Thread: How Many Miles?

On a trip, Powell travelled 40 mph for the first third of the distance and 50 mph for the remainder of the distance. If the whole trip took 4 hours and 20 minutes, how many miles did he travel?

I understand that in this question, we are searching for distance not rate or time.

I know that we need to use D = rt.

I need help setting up the equation for this distance question.

Originally Posted by magentarita

On a trip, Powell travelled 40 mph for the first third of the distance and 50 mph for the remainder of the distance. If the whole trip took 4 hours and 20 minutes, how many miles did he travel?

I understand that in this question, we are searching for distance not rate or time.

I know that we need to use D = rt.

I need help setting up the equation for this distance question.

Well, 4 hours and 20 minutes can be written as 4.33 hours. So

Hence:

Originally Posted by magentarita

On a trip, Powell travelled 40 mph for the first third of the distance and 50 mph for the remainder of the distance. If the whole trip took 4 hours and 20 minutes, how many miles did he travel?

I need help setting up the equation for this distance question.

To learn the basic methodology for this sort of exercise, try here.

Once you've learned the background material, look this over:

. . .first part:
. . . . .distance: d/3
. . . . .rate: 40
. . . . .time: (d/3)/40 = d/120

. . .last part:
. . . . .distance: (2/3)d
. . . . .rate: 50
. . . . .time: [(2/3)d]/50 = d/75

Since the total distance is "d", then the distances covered during each of the two parts add up to "d".

Create the equation, and solve.

Originally Posted by Mush

Well, 4 hours and 20 minutes can be written as 4.33 hours. So

Hence:

I have played with distance questions in terms of seeking for the rate and/or time but never with questions where the actual distance is missing. I am not talking about "Jerry drove 50 miles at 55 miles per hour. Find the distance." That is too simple. The question concerning Powell is tricky because of the wording. Do you agree?

Originally Posted by stapel

To learn the basic methodology for this sort of exercise, try here.

Once you've learned the background material, look this over:

. . .first part:
. . . . .distance: d/3
. . . . .rate: 40
. . . . .time: (d/3)/40 = d/120

. . .last part:
. . . . .distance: (2/3)d
. . . . .rate: 50
. . . . .time: [(2/3)d]/50 = d/75

Since the total distance is "d", then the distances covered during each of the two parts add up to "d".

Create the equation, and solve.

I did the same thing with the exception that I placed 120 for d/120 in the numerator and 75 for d/75 in the numerator. I can now add up the two and find my answer.