SOLUTION: Hello! I was hoping you could help me with a question. "One train leaves city A heading for city B which is 390 miles away. At the same time a second train leaves city B heading fo

Question 261619: Hello! I was hoping you could help me with a question. "One train leaves city A heading for city B which is 390 miles away. At the same time a second train leaves city B heading for city A, Going 15 mph faster then the first train. If they meet in 3 hours and 20 minutes how fast were the trains traveling?
I keep trying to put it in the formula 3.2(x)+ 3.2(x-15)=390 from there I get (simplified) 6.4x= 438
Please tell me where i'm going wrong
Thank you so much.
Vincent
Found 3 solutions by Alan3354, richwmiller, stanbon:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
"One train leaves city A heading for city B which is 390 miles away. At the same time a second train leaves city B heading for city A, Going 15 mph faster then the first train. If they meet in 3 hours and 20 minutes how fast were the trains traveling?
------------------
3:20 is not 3.2 hours, it's 3 1/3 hours.
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d = rt
The speeds are added since they're going opposite directions.
Train A is going x mph
Train B is going x+15 mph.
The sum is 2x+15 mph
The go 390 miles in 3 1/3 hours, so their speed is
390/(10/3) = 117 mph
2x+15 = 117
x = 51 mph, the speed of the 1st train
66 mph = the speed of the 2nd train
Answer by richwmiller(17219)   (Show Source): You can put this solution on YOUR website!

3 hours 20 minutes. is not 3.2 but 3 1/3 or 10/3
10/3*x+10/3(x-15)=390
10/3(2x-15)=390
I get x=66
check
10/3*66+10/3(66-15)=390
220+10/3*(51)=390
220+170=390
ok
check
10/3(2*66-15)=390
10(2*22-5)=390
10*(44-5)=390
10*39=390
ok
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
"One train leaves city A heading for city B which is 390 miles away. At the same time a second train leaves city B heading for city A, Going 15 mph faster then the first train. If they meet in 3 hours and 20 minutes how fast were the trains traveling?
----------------------------
Train A DATA:
rate = x mph ; time = 3 1/3 = 10/3 hrs ; distance = rt = (10/3)x miles
----------------------------
Train B DATA:
rate = x+15 mph ; time = (10/3) hrs ; distance = rt = (10/3)(x+15) miles
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Equation:
A distance + B distance = 390 miles
(10/3)x + (10/3)(x+15) = 390
10x + 10(x+15) = 1170
20x + 150 = 1170
20x = 1020
x = 51 mph (Train A rate)
x+15 = 66 mph (Train B rate)
=================================
Cheers,
Stan H.

I keep trying to put it in the formula 3.2(x)+ 3.2(x-15)=390 from there I get (simplified) 6.4x= 438
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