SOLUTION: A freight train leaves a station and travels north at the same time a passenger train leaves the same station and travels in the opposite direction. The speed of the freight train
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Question 1056128: A freight train leaves a station and travels north at the same time a passenger train leaves the same station and travels in the opposite direction. The speed of the freight train is 18 miles per hour slower than the speed of the passenger train. After 2.5 hours the two trains are 205 miles apart. Find the speed of the two trains. THANK YOU!!!! Found 2 solutions by ikleyn, MathTherapy:Answer by ikleyn(52775) (Show Source):
You can put this solution on YOUR website! .
A freight train leaves a station and travels north at the same time a passenger train leaves the same station
and travels in the opposite direction. The speed of the freight train is 18 miles per hour slower than the speed
of the passenger train. After 2.5 hours the two trains are 205 miles apart. Find the speed of the two trains. THANK YOU!!!!
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1. = 82 .
It is the sum of the two rates, and .
So, your first equation is
+ = 82.
2. Your second equation is
- = 18. ("The speed of the freight train is 18 miles per hour slower than the speed of the passenger train.")
3. Combine these two equation to the system:
+ = 82.
- = 18.
4. Now add the two equations. You will get.
= 82 + 18 = 100.
Hence, = 100.
Then = = 50 is the speed of the passenger train.
Then the speed of the freight train is 50-18 = 32 .
Answer. 50 km/h for the passenger train and 32 km/h for the freight train.
You can put this solution on YOUR website!
A freight train leaves a station and travels north at the same time a passenger train leaves the same station and travels in the opposite direction. The speed of the freight train is 18 miles per hour slower than the speed of the passenger train. After 2.5 hours the two trains are 205 miles apart. Find the speed of the two trains. THANK YOU!!!!
Let speed of freight (slower) train be S
Then speed of passenger (faster) train = S + 18
We then get the following DISTANCE equation: 2.5S + 2.5(S + 18) = 205
2.5S + 2.5S + 45 = 205
5S = 205 - 45
5S = 160
S, or speed of freight (slower) train = , or
Speed of passenger (faster) train = 32 + 18, or