Question

Jason ran for 1.5 hours with a 2 mph wind at his back. It took him 2.5 hours to get home with the same wind in his face. If he ran in the same manner and effort, how fast would Jason have run without the wind?

a. 5 mph
b. 10 mph
c. 15 mph
d. 20 mph

Answer 1

Distance = Rate * Time

J = Jason Speed (Unassisted or deterred)

Trip Out

Distance = (J + 2 mph)*(1.5 hr)

Trip Back

Distance = (J - 2 mph)*(2.5 hr)

Any of that make sense?

Answer 2

Yes, I did that but my answer came out to 8 @tkhunny

Answer 3

How did you get that?

Since the distances are equal:

(J + 2 mph)*(1.5 hr) = (J - 2 mph)*(2.5 hr)

Show steps in solving for J.

Answer 4

@tkhunny
(J + 2 mph)*(1.5 hr) = (J - 2 mph)*(2.5 hr)
1.5J + 3 = 2.5J - 5
-1.5j -1.5j

3= 1J - 5
+5 +5

8= J

Answer 5

Seems reasonable to me. Not every answer set is printed correctly.

Answer 6

What do I do? @tkhunny

Answer 7

Punt? Just show your work and prove your answer.

Maybe we're supposed to assume the wind is only partially effective? If the wind is 62.5% effective, we get j = 5 mph. That might be a fun answer to show.

For k = 0.625

(J + k*(2 mph))*(1.5 hr) = (J - k*(2 mph))*(2.5 hr)

Of course, I'm just goofing off and making up my own problem statement. If you have to mark one, try B. 10 but make sure you go ARGUE it at your earliest convenience.

Answer 8

Okay thank you @tkhunny

Answer 9

Sorry, the World isn't always clean or clear. Sometimes you just have to make stuff up. Show your work and prove what you are doing. A good teacher will give extra credit for creativity and ambition. :-)