SOLUTION: [ A train leaves San Diego at 3:00 pm. A second train leaves the same city traveling in the same direction at 6:00 pm. The second train travels 45 mph faster than the first. If the

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Question 817157: [ A train leaves San Diego at 3:00 pm. A second train leaves the same city traveling in the same direction at 6:00 pm. The second train travels 45 mph faster than the first. If the second train overtakes the first at 10:00 pm, what is the speed of each of the two trains. ]
Found 2 solutions by TimothyLamb, mathgranny:
Answer by TimothyLamb(4379) (Show Source):
You can put this solution on YOUR website!
a = s mph
b = (s + 45) mph
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s = d / t
d = s * t
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when the trains meet at 10 PM they are both at the same distance D from the station, but b has traveled 3 hours less than a
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train a:
D = a(10 - 3)
D = 7a
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train b:
D = b(10 - 6)
D = 4b
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4b = 7a
4(s + 45) = 7s
4s + 180 = 7s
3s = 180
s = 60
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Answer:
train-a speed: 60 mph
train-b speed: 105 mph
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Answer by mathgranny(13) (Show Source):
You can put this solution on YOUR website!

Train 1. d = rt. Train 2 d = rt
r1 means rate of train 1, r2 means rate of train 2
Rate of train 2 is 45 mph faster than train 1
r2 = r1 + 45
Time for train 1 to reach same spot at 10:00 is t1 = 7 hrs.
Time for train 2 to reach same spot at 10:00 is t2 = 4 hrs.
distance coverd by 10:00 is the same
d1 =. d2
d1 = rt = (r1)(7). d2 = ( r2)t = (r1 + 45) 4
So 7(r1) = (r1 + 45) 4
7(r1) = 4(r1)+ 180
Solve for r1 to get r1 = 60 mph
Train 2 is 45 mph faster so rate of train 2 is 105 mph