SOLUTION: rowing with a current keith rowed a canoe 3 miles an hour. on his return trip against the current keith take 1.5hours. how fast does keith row in still water and how fast is the cu

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Question 1154026: rowing with a current keith rowed a canoe 3 miles an hour. on his return trip against the current keith take 1.5hours. how fast does keith row in still water and how fast is the current
Answer by ramkikk66(644) About Me  (Show Source):
You can put this solution on YOUR website!
Let current speed be x mph
Let keith's speed in still water be y mph
Distance travelled in 1 hour = 3 miles
Travelling with current, effective speed is (y + x) mph
So the distance travelled = speed * time
= (y + x) * 1 = (y + x) = 3
Travelling against current, his effective speed is (y - x) mph
At (y - x) mph, he took 1.5 hours for the same distance i.e. 3 miles
Since speed * time = distance, we get the equation
(y - x) * 1.5 = y + x = 3
Which can be broken up as 2 equations in x and y as below
x+%2B+y+=+3 --- (1)
1.5y+-+1.5x+=+3 --- (2)
Multiplying (1) by 1.5
1.5x+%2B+1.5y+=+4.5 --- (3)
Adding (2) and (3)
3y+=+7.5
i.e.
y+=+2.5
x+=+3+-+y+=+0.5
Speed in still water = 2.5 mph
Speed of current = 0.5 mph
Check.
With current, effective speed = 2.5 + 0.5 = 3 mph, so time taken to travel 3 miles = 3 / 3 = 1 hour (check)
Against current, effective speed = 2.5 - 0.5 = 2 mph, so time taken to travel 3 miles = 3 / 2 = 1.5 hours (check)
Hope this helps!