SOLUTION: Rowing with the current of a river, a rowing team can row 25 mi in the same amount of time it takes to row 15 mi against the current. The rate of the rowing team in calm water is 2
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Question 83082This question is from textbook introductory algebra
: Rowing with the current of a river, a rowing team can row 25 mi in the same amount of time it takes to row 15 mi against the current. The rate of the rowing team in calm water is 20 mph. Find the rate of the current.
This question is from textbook introductory algebra
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Rowing with the current of a river, a rowing team can row 25 mi in the same amount of time it takes to row 15 mi against the current. The rate of the rowing team in calm water is 20 mph. Find the rate of the current.
:
Let x = rate of the current
Then:
(20+x) = speed with the current
(20-x) = speed against the current
:
The problems states that the times for the two trips are equal so we can make a
simple time equation. WE know that Time = Distance/speed
:
Time with = Time against
= :
:
We have single fractions on both sides of the equal side so we can cross mult
15(20+x) = 25(20-x)
300 + 15x = 500 - 25x; multiplied what's inside the brackets
15x + 25x = 500 - 300
40x = 200
x = 200/40
x = 5 mph is the current
:
:
Check solution by seeing if the times are equal:
25/(20+5) = 1
15/(20-5) = 1
:
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