SOLUTION: Alika is training for a paddleboard race between Molokai and Oahu. He knows he can average 5 mph with no current. During a recent training run in the ocean, he noticed it took th

Algebra ->  Linear-equations -> SOLUTION: Alika is training for a paddleboard race between Molokai and Oahu. He knows he can average 5 mph with no current. During a recent training run in the ocean, he noticed it took th      Log On


   

Question 1113230: Alika is training for a paddleboard race between Molokai and Oahu. He knows he can average 5 mph with no current. During a recent training run in the ocean, he noticed it took the same amount of time to travel 10 miles with the current as it did to travel 6 miles against the current. What was the rate of the current that day?
thank you!
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
v*t=distance
(5-c)*t=6
(5+c)*t=10, where c is the current
t=6/(5-c)
t also = (10)/(5+c)
Those two are equal since the times are stated to be equal.
cross multiply
6(5+c)=10(5-c)
30+6c=50-10c
16c=20
c=1.25 mph ANSWER
check
3.75 mph against and takes 6/3.75 or 1.6 hours to go 6 miles.
6.25 mph with and takes 1.6 hours to go 10 miles