SOLUTION: Please help me with this problem I have exhausted all of my resources to try and figure it out. On a windy day Yoshiaki found that he could go 16mi downstream and then 4mi back

Question 19478: Please help me with this problem I have exhausted all of my resources to try and figure it out.
On a windy day Yoshiaki found that he could go 16mi downstream and then 4mi back upstream at top speed in a total of 48min. What was the top speed of Yoshiaki's boat if the current was 15mph?

The answer in the back of the book is 25mph, but I can't seem to figure out how they came up with that answer. We are working with quadratic equations. Here is what I have tried:
______________D_____R________T_______
upstream:_____4mi___x-15___4/(x-15)
downstream:___16mi__x+15___16/(x+15)
_____________________________________
4/(x-15) + 16/(x+15) = 48
I cannot seem to come up with the answer 25, I must be doing something wrong in the form of the equation. Please Help!!
Answer by xcentaur(357) (Show Source): You can put this solution on YOUR website!
Let the speed of the boat be x mph
let the speed of the current be y mph

Then while going downstream,
distance downstream=16miles
speed=(x+y) [since the current helps the boat]
time taken=d/s=16/(x+y).......[1]

while going upstream,
distance upstream=4 miles
speed=(x-y) [since the current is in the opposite direction of the boat]
time taken=d/s=4/(x-y).........[2]

Now we have been told that the total time taken is 48 minutes
48 minutes means : (48/60)hour=(4/5)hour

Add up both the times we got (equations 1 and 2)

Now we have been told in the question that the speed of the current (y) is 15 mph
So we get,

opening the brackets

now '-180' cancels off

which can be written as

cancelling out the x we get,

And so we get the top speed of the boat(x) = 25 mph

Hope this helps,
Prabhat

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