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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/ 
Associated Topics:
Elementary Puzzles
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Middle School Word Problems

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