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Rate Word Problems  Problem 2
Alissa Fong
Alissa Fong
MA, Stanford University
Teaching in the San Francisco Bay Area
Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts
Now this kind of problem is sure to show up if not in your algebra class, then you when you'll do in your algebra 2 or maybe even your precalculus class. And what this is we have a boat, sometimes it’s a plane, who goes one speed in the river against the current and another speed when he is travelling with the current.
So here we go. A boat travelled 288 miles each way downstream and back. The trip down stream with the water current took 12 hours the trip upstream against the water current took 36 hours, how fast will the boat be travelling in still water? How fast is the current?
Okay so and if you guys have been in a boat travelling up or downstream know what this is about. If you are going against the water you go a lot more slowly. What we’re going to be thinking about in this problem is the equation distance is equal rate times time and we want to think about what is the scene in both of my situations does he go the same distance, does he sit in the boat with the same time or does he travel the same speed.
Well we know from the first sentence he goes the same distance. He goes 288 upstream and downstream. What that means is I’m going to have two different problems where rate times time is equal to rate times time. This means like that will be his rate times time downstream, and this will be his rate times time upstream and I got rid of the D because the D or distance was the same in both directions.
Okay let's talk about downstream how fast or what is his rate when he travelled downstream let's see the trip down stream with the current took 12 hours. I'm going to write that as his rate plus the current times 12. That's how fast he is going plus the current, it’s helping him along it’s making him go faster. Then the other thing we have is that the trip upstream against the current so now I’m going to call that R take away C, it’s against the current, took him 36 hours. Rate times time, only now the rate became kind of a package deal. It’s this whole quantity that involves not only how fast the boat is going, but how fast his boat is when you factor in the whether and not the current is helping him or hurting him.
Once I have this I can go through and solve for either of the variables that I want to, let's see 12 times R when I distribute kind of backwards I’ll have 12R plus 12C equals 36R take away 36C. Let's go through and get one of the variables isolated so let's see. If I were to subtract 12 from both sides then it would look like this 36 take away 12 is 24 Rs and then if I were to add 36C's to both sides then I would have 48C equals 24R.
And the last thing I’m going to do is right this as some kind of ratio involving R and C. I can say that r is equal to divide by 24, 48 divided by 24 is 2. R is equal to 2C. What that tells me is however fast his boat is going is twice a fact as how fast the current is going. Let me say that one more time however fast R is going, however fast his boat goes is equal to two times the current speed. What I’m going to do is go back and find an equation I can substitute this into in order to solve for my variable.
I know that for example when he went downstream it was equal to 288, so 288 is equal to R plus C times 12. I'm going to substitute in what I found about R. 288 is equal to, instead of R I’ll use 2C. 2C plus C is 3C times 12, go through and do that multiplying 288 is equal to 36C. Divide both sides by 36 and you get that the current speed is 8 miles per hour. Halfway there but the rest is really, actually more than half way.
I only know how fast the current is going and the current is going 8 miles per hour and I know his boat, his R is equal to 2 times the current so I can say his rate in still water if there was no current attached to it would be 2 times 8 or 16 miles per hour.
So again guys what I did in the very beginning is the most difficult step for most students. You need to think about what's the same in both of your equations in our situation they told us he went 288 miles each way 288 downstream and 288 upstream. That's how I set up this original equation that helped me find my answer.
So good luck with these you guys you can do it just try to keep in mind what stays the same and what's different in each of the two different equations.
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Alissa Fong
M.A. in Secondary Mathematics, Stanford University
B.S., Stanford University
Alissa has a quirky sense of humor and a relatable personality that make it easy for students to pay attention and understand the material. She has all the math tips and tricks students are looking for.
Your tutorials are good and you have a personality as well. I hope you have more advanced college level stuff, because I like the way you teach.”
Thanks alot for such great lectures... I never found learning this easier ever before... keep up the great work.... :)”
You seem so kind, it's awesome. Easier to learn from people who seem to be rooting for ya!' thanks”
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