Class 6 – Tuesday, February 15, 2011

From Maura:

Observations from grading homework 2:

As a result of these observations, we reviewed units again.  I gave them the Harry Potter problem (exercise 2.4) and we focused on the last two parts: find the number of Galleons in a dollar and convert the $34.99 book price to Galleons.  It took the class a good half-hour to work through.  When we came back together to discuss it, I asked a question that many of them had asked me: where do I start?  We talked through a bit of problem-solving strategy.  Number 1:  don’t panic (or panic but not for too long – remember that you still have to solve the problem); Number 2:  write what you know. This is actually quite helpful in terms of getting over the panic – it pushes the brain to relax and focus on the problem.  I told the class that I don’t know how to solve my math problems right away and that I often see connections when I start to write down what I know.  Number 3:  write down what you need to find out.  That is, keep in mind where you need to end up.  We followed that strategy with the Harry Potter problem and I hope it’s helpful for other problems.

We spent the rest of the time going through Chapter 3 material. We reviewed the “1 + ” technique then tried an example of working backwards.  This is where the “1 + ” technique makes the work much faster.  I reminded them that a percentage is always attached to a number.  We can’t just say 7%, just as we can’t ignore the units.  We have to say 7% OF something.  Once you pay attention to that you will have a better idea of how to do the arithmetic.

Then we looked at discounts.  We started with a 30% discount off of a $59.99 pair of shoes.  Most people calculated the discount first, then subtracted that to get the sale price.  But there’s a trick with 1 here too, except it’s a “1 – ” technique:  30% savings means you pay 70% of the price (or 100% – 30%).  So the sale price is (0.7)(59.99) = $42.

Then we looked at successive discounts.  Suppose you open a charge account and save an additional 10%.  What is your final price?  It’s tempting to add the percentages, but I think the class is cautious about that now.  We can use the “1 – ” to get the answer quickly:

(0.9)(0.7)59.99 = (0.63)(59.99) = $37.79

We pay 63% of the original price which means we save 37% of the original price.   Someone suggested that we then calculate the tax. We didn’t have time to do that but it’s a good question.  Maybe we’ll start there on Thursday.

From Ethan:

Returned homework 2, with grades of check, + and -. Most of the class was checked – moving along, making progress. There were three or four papers that were just a pleasure to read (of course I can’t tell whether they learned what I was teaching, or would have done as well without me …). Three or four are in potential trouble, either because they’re not paying attention or they don’t (yet) know how to pay the kind of attention needed to do these problems. In general, students who tried to follow my instructions to write more words did well even when they’d made mistakes, because I could follow what they were thinking and help them when they got off track.  (I told the class pretty much what I’ve just written here.)

I asked for questions about the homework. The first was about the meaning of absolute vs relative change. I realized that I’d never actually said the words in class, so we looked at the Red Sox ticket prices (Figure 3.1 in Section 3.2). I intended that as a lead in to inflation at the end of the class, but didn’t have time for that.

The hardest homework problem was figuring out the effect of a 10% improvement in fuel economy on gas consumption. I used that problem as a chance to discuss the strategy of inventing some numbers that seemed to be missing and necessary. In this case, what happens if fuel economy is 20 miles/gallon and you drive 100 miles. That takes 5 gallons. If fuel economy improves by 10%, to 22 miles/gallon, then you need 100/22 = 4.545… gallons to drive that distance, for a relative change of 4.545…/5 = 0.0909… = 90.9%, which is a savings of only about 9%, not 10%. The point is even clearer if you have a 100% increase in miles/gallon, to 40 miles/gallon. That doesn’t mean a 100% decrease in gas consumption, to 0, just a 50% decrease.

It was then easy to see that when measured in gallons per mile, a percentage decrease in the fuel economy corresponded exactly to the percentage decrease in the amount of gas used.

I tried to reinforce the method – pick some good numbers when none are given.

One student asked why I couldn’t just give them templates (my word) for the kinds of problems we’d encounter. So I said (again) that there were no templates for QR in real life, so that if we were to do anything in this class that would serve them after the class was over we had to think about how to think. Templates would continue the fraud in all too many math classes, where the exam problems are homework problems “with different numbers”. Then students can pretend they learned something and teachers can pretend they taught something.

The next problem they wanted help with was figuring out the change in the cost of wine when bottles went from 0.2 gallons to 0.75 liters. I had them work that one out in groups. (Group work always works well and I don’t do enough of it.) We came to the collective conclusion that the cost went up by just about 1%.

I just had time at the end for successive discounts: 25% then 15% isn’t 40%. The “pick a good starting point” strategy worked – they all realized that the problem was easiest if you imagined a $100 list price. We actually saw at the end that you got the answer from

(1-0.25)(1-0.25)=0.75*0.85=0.6375=(1-0.3625)

which means a 36.25% discount.

(At some point during the class I digressed to discuss “=” and the common misuse in solving “add 2 and 3, then multiply by 5″ by writing 2+3=8*5=40. The answer is indeed 40, but along the way you’ve said 2+3=8*5, which is just plain wrong.)

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