A boat goes upstream from Carter to Hixville in 6 hrs and back again in 2 hrs. It's 24 miles from Carter to Hixville. What is the rate of the boat in still water, and what is the rate of the current?
2006-10-29 07:39:34 · 3 answers · asked by shy l 1 in Science & Mathematics ➔ Mathematics
Let s be the rate in still water, and c be the rate of the current. The rate upstream is s - c, and the rate down is s + c.
s - c = 24/6 = 4, and s + c = 24/2 = 12, so now we solve the system of equations
s - c = 4
s + c = 12
Add the two equations: 2s = 16, so s = 8 mph, and c= 4 mph.
2006-10-29 07:51:39 · answer #1 · answered by James L 5 · 0⤊ 0⤋
b - c = 24/6 = 4
b + c = 24/2 = 12
b = (4 + 12)/2 = 8
c = 12 - 8 = 4
2006-10-29 16:13:30 · answer #2 · answered by Helmut 7 · 0⤊ 0⤋
Let x = rate in still water
Let y = rate of current
x-y =rate going upstream
x+y = rate going downstream
Since ratte times time = distance, we have
(x-y)*6 =24, and
(x+y)*2 = 24
So 6x-6y=24 [EQ I], and
2x+2y =24 [EQ II]
Multiply [EQ II] by3 to get:
6x + 6y = 72 [EQ III]
Now add [EQ I]to [EQ III] to get:
12x = 96
x = 8 mph in still water
Now substitute x=8 into [EQ II] and get:
2*8 + 2y = 24
2y = 8
y= 4 mph rate of current
2006-10-29 16:08:17 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋