SOLUTION: A MAN ROWED 10 KM UP A RIVER IN 5 HOURS AND BACK IN 21 AND A HALF HOURS. FIND THE RATE OF THE CURRENT AND HIS RATE OF ROWING IN STILL WATER.

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Question 162663: A MAN ROWED 10 KM UP A RIVER IN 5 HOURS AND BACK IN 21 AND A HALF HOURS. FIND THE RATE OF THE CURRENT AND HIS RATE OF ROWING IN STILL WATER.
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let his rate in still water = s
Let the rate of the current = c
His rate against the current is s+-+c
His rate with the current is s+%2B+c
The distance both ways is 10km
distance = rate * time
(1) 10+=+%28s+%2B+c%29%2A5 mi (with the current)
(2) 10+=+%28s+-+c%29%2A21.5 mi (against the current)
----------------------------------------------------
(1) 10+=+5s+%2B+5c
(2) 10+=+21.5s+-+21.5c
Multiply both sides of (2) by 10
(2) 100+=+215s+-+215c
Multiply both sides of (1) by 43
(1) 430+=+215s+%2B+215c
(2) 100+=+215s+-+215c Now add the equations
(3) 530+=+430s
s+=+1.233km/hr
(1) 10+=+5s+%2B+5c
5c+=+10+-+5%2A1.233
5c+=+10+-+6.163
5c+=+3.837
c+=+.7674km/hr
His rate in still water is 1.233 km/hr
The rate of the current is .7674 km/hr
check answer:
(1) 10+=+%28s+%2B+c%29%2A5
(2) 10+=+%28s+-+c%29%2A21.5
--------------------------
(1) 10+=+%281.233+%2B+.7674%29%2A5
10+=+2.0004%2A5
10+=+10.002 error due to rounding off
(2) 10+=+%28s+-+c%29%2A21.5
10+=+%281.233+-+.7674%29%2A21.5
10+=+.4656%2A21.5
10+=+10.010 (error due to rounding)
OK