Question 1166744
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Did it not occur to you to ask a question about this...?<br>
It's obvious what we/you are supposed to do with the given information... but you should make that part of your post.<br>
The downstream and upstream rates are distance divided by time:
rate downstream: D = 33.25/3.5 ( = x+y)
rate upstream: U = 33.25/9.5 ( = x-y)<br>
Algebraically, the canoe speed x is found by adding the upstream rate and the downstream rate and dividing by 2:
((x+y)+(x-y))/2 = 2x/2 = x<br>
Without algebra, logical reasoning tells us the canoe speed is the average of the upstream and downstream speeds.<br>
rate of canoe: x = (D+U)/2<br>
The speed of the current is then the difference between the downstream speed and the canoe speed ((x+y)-x = y), or the difference between the canoe speed and the upstream speed (x-(x-y) = y).<br>
rate of river current: y = D-x, or y = x-U<br>
You can do the calculations.<br>
Note the rate of the current is also equal to half the difference of the upstream and downstream speeds: y = ((x+y)-(x-y))/2<br>
But it's hard to see logically why that calculation gives the speed of the current.  It makes much more sense to find the rate of the current as the difference between the canoe speed and either the upstream or downstream speed.<br>