Heads and LegsDate: 07/25/2001 at 13:51:09 From: Joseph Brazell Subject: Algebra Joe counts 48 heads and 134 legs among the chickens and dogs on his farm. How many dogs and how many chickens does he have? Date: 07/25/2001 at 16:29:37 From: Doctor Greenie Subject: Re: Algebra Hi, Joseph - This is a classic problem for beginning algebra students, but if you are in fact 11 years old (as you said you were), then you probably are not yet using algebra to solve problems. So here is another approach I like to use with this type of problem. (In many cases, algebraic techniques for solving a problem greatly simplify the amount of work required to get an answer. In this type of problem, I find the algebraic approach actually takes more effort, so this alternate method of solution is in fact the one I usually use.) Each chicken and each dog has one head, so the number of chickens and dogs together is 48. Each chicken has two legs and each dog has four; together the 48 animals have 134 legs. Let's first check to see if the problem even makes sense. The minimum number of legs possible would be if all 48 animals were chickens; that would be 48*2 = 96 legs. The maximum number of legs possible would be if all 48 animals were dogs; that would be 48*4 = 192 legs. Since 134 is between 96 and 192, the problem should have a solution. Now let's solve the problem. Start by guessing that all 48 animals are chickens. We have already determined that this would give us 96 legs. We want 134 legs, so we are short by 134-96 = 38 legs. Now suppose we replace one chicken with a dog. The replaced chicken had two legs, and the dog has four - so we now have two legs more than before. And each time we replace another chicken with another dog, we get two more legs. So I started with a "guess" of 48 chickens and no dogs, which left me 38 legs short of what I wanted. And I found that each time I replaced a chicken with a dog I gained two legs. If my first guess left me 38 legs short, and if I gain two legs each time I replace a chicken with a dog, then the number of chickens I need to replace with dogs to get the other 38 legs I need is 38/2 = 19. So I should replace 19 chickens with dogs, leaving me with 19 dogs and 48-19 = 29 chickens. Whatever method you use to get the answer to the problem, you should check your answer: 19 dogs with 4 legs each = 19*4 = 76 legs + 29 chickens with 2 legs each = 29*2 = 58 legs ----------------------------------------------- 48 animals total 134 legs total In the remainder of my response, I will show one algebraic approach to the solution of this problem (there are many others...), just in case you know enough algebra to understand the method. It is interesting to compare this algebraic approach to the method I demonstrated above. Let c be the number of chickens and d the number of dogs. Then... (1) the number of chickens and dogs together is 48: c + d = 48 (1) (2) the number of legs, with 2 legs for each chicken and 4 for each dog, is 134: 2c + 4d = 134 (2) So now I have two equations relating the numbers c and d: c + d = 48 (1) 2c + 4d = 134 (2) Now I want to "play" with these equations algebraically and then compare them to find the value of one of the numbers. One way I can do this is to imagine that I double the size of my farm, so that I have twice as many chickens and twice as many dogs as before; to see the resulting number of animals and heads, I can "double" equation (1): 2c + 2d = 96 (3) Then I have two equations 2c + 4d = 134 (2) 2c + 2d = 96 (3) When I "compare" these equations to see how they differ, I see that the number of chickens is the same in both, but the number of dogs is different. That means the difference in the numbers on the right must be due to the difference in the numbers of dogs in the two equations. So I "subtract" the two equations to find the "difference" between them: 2c + 4d = 134 - (2c + 2d = 96) ----------------- 2d = 38 This tells me that twice the number of dogs is 38, so the number of dogs is 19. Then from my original equation (1) I know that the number of chickens is 48-19 = 29. For a similar problem in our archives, see: 60 Eyes and 86 Feet http://mathforum.org/library/drmath/view/57957.html - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/ |
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