<PRE><B>Question 1159706</B>
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This problem can be worked in several different ways, including many that would not be appropriate for 5th grade.&lt;br&gt;
If I were helping a 5th grader with this problem, I would try to get her to participate in solving the problem -- rather than simply trying to explain it to her.&lt;br&gt;
Here are a couple of things that are involved in working the problem that might be appropriate for 5th grade.&lt;br&gt;
(1) If the cans have to be bought in the given groups, then we can&#39;t spend the whole $50.  The $8 prices are each double the $4 price, so adding any number of $4 prices and $8 prices will always give a total that is a multiple of $4.  $50 is not a multiple of $4; so the most we can spend is 12 times $4 = $48.&lt;br&gt;
So we will think about making purchases with a total of $48 instead of $50.&lt;br&gt;
(2) We can look at the average cost of the total purchase and the average cost of each item to get an idea of how many of each item might give us a solution.&lt;br&gt;
fraction of a dollar for each can of beans: 8/12 = 2/3
fraction of a dollar for each can of chili: 8/10 = 4/5
fraction of a dollar for each can of soup: 4/8 = 1/2&lt;br&gt;
fraction of a dollar for each can of the whole purchase: 50/75 = 2/3&lt;br&gt;
Since the 1/2 dollar per can price of the soup cans is less than the overall average of 2/3 dollar per can, and since the 4/5 dollar per can price of the chili cans is more than the overall average of 2/3 dollar per can, those fractions give us the idea that we should have relatively few cans of chili and a relatively large number of cans of soup.&lt;br&gt;
Now we need to ask ourselves about the real meaning of the problem.  Since we are given the prices of beans, chili, and soup, it might seem to imply that we need to buy some of each.&lt;br&gt;
But if that is not the case, and we only need to buy 75 or more cans, then we can get the most for our money if we spend it all on the least expensive soup.  At $4 for 8 cans, $48 dollars (= 12 times $4) will buy us 12 times 8 = 96 cans.&lt;br&gt;
And quite possibly that was the intention of the problem, as presented to a 5th grader -- reasoning that you can buy the largest number of cans by buying only the item that costs the least per can.&lt;br&gt;
To get more mental exercise with the problem, however (and of course that is the primary reason for education), you and your niece can look for solutions that include all three types of cans.&lt;br&gt;
Starting with 12 cans of beans for $8, 10 cans of chili for $8, and 8 cans of soup for $4, that is a total of 30 cans for $20.  So we have $28 left to buy 45 or more cans.&lt;br&gt;
At this point, you and your niece can simply try different combinations of the three types of cans that will give you a combination that satisfies the requirements.&lt;br&gt;
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