SOLUTION: Hello, I am having trouble understanding and solving this problem: A kayaker can paddle 12 miles in 2 hours moving with the river current. Paddling at the same pace, the trip ba

Question 983517: Hello, I am having trouble understanding and solving this problem:
A kayaker can paddle 12 miles in 2 hours moving with the river current. Paddling at the same pace, the trip back against the current takes 4 hours. Assume that the river current is constant. Find what the kayaker's speed would be in still water.

-Thank you.
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39617) (Show Source): You can put this solution on YOUR website!
r, the speed in still-water
c, speed of the river current
d=12
u=2, time with the current
b=4, time against the current

Basic uniform rates for travel, RT=D to relate speed, time, distance

"Paddling at the same pace", must mean kayaker still uses his own speed r, but that the speed of current, c, changes the effective rate according to which direction he paddles.

speed time distance With current, up r+c u d Against current, back r-c b d

Unknown variables are r and c; but you are interested in answering for solving r.

Elimination Method can be used (but you could use substitution instead; your choice). Best to first eliminate r. I will instead use substitution.

-
, subst into second equation. Notice this equation uses only the unknown variable r, to be found as asked.

Substitute for d, u, b, and evaluate r.

Answer by MathTherapy(10552) (Show Source): You can put this solution on YOUR website!
Hello, I am having trouble understanding and solving this problem:
A kayaker can paddle 12 miles in 2 hours moving with the river current. Paddling at the same pace, the trip back against the current takes 4 hours. Assume that the river current is constant. Find what the kayaker's speed would be in still water.

-Thank you.

Let speed of kayak, in still water be S, and speed of current, C
Total speed moving with the current: , or 6 mph
Total speed moving against the current: , or 3 mph
We then get: S + C = 6 ------- eq (i)
S - C = 3 ------- eq (ii)
2S = 9 ------- Adding eqs (ii) & (i)
S, or speed in still water = , or mph
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