SOLUTION: A boy rows his boat 8 km downstream in 1 hour and 4 min, and returns in
2 2/7 hours. At what rate will he row in still water? What is the rate
of the current?
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Question 152450: A boy rows his boat 8 km downstream in 1 hour and 4 min, and returns in
2 2/7 hours. At what rate will he row in still water? What is the rate
of the current?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A boy rows his boat 8 km downstream in 1 hour and 4 min, and returns in
2 2/7 hours. At what rate will he row in still water? What is the rate
of the current?
:
Let x = his rowing rate in still water
Let y = rate of the current
then
(x+y) = his rate down stream
(x-y) = his rate up stream
:
Convert 4 min to hrs: 4/60 = 1/15 hr
:
write a distance equation for each trip; dist = time * rate
;
1(x+y) = 8
2(x-y) = 8
Convert to improper fractions:
(x+y) = 8
(x-y) = 8
:
Multiply each equation by the denominator to get rid of the denominators
16(x+y) = 15(8)
16(x-y) = 7(8)
:
16x + 16y = 120
16x - 16y = 56
--------------------adding eliminates y, find x
32x + 0y = 176
x =
x = 5 km/hr is the rowing rate
:
Find the current using the simplified equation: 16x + 16y = 120; (x=5.5)
16(5.5) + 16y = 120
88 + 16y = 120
16y = 120 - 88
16y = 32
y =
y = 2 km/hr is the current rate
:
:
Check our solutions in the 2nd original equation; (x-y) = 8
(5 - 2) = 8
*3 = 8
* = 8
= 8; confirms our solutions
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