Guess and Check - How Many Coins?Date: 10/03/2002 at 23:22:00 From: Christine Subject: Guess and test problem solving? Jeremy and Hanna collect coins. Jeremy has seven fewer coins than Hanna. If together they have 83 coins, how many does Hanna have? My daughter is being taught to "Guess and Test" to solve this problem. Isn't there a better, less time consuming way to solve this? My daughter solved this problem by "guessing" a number of coins that Hanna might have. Say 60. Then she subtracted 7, and checked to see if the sum was 83. Her guessing went on and on... I hope there is a better way. Thanks! Date: 10/04/2002 at 01:57:07 From: Doctor Ian Subject: Re: Guess and test problem solving? Hi Christine, There is a method to the madness, actually. It's only when you've spent a lot of time trying to guess the answer to problems that you begin to appreciate the power that mathematics has to offer. But even when you know a lot of math, it turns out that guessing is often an excellent way to start approaching a problem, so long as you don't just keep guessing blindly: Guess and Check http://mathforum.org/library/drmath/view/60819.html Looking at your specific problem, one place to start is with the fact the the total number of coins must add up to 83. So there is a limited number of possibilities: 1 + 82 2 + 81 3 + 80 . . 41 + 42 Now, on the one hand, you might systematically fill in all the missing pairs until you come to one where the difference between the two numbers is 7. Or you might keep track of the differences, 1 + 82 = 83 82 - 1 = 81 2 + 81 81 - 2 = 79 3 + 80 80 - 3 = 77 After just a few of these, it becomes pretty clear that the difference decreases by 2 each time you go one step down the list. So you might just start skipping pairs, hoping to get close to the answer: 1 + 82 = 83 82 - 1 = 81 2 + 81 81 - 2 = 79 3 + 80 80 - 3 = 77 20 + 63 63 - 20 = 43 30 + 53 53 - 30 = 23 This is quicker... but you might also ask yourself: How many times would I have to drop 2, to get from 77 to 7? That's a total drop of 70, which would require 35 steps. So if you add 35 to 3, you get 38; and 38 and 45 add up to 83. So this is one way to go. Another would be to start from the fact that there must be a difference of 7 coins. Again, there are limited possibilities: 1 + 8 = 9 2 + 9 = 11 3 + 10 = 13 and so on. Now we're back in the same boat as before, and we can proceed in the same ways: (1) keep filling in pairs systematically, (2) skip pairs until we get close, and then fill them in systematically, or (3) note that each time we bump the lower number up by 1, the total is bumped by 2, and use that to compute the number of bumps we need directly. The quickest way of all is to use an equation, which looks like this. We can use '?' to represent the smaller number, and '(?+7)' to represent the larger number. Then ? + (?+7) = 83 ? + ? + 7 = 83 2*? + 7 = 83 Now, if you add 7 to something and end up with 83, then the something must be equal to 76, right? So 2*? = 76 Now, by the definition of division, this means that ? = 76/2 = 38 So the smaller number must be 38, and the larger one must be 45. This isn't to say that setting up an equation is always preferable to guess-and-check. In many cases, it's not; and in many other cases, making and checking a few guesses is the quickest way to figure out what equations you should be using. The important thing is to avoid guessing randomly or blindly. You want to keep track of your guesses, and look for patterns that you can use to keep improving your guesses. Even if there are a million possible answers to a problem, if each guess can eliminate half the remaining possibilities, you're guaranteed to find the answer in only 20 steps. Does this help? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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