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.
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?
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
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.
To solve this problem, we can start by converting the given information into a mathematical equation. Let's assume the distance Paul travels to work is d miles.
We are given that Paul's driving time is 40 minutes, which is equal to 40/60 = 2/3 hours. So, the driving speed can be calculated as d/(2/3) = (3d/2) miles per hour.
The train ride takes 30 minutes, which is equal to 30/60 = 1/2 hours. We are also given that the train travels 15 mph faster than Paul's driving speed. So, the train's speed can be calculated as (3d/2) + 15 miles per hour.
Now, we know that distance = speed × time. Since the train and the car are traveling the same distance, we can equate the distances traveled by both to get the equation:
(d/(2/3)) × (2/3) = ((3d/2) + 15) × (1/2)
Simplifying the equation:
3d = 3d + 30
We can see that the distance term (d) cancels out on both sides of the equation. This implies that the distance Paul travels is irrelevant to the answer. Therefore, there is no unique solution for the distance Paul travels to work.