Finding the Cost of Pets in a Pet ShopDate: 2/26/96 at 13:27:46 From: Larry Braun Subject: Pet store word problem I have tried for several days to help my son with this problem. Now I am seeking your help. A Cub Scout troop went to a very large pet store. The troop leader wanted to find out how much the pets were, because some of the scouts said they wanted one. He noticed some signs at various places in the store. The sign above the fish tank said, "buy sixteen fish and eight cats for the price of seven dogs." The sign above the cat cage said, "Buy eleven cats and seven dogs for the price of nine fish." The sign above the dog cage said, "Buy three dogs and nine fish for the price of eight cats." The troop leader ended up helping one of his scouts buy a fish and a dog. The fish cost $42 less than the dog. At this time the Scout Master discovered that exactly one of the three signs was incorrect. Which sign was wrong, and how much did the cat cost? Date: 2/26/96 at 14:22:1 From: Doctor Steve Subject: Re: Pet store word problem I suppose you did something like what I did at first, set up a series of equations to represent the facts symbolically. In general, my approach to such problems is to see if I can combine the different pieces of information in a way that makes equations with fewer unknowns. We start out with three unknowns here: the cost of a dog, the cost of a cat, and the cost of a fish. >The sign above the fish tank said, "buy sixteen fish and eight cats >for the price of seven dogs." 16f + 8c = 7d >"The sign above the cat cage said, "Buy eleven cats and seven dogs >for the price of nine fish." 11c + 7d = 9f >The sign above the dog cage said," Buy three dogs >and nine fish for the price of eight cats." 3d + 9f = 8c >The troop leader ended up >helping one of his scouts buy a fish and a dog. >The fish cost $42 less than the dog. f = d-$42 Right away I'm suspicious of the second sign (11c + 7d = 9f). According to the first one, seven dogs equals sixteen fish and eight cats so the idea of 9 fish, which are much less expensive than the dogs, equaling eleven cats and seven dogs makes me wonder what's going on. To solve further I would combine the information from the equations that I have more confidence in. For instance, notice that the third equation tells us what 8c is in terms of fish and dogs and 8c is also used in the first equation. What happens if I use 3d + 9f in the first equation instead of 8c, since they are the same? This way I won't have any "cats" in the equation; one less unknown to think about for now. 16f + (3d + 9f) = 7d This tells me 25f=4d or one dog costs 25/4ths (six and a quarter times) what a fish costs (d = 25/4 f). I can then combine this information with the other bit of information I haven't used yet, namely f = d - $42 and find out what a fish costs. Do you see how to do that, similar to what I did with 8c above? I won't have any dogs in my new equation, only fish and that'll tell me how much the fish cost. You might want to see if you can solve the rest before reading on. f = (25/4 f) - $42 (I'll let you work out the calculations here. Please ask if you want help.) $8 = f So, if we know how much a fish is, we can figure out what a dog is, since it costs $42 more. Then we can figure out what a cat costs. And finally, we should be able to see if our answers work for every sign but the second. If they don't then we'll have to choose one of the other signs as the culprit. Please let us know what you come up with or where you need further explanation. -Doctor Steve, The Math Forum Date: 2/26/96 at 14:37:39 From: Doctor Byron Subject: Re: answer to word problem Hi Larry, I think the best thing to do in the case of this pet store question is to start by putting the equation in algebraic form. First, we have the assertions made by the three signs, which may or may not be true: 16f + 8c = 7d 11c + 7d = 9f 3d + 9f = 8c Where c, d, and f are the prices of cats, dogs, and fish, respectively. Rearranging these into a slightly more standard form we have: 8c - 7d + 16f = 0 11c + 7d - 9f = 0 -8c + 3d + 9f = 0 We also know that the following equation _must_ be true: f = d - 42 Now, since we know f in terms of d, we can substitute this expression for d into all the places where f appears in the above system of equations. We then get the following set of three equations in only two variables: 8c - 7d + 16(d - 42) = 0 11c + 7d - 9(d - 42) = 0 -8c + 3d + 9(d - 42) = 0 8c + 9d = 672 11c - 2d = -378 -8c + 12d = 378 Now, if all of these equations were really true, we could pick any two of them and get a reasonable answer to the problem. For example, adding the first and third equations eliminates the variable c, and allows us to find that d = $50.00 and c = $27.75. These two seem to give a good answer to the problem, as we might expect. At this point, we should be suspicious that the second equation may not be true. You can confirm this by solving the remaining two sets of equations. You will see that either 1 and 2 or 2 and 3 together yield a negative solution for c. The sign above the cat cage is therefore false, and the pets have the following prices: Dog - $50.00 Cat - $27.75 Fish - $8.00 (This is all assuming, of course, that the shopkeeper doesn't intend to pay people to take his pets away.) -Doctor Byron, The Math Forum |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/