Question 1171616
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Here is a highly unusual way for solving this kind of problem that I personally find easy to use.<br>
Take a look at it and see if it "works" for you.  If not, there are of course many other ways to solve the problem.<br>
The times are the same, and the distances are in the ratio 10:15 = 2:3.  That means the speeds are in the ratio 2:3.<br>
If the speed of the boat in still water is x, then the upstream speed is x-3 and the downstream speed is x+3.<br>
Solve for x by writing the ratio 2:3 as an equivalent ratio in which the difference between the two numbers is the difference between the upstream and downstream speeds:<br>
{{{2/3 = (x-3)/(x+3)}}}<br>
On the left, the difference between numerator and denominator is 1; on the right is is 6.  To get the desired equivalent ratio, multiply numerator and denominator on the left by 6.<br>
{{{2/3 = 12/18 = (x-3)/(x+3)}}}<br>
So x-3 is 12 and x+3 is 18; that means x is 15.<br>
ANSWER: The speed of the boat in still water is 15 mph.<br>