SOLUTION: Jim can run 5 miles per hour on level ground on a still day. One windy day, he runs 15 miles with the wind and in the same amount of time runs 4 miles against the wind. What is
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Question 29478: Jim can run 5 miles per hour on level ground on a still day. One windy day, he runs 15 miles with the wind and in the same amount of time runs 4 miles against the wind. What is the rate of the wind?
This is what I have
x= rate of wind
5 + x = running with the wind
5 - x = running against the wind
then I have
15/5+x = distance on a windy day running with the wind
4/5-x = distance on a windy day running against the wind
Am I on the right track? When I solve for x I get a decimal of 1.89 or 1.90
thanks for your help Answer by josmiceli(19441) (Show Source):
You can put this solution on YOUR website! You were almost there
You are given two situatiuons and the common element is time
let r = runners rate on a still day
let w = rate of the wind
let t = time
running with the wind the combined rate is
r + w
running against the wind the combined rate is
r - w
distance = rate x time
[1] 15 = (r + w) x t
[2] 4 = (r - w) x t
divide both sides of [1] by (r + w)
divide both sides of [2] by (r - w)
15 / (r + w) = t
4 / (r - w) = t
since they're both equal to t, set them equal to eachother
15/(r + w) = 4 / (r - w)
from the problem description, r = 5
15 / (5 + w) = 4 / (5 - w)
cross-multiply
15(5 - w) = 4(5 + w)
75 - 15w = 20 + 4w
19w = 55
w = 2.895
check
15(5 - 55/19) = 4(5 + 55/19)
15(95 - 55)/19 = 4(95 + 55)/19
15 x 40 / 19 = 4 x 150 / 19
multiply both sides by 19
divide both sides by 150
4 = 4
OK
where you went wrong was here
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15/5+x = distance on a windy day running with the wind
4/5-x = distance on a windy day running against the wind
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sould have read
"TIME on a windy day running with the wind" and
"TIME on a windy day running against the wind"
