SOLUTION: I'm sorry to ask this questions. I can't figure any formula to solve this questions.This problem is from College Algebra, a private book from one of the universities here in the Ph

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Question 157468: I'm sorry to ask this questions. I can't figure any formula to solve this questions.This problem is from College Algebra, a private book from one of the universities here in the Philippines.Thank you for any answer.
Here are the problems.
1) The current of water is 8 KPH. The boat travels 2/3 of the time downstream from point C to D.The boat travels downstream from point D to C. Determine the rate of the boat in still water in KPH.
I think the question is wrong.the part of question "the boat travels downstream from point D to C made me really confused.I will appreciate any help from you.
2) A boatman rows to a place 48 miles distant and back in 14 hours, but find that he can row 4 miles with the stream in the same time as 3 miles against the stream.Find the rate of the stream.

Answer by gonzo(654) About Me  (Show Source):
You can put this solution on YOUR website!
it appears that they meant that 2/3 of the time the boat is going upstream against the current of the river.
i would suggest a rewording of the problem as follows:
The current of water is 8 KPH. The boat travels 2/3 of the time upstream (against the current of the water) from point C to D. The boat travels downstream from point C to D. Determine the rate of the boat in still water in KPH.
working the problem, i get the following:
rate of water is 8KPH.
D = Distance upstream and also equals Distance downstream.
Rate * Time = Distance.
Time to get upstream = is 2/3 of the total time while the time to get downstream is 1/3 of the total time.
B = rate of the boat.
B-8 = rate of the boat against the stream.
B+8 = rate of the boat with the stream.
going upstream using rate * time = distance, the formula is 2/3*(B-8) = D
going downstream using rate * time = distance, the formula is 1/3*(B+8)=D
since both upstream and downstream travel the same distance, then R*T upstream = R*T downstream.
the formula then becomes 2/3*(B-8) = 1/3*(B+8)
multiplying by 3 to get rid of the fractions we get 2*(B-8) = 1*(B+8)
this becomes 2*B-16 = B+8
this becomes B = 24
rate of the Boat is 24 KPH.
rate of the boat against the stream is (24-8) = 16 KPH.
rate of the boat with the stream is (24+8) = 32 KPH.
answer to number 1 is 24 KPH is the rate of the boat in still water.
confirming the formula we get 2/3*(24-8) = 1/3*(24+8) making 2/3*16 = 1/3*32 which becomes 32 = 32 proving the formula is sound.
as for number 2, the solution follows:
---------------------------------------------
boatman rows 48 miles and back in 14 hours.
he can row 4 miles with the stream in the same time as 3 miles against the stream.
find the rate of the stream.
R*T = D
48 = D
let R1 = rate of the boat with the stream.
let R2 = rate of the boat against the stream.
use T = D/R to derive a formula for the different rates.
the formula T = 3/R2 describes traveling 3 miles going against the stream in T hours at the rate of R2.
the formula T = 4/R1 describes traveling 4 miles going with the stream in T hours at the rate of R1.
since they both = T they are equal to each other.
so 4/R1 = 3/R2.
multiplying by R1*R2 to remove the denominators, we get
4*R2 = 3*R1
which becomes R1 = 4/3*R2 which means that the rate with the stream is 4/3 * the rate against the stream.
plugging this in the equation to find the rate upstream and the rate downstream we solve for T1 and T2.
the formula is T = D/R.
48 miles is the distance
rate going upstream is R2.
rate going downstream is 4/3*R2.
total time is 14 hours.
formula is: %2848%2F%284%2F3%2AR2%29%29%2B%2848%2FR2%29+=+14
multiplying both sides of the equation by 12*R2 to get rid of the denominators, we get
%2812%2AR2%2A48%29%2F%28%284%2F3%2AR2%29%29%2B%2848%2A12%2AR2%2FR2%29=14%2A12%2AR2%29
since a/(b/c) is the same as (a*c/b) the formula becomes
%2812%2AR2%2A48%2A3%29%2F%28%284%2AR2%29%29%2B%2848%2A12%2AR2%2FR2%29=14%2A12%2AR2%29
simplifying, we get
%283%2A3%2A48%29%2B%2848%2A12%29=%2814%2A12%2AR2%29
this becomes
432%2B576=168%2AR2
which becomes
1008=168%2AR2
which becomes R2+=+6.
if R2 = 6 then R1 = 4/3*R2 becomes 8.
if B + S is going with the stream then B+S = 8.
if B - S is going against the stream then B-S = 6
if B+S = 8 then S = 8 - B
likewise if B-S = 6, then S = 6+B
since they both = S, then the right side of the equations are equal to each other so
8-B = 6+B
making 2*B = 2 making B = 1.
rate of the stream is 1.
solving in the original equation, boatman travels at 8 miles per hour downstream and 6 miles per hour up stream doing the whole thing in 14 hours.
T = D/R
48/8 + 48/6 = 14 hours
takes boatman 6 hours to go downstream and 8 hours to go upstream.
6 + 8 = 14.
8 = 4/3 * 6.
formula is confirmed.
rate of the stream is 1 KPH or MPH whatever they were using.