SOLUTION: A paddle boat can move at a speed of 12 km/h in still water. The boat is paddled 6 km downstream in a river in the same time it takes to go 3 km upstream. What is the speed of the

Click here to see ALL problems on Travel Word Problems

Question 130942: A paddle boat can move at a speed of 12 km/h in still water. The boat is paddled 6 km downstream in a river in the same time it takes to go 3 km upstream. What is the speed of the river?
Found 2 solutions by checkley71, TakeATuition.com:
Answer by checkley71(8403) (Show Source):
You can put this solution on YOUR website!
6/(12+X)=3/(12-X) NOW CROSS MULTIPLY.
6(12-X)=3(12+X)
72-6X=36+3X
-6X-3X=36-72
-9X=-36
X=-36/-9
X=4 KPM IS THE SPEED OF THE CURRENT.
PROOF:
6/(12+4)=3/(12-4)
6/16=3/8
3/8=3/8

Answer by TakeATuition.com(57) (Show Source):
You can put this solution on YOUR website!
Solution-
The format of these type (BOAT ,UPSTREAM AND DOWNSTREAM)is given below.you just have to substitute the values in the given format and solve it according to the condition. Here is the format
let the speed of the boat be x km/h
and the speed of the river/stream be y km/h
speed of boat in downstream = (x+y)km/h
speed of boat in upstream = (x-y)km/h
time taken by boat in downstream = distance for downstream/(x+y)
time taken by boat in upstream = diatance for upstream/(x-y)km/h
now solve according to the condition

HERE IS THE SOLUTION OF YOUR QUESTION
the speed of the boat is 12km/h
let the speed of the river be y km/h
speed of the boat in downstream = (12+y)km/h
speed of the boat in upstream =(12-y)km/h
time taken by boat in downstream = distance for downstream/(12+y)
= 6/(12+y)
time taken by boat in upstream = distance for upstream/(12-y)
= 3/(12-y)
now according to the condition
time taken by boat in downstream = time taken by boat in upstream
6/(12+y)= 3/(12-y)
now by cross multiplication we have
6(12-y)=3(12+y)
72-6y = 36+3y
shifting the like terms on one side we have
-6y-3y = 36-72
-9y = -36
y = 4
Therefore the speed of the river is 4 km/h
for more help in math and physics visit: www.TakeATuition.com