SOLUTION: A boat can travel 60 km/h in calm water. The boat travels 200 km against the current and then returns to the location where it started. If the trip takes 10 total hours, what is
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Question 910315: A boat can travel 60 km/h in calm water. The boat travels 200 km against the current and then returns to the location where it started. If the trip takes 10 total hours, what is the speed of the current?
I started with t = dr to r = t/d
(1/60) x 10 = 10/600
10/200 = 10/600
cross multiply = 6000/2000=3
I don't know if 3 would be the answer or if I have to do more, or if its the answer I don't know if its the right answer. Found 2 solutions by rothauserc, richwmiller:Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website! rate(r) * time(t) = distance(d)
let r be the speed of the current and t1 + t2 = 10 and d = 200, then
let t1 be time with current and t2 time against current, then we have
t1 + t2 = 10
time with current is
t1 = 200 / (60+r)
time against current is
t2 = 200 / (60-r)
now we can add the two times
200/(60+r) + 200/(60-r) = 10
our lowest common denominator is (60+r)*(60-r)
200*(60-r) + 200*(60+r) = 10 * (60+r) * (60-r)
we can divide both sides of = by 10
20*(60-r) + 20*(60+r) = 3600 - r^2
simplify and combine like terms
1200 -20r + 1200 +20r = 3600 - r^2
2400 = 3600 -r^2
r^2 = 1200
r = square root of (1200) = 34.6 mph
speed of current is 34.6 mph
You can put this solution on YOUR website! r*t=d
200/(60-c)+200/(60+c)=10/1
Multiply thru by 1*(60-c)(60+c)
200*1*(60+c)+200*1(60-c)=10(c^2-3600)
200*60+200*c+200*60-200*c=10*c^2-10*3600
12000+200*c+12000-200*c=10*c^2-36000
0=10*c^2-36000+(24000)+(0)*c
10*c^2+-12000=0
10*c^2+-12000=0
c=34.641
