Question 253318: An Everglades air boat travels 78 miles downstream in the same amount of time it requires to move upstream a distance of 48 miles. The boat's engines drive in still water at an average rate of 16 mph greater than the rate of the current. Find the speed of the boat and the water current. (Find the system of two equations and use the system to find the speed of the boat and the current).
Answer by palanisamy(496)   (Show Source):
You can put this solution on YOUR website! Let the speed of the boat in still water be x miles/h
Let the speed of the current be y miles/h
Given, x-y = 16 ...(1)
Given, An Everglades air boat travels 78 miles downstream in the same amount of time it requires to move upstream a distance of 48 miles.
So, 78/(x+y) = 48/(x-y) [Time = distance/speed)
78(x-y)= 48(x+y)
78*16= 48(x+y)
1248= 48(x+y)
x+y = 1248/48
x+y = 26 ...(2)
(1)+(2): 2x = 16+26
2x = 42
x = 21
(2)=> 21+y = 26
y = 26-21
y = 5
Therefore the speed of the boat in still water is 21 miles/h
and the speed of the current is 5 miles/h
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