Asked by ty
A jogger ran 6 miles, decreased her speed by 1 mile per hour and than ran another 7 miles. If total jogging time was 2 1/42, find speed for each part of her run.
Answers
Answered by
Reiny
first speed --- x
second speed --- x-1
time at first speed = 6/x
time at 2nd speed = 7/(x-1)
6/x + 7/(x-1) = 2 1/42 = 85/42
times: 42x(x-1)
6(42)(x-1) + 7(42)x = 85x(x-1)
252x - 252 + 294x = 85x^2 - 85x
85x^2 - 631x + 252 = 0
x = (631 ± √312481)/170
= 7 or ...
aha , it facored
(x-7)(85x - 36) = 0
x=7 or x = 36/85
but 36/85 would make his 2nd speed negative, not very likeyly, so
x = 7
for the first leg he ran at 7 mph, for the 2nd leg he ran at 6 mph
check: 6/7 + 7/6 = (36+49)/42 = 85/42 = 2 1/42
second speed --- x-1
time at first speed = 6/x
time at 2nd speed = 7/(x-1)
6/x + 7/(x-1) = 2 1/42 = 85/42
times: 42x(x-1)
6(42)(x-1) + 7(42)x = 85x(x-1)
252x - 252 + 294x = 85x^2 - 85x
85x^2 - 631x + 252 = 0
x = (631 ± √312481)/170
= 7 or ...
aha , it facored
(x-7)(85x - 36) = 0
x=7 or x = 36/85
but 36/85 would make his 2nd speed negative, not very likeyly, so
x = 7
for the first leg he ran at 7 mph, for the 2nd leg he ran at 6 mph
check: 6/7 + 7/6 = (36+49)/42 = 85/42 = 2 1/42
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