SOLUTION: A Boat goes upstream 40 miles and downstream 86 miles in 8 Hours. If it goes upstream 24 miles and down the same distance in 5 hrs, what is the boat speed?
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Question 269229: A Boat goes upstream 40 miles and downstream 86 miles in 8 Hours. If it goes upstream 24 miles and down the same distance in 5 hrs, what is the boat speed? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! A Boat goes upstream 40 miles and downstream 86 miles in 8 Hours.
If it goes upstream 24 miles and down the same distance in 5 hrs, what is the boat speed?
:
Let x = boat speed in still water
Let y = speed of the current
Then
(x+y) = speed downstream
(x-y) = speed upstream
:
Write a time equation for each statement
:
"A Boat goes upstream 40 miles and downstream 86 miles in 8 Hours." + = 8
Multiply by (x-y)(x+y) to get rid of the denominators
:
40(x+y) + 86(x-y) = 8(x+y)(x-y)
40x + 40y + 86x - 86y = 8(x^2 - y^2)
126x - 46y = 8x^2 - 8y^2
Simplify, divide by 2
63x - 23y = 4x^2 - 4y^2
4x^2 - 63x + 23y - 4y^2 = 0
:
" If it goes upstream 24 miles and down the same distance in 5 hrs," + = 5
Multiply by (x-y)(x+y) to get rid of the denominators
:
24(x+y) + 24(x-y) = 5(x+y)(x-y)
24x + 24y + 24x - 24y = 5(x^2 - y^2)
48x = 5x^2 - 5y^2
5x^2 - 48x - 5y^2 = 0
:
Try to eliminate some of this, mult the 1st by 5, the above by 4
20x^2 - 315x + 115y - 20y^2 = 0
20x^2 - 192x + 0y - 20y^2 = 0
-------------------------------------subtraction eliminates x^2 and y^2
-123x + 115y = 0
115y = 123x
:
I don't think this is possible, the current exceeds the boat speed