SOLUTION: A boat took 1 hour and 50 minutes to go 55 km downstream and 3 hours, 40 min to return . Find a boat's rate in still water and the rate of the current. down stream data ok so

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: A boat took 1 hour and 50 minutes to go 55 km downstream and 3 hours, 40 min to return . Find a boat's rate in still water and the rate of the current. down stream data ok so       Log On


   



Question 188816: A boat took 1 hour and 50 minutes to go 55 km downstream and 3 hours, 40 min to return . Find a boat's rate in still water and the rate of the current.
down stream data
ok so distance = 55 km ; time =11/6 hr. ; rate =d/t= 55 ( 11/6)=30

upstream data
distance = 55km; time= (11/3)hr.; rate= d/t=55 (11/3)=15
equations b+ c =30 kph
b-c=15kph
2b=45 kph
b=22.5 kph ( speed of the boat in still water
22.5+c=30
c=30 -22.5=7.5 kph (spped of the current)

question: how did one hour and 60 minutes turn into 11/6 hr?
how did 3 hours and 40 minutes tunr into 11/3hr?
how did d/t= 55/(11/6) turn into 30 kph?
how did d/t=55/(11/3) turn into 15 kph?
please help !

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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A boat took 1 hour and 50 minutes to go 55 km downstream and 3 hours, 40 min to return . Find a boat's rate in still water and the rate of the current.
down stream data
ok so distance = 55 km ; time = 11/6 hr. ; rate = d/t = 55 ( 11/6)=30
upstream data
distance = 55km; time= (11/3)hr.; rate= d/t=55 (11/3)=15
equations b+ c =30 kph
b-c=15kph
2b=45 kph
b=22.5 kph ( speed of the boat in still water
22.5+c=30
c=30 -22.5=7.5 kph (speed of the current)
question: how did one hour and 60 minutes turn into 11/6 hr?
:
Think you mean,"how did one hour and 50 minutes turn into 11/6 hr?"
We want to work in hrs here since we are using km/hr, to convert 50 min to hrs:
50/60 = 5/6 hrs, plus the 1 hr which can be written 6/6 hrs, 6/6 + 5/6 = 11/6 hr
:
:
how did 3 hours and 40 minutes turn into 11/3hr?
:
Convert 40 min to hrs: 40/60 reduces to 4/6 reduces to 2/3 hr
Add the 3 hr which can be written 9/3 for a total: 9/3 + 2/3 = 11/3 hrs
:
:
how did d/t= 55/(11/6) turn into 30 kph?
:
Remember rule of dividing fractions, "Invert the dividing fraction & multiply"
55 * 6/11, cancel 11 into 55: 5 * 6 = 30
:
:
how did d/t=55/(11/3) turn into 15 kph?
:
Like-wise, invert the dividing fraction and multiply 55 * 3/11
Cancel 11 into 55 and you have 5 * 3 = 15
;
:
Did this help explain things here? Any questions?