town and arrives at the same time as th

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Question 33543: A train traveling at 40 miles per hour leaves for a certain town. One hour later, a bus traveling at 50 miles per hours leaves for the same town and arrives at the same time as the train. If both the train and the bus traveled in a straight line, how far is the town from where they started?
Please help me I think the answer is 200 but I don't know how to set the problem up
Found 2 solutions by Cintchr, Earlsdon:
Answer by Cintchr(481) About Me  (Show Source):
You can put this solution on YOUR website!
+rt=d+
In this case, the distance for both is the same, so for our convenience the train's variables are in caps.
+RT+=+rt+
r=50
t= T-1 (it started an hour later)
R=40
T= this is what we will solve for
+%2840%29T+=+%2850%29%28T-1%29+
Solve for T
+40T+=+50T-50+
Subtract 50T
+-10T=-50+
divide by -10
+T=5+
So if the Train takes 5 hours, the bus takes 4.

CHECK
+RT+=+rt+
+40%285%29+=+50%284%29+
+200+=+200+
and this IS the distance.
This is a duplicate of problem 33542

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
You can use the distance formula:
d+=+rt where: d = distance travelled, r = rate of travel (speed), and t = time of travel.
For the train:
d1+=+r1%28t1%29
1) d1+=+40%28t1%29
For the bus:
d2+=+r2%28t2%29
2) d2+=+50%28t2%29
The distance is the same in each case, so d1 = d2, therefore, we can set:
40%28t1%29+=+50%28t2%29
But the train travels one hour longer than the bus, so t1 = t2+1
Making this substitution, we get:
40%28t2%2B1%29+=+50%28t2%29 Simplify and solve for t2.
40%28t2%29%2B40+=+50%28t2%29 Subtract 40(t20 from both sides of the equation.
40+=+10%28t2%29 Divide both sides by 10.
4+=+t2 Now substitute this into equation 2) and solve for d2:
d2+=+50%284%29
d2+=+200miles.
The town is 200 miles from the starting point.