This is the solution to the algebra time and distance word problem # 10 which was submitted under the ask a question post anonymously which read..

"Junior's boat will go 15 miles per hour in still water. If he can go 12 miles downstream in the same amount of time as it takes to go 9 miles upstream then what is the speed of the current?"

distance = rate X time ... d = r*t we know the speed of the boat in still water 15 mph

we know he can go 12 miles downstream in the same time he can go 9 miles upstream-- so times are equal --- so since time is equal we can set the distances and rates equal to one another...

d1/r1 = d2/r2

The rate downstream will be 15 + x and upstream will be 15 - x so we have:

12/(15+x) = 9 /(15-x)

cross multiply and you get 12(15-x) = 9 (15+x)

180 - 12x = 135 + 9x

45 = 21x

x = 45/21 = 15/7 This is the answer the current is 15/7 or 2 and 1/7ths mph..

## Saturday, September 12, 2009

### Time and Distance Problem 10 Solution

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The rate downstream will be 15 + x and upstream will be 15 - x right?

ReplyDeleteif im not mistaken it will be

12/(15+x)=9/(15-x) coz the 12 is the downstream and the 9 is the upstream

and if there's division outside the parenthesis it must be cross multiply so;

12/(15+x)=9/(15-x) 12/(15+x)=9/(15-x)

|_____x______| |_x__|

135+12x=180-9x

45=21x

=2.14mph

Ok the exact same solution is above but good job if you were covering the screen when you solved that.

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