SOLUTION: A passenger train traveled 110 miles in the same amount of time it took a freight train to travel 90 miles. The rate of the freight train was 10 miles per hour slower than the rate

Question 924350: A passenger train traveled 110 miles in the same amount of time it took a freight train to travel 90 miles. The rate of the freight train was 10 miles per hour slower than the rate of the passenger train. Find the rate of the passenger train.
Found 2 solutions by TimothyLamb, MathTherapy:
Answer by TimothyLamb(4379) (Show Source): You can put this solution on YOUR website!
x = speed of passenger train
y = speed of freight train
---
s = d/t
t = d/s
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x = 110/t
y = 90/t
y = x - 10
---
y = 90/t
x - 10 = 90/t
x = 90/t + 10
---
x = 90/t + 10
x = 110/t
90/t + 10 = 110/t
20/t = 10
t = 2 hours
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answer:
x = 110/t
x = 110/2
x = speed of passenger train = 55 mph
---
y = 90/t
y = 90/2
y = speed of freight train = 45 mph
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Answer by MathTherapy(10552) (Show Source): You can put this solution on YOUR website!
A passenger train traveled 110 miles in the same amount of time it took a freight train to travel 90 miles. The rate of the freight train was 10 miles per hour slower than the rate of the passenger train. Find the rate of the passenger train.

Let passenger train's speed be S
Then freight train's speed = S - 10
Therefore,
110(S - 10) = 90S ------ Cross-multiplying
110S - 1,100 = 90S
110S - 90S = 1,100
20S = 1,100
S, or passenger train's speed = , or mph
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