SOLUTION: Chuck kayaked 4 miles upstream in
the same time it took him to kayak 10
miles downstream. If the speed of the
current is 3 mph, what is Chuck’s speed in
still water?
Algebra.Com
Question 1056082: Chuck kayaked 4 miles upstream in
the same time it took him to kayak 10
miles downstream. If the speed of the
current is 3 mph, what is Chuck’s speed in
still water?
Found 2 solutions by Boreal, jorel555:
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
v1*t1=distance
(x-3)*t1=4
t1=4/(x-3)
v2*t1=10
(x+3)*t1=10
t1=10/(x+3)
set the two times, which are the same, equal to each other.
4/(x-3)=10/(x+3)
Cross-multiply
10x-30=4x+12
6x=42
x=7 mph ANSWER
downstream at 10 mph, counting current, and it takes him 1 hour to do 10 miles
upstream at 4 mph, counting current, and it takes him 1 hour to do 4 miles.
Answer by jorel555(1290) (Show Source): You can put this solution on YOUR website!
Let n be Chuck's rate in still water. Then:
4/n-3=10/n+3
4n+12=10n-30
6n=42
n=7 mph as Chuck's rate in still water.☺☺☺☺
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