Learning math with a fifth grader

My younger son takes some ELP classes at school. (I think it stands for extended learning program but I’m not sure.) He’s done different subjects, including reading, writing, and science, but the one he has done most is math. He regularly brings home logic puzzles and brainteasers of various sorts. When he gets stuck, I help out, and sometimes the problems take some serious thinking even for me.

Recently he has been bringing home word problems, which come with an explanation of strategies on how to solve the problems more easily. I know a lot of people dread word problems, and while I’ve never had much difficulty doing them, I do find it hard to explain how to go about figuring them out.

When he brought home the “Think 1” set of problems, we went through them together, and I was impressed how easy the Think 1 technique made problems that might otherwise have seemed quite daunting. They were all along the lines of “Bill can paint a house in 1 hour and Bob can paint a house in 2 hours. If they work together, how long will it take them to paint 3 houses?”

I don’t know how I would have gone about doing it before, but the Think 1 strategy says to start by figuring out how much Bill and Bob will get done in 1 hour. It’s not hard to figure out that Bill will paint one house and Bob will paint half a house, so together they’ll paint one and a half houses in one hour. And from there it’s easy to see that in two hours they’ll finish the three houses, since two times one and a half is three.

Even if the numbers get a bit more complicated, the same method works. On one of the more advanced problems, we ended up with a fraction of an hour that wouldn’t even work out to an even number of minutes, but it did work out to an even number of minutes and seconds. I don’t have to try to set up the problem using algebra now, just use the Think 1 method.

Tonight I learned the 2-10 method with him. The idea is that a lot of times the word problems would be easy if they had nice numbers like 2 and 10, instead of fractions like 5/7 or big numbers like 12,000. So you restate the problem using 2 and 10 in place of those numbers, and figure out what you need to do – add, subtract, multiply, and/or divide. Then once you know what to do with the numbers 2 and 10, you substitute the original numbers back in and do the same arithmetic operations.

It does still require knowing how to deal with fractions and decimals, which I know I hadn’t learned as a fifth grader – and my son is a bit shaky on. So I helped him with that part. And there was one problem where I didn’t find the 2-10 method nearly as helpful as noticing that $1.28 and $1.92 were both multiples of $0.64. (Otherwise he would have had to divide 1.28 by 1.92, which he could have done, but why go to that trouble if you don’t have to?)

Anyway, if you or your child could use some help of that nature with word problems, you might want to check out Becoming a Problem Solving Genius by Edward Zaccaro. As the reader comments at amazon.com point out, it works well for a variety of ages and skill levels, because the explanations are simple, using cartoon illustrations, and there are various levels of problem difficulty.

I wonder what I’ll learn next from Zaccaro.


7 Responses to Learning math with a fifth grader

  1. Karen O says:

    Is this book for younger grades or would it also be appropriate for high school?

  2. Margaret says:

    I’m sure I learned about fractions and decimals before 5th grade, but I guess they had changed it by the time you reached that level 6 years later. And if you ever need your house painted, better hire Bob. He is very fast at house painting, but Bill is so fast that he probably spills a lot of paint.

  3. Pauline says:

    Karen O,
    My impression is that the primary audience is middle school. Younger students can use it with some help. I think it would work fine for high school students also. They simply would grasp the initial concepts more quickly and go right on to the problems – some of which are plenty challenging (the last two sets in each section are “Einstein” level and “Super Einstein”) even for me. I’ve seen high school textbooks (most foreign language books since that’s what I taught) that use cartoons to teach, and I don’t think high schoolers are put off by those.

    I oversimplified the problem to make it easier to explain it in two or three sentences instead of two pages including illustrations. I think in the original it was 2 hours and 3 hours, and maybe it was a car instead of a house.
    But we’ll say they were doghouses – which means Bill is probably a bit slow and Bob should be in the doghouse.
    I knew something of fractions before fifth grade but I’m pretty sure I didn’t know how to divide by a fraction, which is conceptually a bit harder. I remember in seventh grade pre-algebra (independent study for advanced students), having trouble with problems involving fractions and decimals, and Mrs. Rouse had to give me some remedial materials to catch up. Apparently that would have been covered in sixth grade, which I had skipped.

  4. Margaret says:

    As a math major, I would be interested to see one or two of those Einstein level problems.

    • Pauline says:

      It’s the Super Einstein level problems that are the hardest. This week’s packet doesn’t have a Super Einstein level, and the whole packet was easier than the last two. Its theme is Sometimes You Must Subtract, where you get the answer by figuring out a large amount and then subtracting a smaller amount to find the difference – such as subtracting the area of a smaller rectangle within a larger one to find the area of the border.

      Some of the problems deal with probability, and the ones where you have to figure out the probability that something doesn’t happen aren’t too hard – just subtract the probability that it does happen from 1. But then under the Einstein level it has a problem where there are two probabilities to deal with, and my tired brain refuses to tell me what I’m supposed to subtract from what in this case.

      It is known that 45% of men snore and 25% of women snore. If you put a man and a woman together, what is the probability that someone will be snoring?

  5. Karen O says:

    100% 🙂

    I don’t think I’ve ever known a couple where one didn’t snore.

  6. Margaret says:

    I like Karen’s reply. Seriously, though, I believe what you have is a 55% chance that the man will not snore, and a 75% chance that the woman will not snore. Therefore a 55×75% (41.25%) chance that neither one will snore. And subtract that from 1 for the probability that one will be snoring.

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