Hi! I was doing quadratic and exponential word problems on Khan Academy, and got this. Is it possible that this type of problem will be on the SAT, and if it is, please explain it.

Question(reworded): A guy travels up a river at v mph for 2 miles. He goes the same distance back, but 5 mph slower than he did on the way up. If he spent 3 hours traveling up and back, which could be used to determine his speed on the way up the river?

a) 2v^2-13v+17=0
b)3v^2-15v-4=0
c)3v^2-11v+10=0
d)3v^2-19v+10=0

Someone can jump all over me in the comments if my memory is faulty but no, I can’t think of any real test questions that track this format. Since you’re here, though, here’s how I’d set it up:

When you have miles and miles per hour, you can divide to get hours, right? So the time it took him to go 2 miles up river is \dfrac{2\text{ miles}}{v\text{ miles per hour}}=\dfrac{2}{v}\text{ hours}. Similarly, the time it took him to return is \dfrac{2}{v-5}\text{ hours}. We know how long it took him altogether, so now we can write a one-variable equation we can use to solve for v:

\dfrac{2}{v}+\dfrac{2}{v-5}=3

With some algebra you can simplify that to an answer choice:

\dfrac{2}{v}+\dfrac{2}{v-5}=3\\\\

 \dfrac{2}{v}+\dfrac{2}{v-5}=3\\\\ \dfrac{2(v-5)}{v(v-5)}+\dfrac{2v}{v(v-5)}=3\\\\ \dfrac{2v-10}{v^2-5v}+\dfrac{2v}{v^2-5v}=3\\\\ \dfrac{4v-10}{v^2-5v}=3\\\\ 4v-10=3v^2-15v\\ 0=3v^2-19v+10

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