Inconsistent System of Equations? Look Closer!Date: 11/17/2004 at 12:25:37 From: Randy Subject: Fith Grade Story Problem that has several adults stumped. When Alexis, Chelsea, and Kammi had lunch together, Alexis spent $1.60 for two small hamburgers, a drink and one order of fries. Chelsea's two orders of fries, two drinks and one small burger cost $1.40 altogether. How much does Kammi owe for a small burger, one order of fries and one drink? It seems as though there is not enough information. I tried looking at several of the variables and considered fractions or ratios or percentages. However, I struggle with the fact that the prices of each could be any non-related amount. Date: 11/17/2004 at 12:57:14 From: Doctor Douglas Subject: Re: Fith Grade Story Problem that has several adults stumped. Hi Randy. For a fifth-grader, I would do the following: put Alexis and Chelsea together--they have a total of three hamburgers, three drinks, and three fries for a total of $3.00. Since Kammi has one burger, one drink, and one order of fries, she should owe one third of this, or $1.00. At a slightly more advanced level, we would probably use algebra to solve a problem of this type. Let H, D and F be the cost of a hamburger, drink, and fries respectively. Alexis: 2H + D + F = 160 Chelsea: H + 2D + 2F = 140 Kammi: H + D + F = ? You can probably see that the shortest path to the answer to the question is essentially encapsulated in the fifth-grade explanation above. Note that the question asks for the combination H + D + F, and not the individual prices of each hamburger/drink/fries. In fact, there is not enough information given in the problem to completely determine the individual costs. For example, any set of {H,D,F} satisfying the equations labelled Alexis and Chelsea will work: H = 60, D = 30, F = 10 H = 60, D = 20, F = 20 H = 60, D = 80, F = -40 (of course this is unreasonable!) Algebra allows us to prove that the cost of the hamburger must be 60 cents: 2*Alexis - Chelsea: 4H + 2D + 2F - H - 2D - 2F = 2*160 - 140 3H = 320 - 140 = 180 H = 60 and what remains is the relationship D + F = 160 - 2H = 40. Any pricing combination such that the drink and the fries together add up to 40 cents is consistent with the problem statement. All of the solutions above are consistent with this. I hope that this helps. Please write back if you have more questions about this. - Doctor Douglas, The Math Forum http://mathforum.org/dr.math/ Date: 11/17/2004 at 13:07:28 From: Randy Subject: Thank you (Fith Grade Story Problem that has several adults stumped.) Thank you Doctor Douglas for your quick response. I have a bachelors in Business Admin but have realized how much is lost over time. I do appreciate the services provided at this site. Now maybe I can get caught up with my own reasoning skills and my 11 year old son might think I actually know something! |
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