Q: When Paul drives to work his trip takes 40 minutes, but when he takes the train it takes 30 minutes. Find the distance Paul travels to work if the train travels an average of 15 mph faster than his driving. Assume that the train travels the same distance as the car. Paul travels ____ miles to work.

Question

I don't know where to begin in solving this problem. If someone could explain or help me figure out where to start that can lead me to the answer, I'd really appreciate it.
Information:
Driving = 40 minutes
Train Ride = 30 minutes
Train is 15 mph FASTER than driving.
10 minutes saved.
Same distance traveled.
What's the distance in miles?

Answer 1

well,let's start with "the train travels an average of 15 mph faster than his driving"
Let the driving speed be x mph
then the train's speed is x+15 mph

remember: distance = rate x time
distance gone by train = (30/60)(x+15)
distance gone driving = (40/60)x
but they are the same, so ......
(30/60)(x+15) = (40/60)x
(1/2)(x+15) = 2/3 x
multiply by 6 , the LCD
3(x+15) = 4x
3x + 45 = 4x
x = 45

put back into one of the distance definitions and you are done

Answer 2

Paul: d1 = r*T.
Train: d2 = (r+15)*T.
d1 = d2
r*T = (r+15)*T
r(40/60) = (r+15)30/60
40r = (r+15)30
40r = 30r+450
r = 45 mi/h
d1 = r*T= 45 * 40/60 = 30 miles = Paul's distance.