SOLUTION: I have spent hours on this one and been through every one I know Can you help? Please? Upstream Downstream Jr's boat will go 15 miles per hour in still water. If he can go 12

Question 152187: I have spent hours on this one and been through every one I know Can you help? Please?
Upstream Downstream
Jr's boat will go 15 miles per hour in still water. If he can go 12 miles down stream in the same amount of time as it takes to go 9 miles upstream, then what is the speed of the current. T= D/R... D = RT .. R = D/T

Found 2 solutions by ankor@dixie-net.com, mducky2:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Jr's boat will go 15 miles per hour in still water. If he can go 12 miles down stream in the same amount of time as it takes to go 9 miles upstream, then what is the speed of the current.
:
Let x = speed of the current
then
(15+x) = boat speed downstream
and
(15-x) = boat speed up stream
:
Times are equal so write a time equation; Time = Dist/speed
:
Downstream time = Upstream time
=
Cross multiply and you have:
9(15+x) = 12(15-x)
:
135 + 9x = 180 - 12x
:
9x + 12x = 180 - 135
:
21x = 45
x =
x = 2.143 mph speed of the current
;
:
Check solution by finding the time of each
(12/17.143) = (9/12.857)
.69999 ~ .70000; because x actually = 2.142877143

Answer by mducky2(62) (Show Source): You can put this solution on YOUR website!
If we think of the stream as a conveyor belt, the current is carrying the boat at the same time that the boat is going 15 mph. In other words, the upstream distance really means the distance that the boat traveled plus the distance the water traveled. The same logic goes for downstream:
D_boat + D_current = 12
D_boat - D_current = 9

If we subtract the bottom equation from the top equation:
D_boat - D_boat + D_current + D_current = 12 - 9
2D_current = 3
D_current = 1.5

Plugging this back into the first equation:
D_boat + 1.5 = 12
D_boat = 10.5

Since we know that the current and the boat traveled a certain distance in the same amount of time, we can equate the time of the boat to the time of the current:
T = (D_boat/R_boat) = (D_current/R_current)

We can plug in the numbers that we know:
D_boat = 10.5 miles
R_boat = 15 mph
D_current = 1.5 miles
R_current = C

Now we can solve for the rate of the current
(D_boat/R_boat) = (D_current/R_current)
10.5/15 = 1.5/C
C = (1.5*15)/10.5
C = 2.143

Therefore, the speed of the current is 2.143 mph.

RELATED QUESTIONS
Karen�s boat has a top speed of 10 miles per hour in still water. While traveling on a... (answered by josmiceli)
hello there i have a word problem which i have been working on for the past couple of... (answered by KMST)
I have a bad time with word problems. Please help me with this: Junior's boat will go (answered by Greenfinch,richwmiller)
I am having a hard time figuring this word problem out. Can you please help. Upstream,... (answered by Fombitz,stanbon)
Good Evening Sir , I have an doubt can you please help me . the question is as following (answered by checkley71)
a tugboat can pull a boat 24mi downstream in 2h. going upstream, the tugboat can pull the (answered by josgarithmetic)
See I'm trying to help my son who's always been in special ed in elementary and has had... (answered by ankor@dixie-net.com)
One more question please ! Someone please help. A boat can travel 10 miles downstream... (answered by josmiceli)
I can row upstream at 2mph and downstream at 8mph. How far can I row to get back to... (answered by mananth)