Solving Problems by Making Organized ListsDate: 04/09/2008 at 18:36:16 From: Remy Subject: not enough room to explain this is a problem on my math homework. it is the hardest one on my page and i did all the others ones and my parents cant even help me. so here is the problem. as Jamey bounces on the trampoline 11 coins fall out of his pocket totaling 1.37. what are these coins? what about 14 coins and 2.10? i thought maybe starting out with quarters but that did not work so i knew it had to be a lot of different type of coins. i started doing 11 divided by 137 but then the period ended so i could not get to it until 5 minutes ago. Date: 04/10/2008 at 09:44:08 From: Doctor Ian Subject: Re: not enough room to explain Hi Remy, You need 11 coins that add up to 137 cents. To finish with 7 cents, there are two possibilities: 1) 2 of them are pennies. In that case, you need to find 9 coins that add up to 135 cents. 2) 7 of them are pennies. In that case, you need to find 4 coins that add up to 130 cents. So now you have two smaller problems that you can work with. We can do the same thing again, with quarters. 1) Find 9 coins that add up to 135 cents. 1a) If 1 is a quarter, find 8 coins that add up to 110 cents. 1b) If 2 are quarters, find 7 coins that add up to 85 cents. 1c) If 3 are quarters, find 6 coins that add up to 60 cents 1d) If 4 are quarters, find 5 coins that add up to 35 cents. 1e) If 5 are quarters, find 4 coins that add up to 10 cents. Note that, as an added bonus, since I've already considered pennies and quarters, I don't even have to think about them when looking at (1a) through (1e). I could really write them as 1) Find 9 quarters, dimes, and nickels that add up to 135 cents. 1a) If 1 is a quarter, find 8 dimes and nickels that add up to 110 cents. 1b) If 2 are quarters, find 7 dimes and nickels that add up to 85 cents. 1c) If 3 are quarters, find 6 dimes and nickels that add up to 60 cents 1d) If 4 are quarters, find 5 dimes and nickels that add up to 35 cents. 1e) If 5 are quarters, find 4 dimes and nickels that add up to 10 cents. And when I write it this way, it reminds me that I have to consider the possibility that there are NO quarters: 1) Find 9 quarters, dimes, and nickels that add up to 135 cents. 1a) If 0 are quarters, find 9 dimes and nickels that add up to 135 cents. 1b) If 1 is a quarter, find 8 dimes and nickels that add up to 110 cents. 1c) If 2 are quarters, find 7 dimes and nickels that add up to 85 cents. 1d) If 3 are quarters, find 6 dimes and nickels that add up to 60 cents 1e) If 4 are quarters, find 5 dimes and nickels that add up to 35 cents. 1f) If 5 are quarters, find 4 dimes and nickels that add up to 10 cents. And now I can see how to answer the question! Because I know how to get 6 coins that add up to 60 cents. I just use 6 dimes. Working backwards, 6 dimes give me 60 cents. 3 quarters give me 75 cents. 2 pennies give me 2 cents. --- 137 cents But if I didn't see that, I could keep going down that path: 1d) If 3 are quarters, find 6 dimes and nickels that add up to 60 cents 1c1) If 0 are dimes, find 6 nickels that add up to 60 cents. 1c2) If 1 is a dime, find 5 nickels that add up to 50 cents. 1c3) If 2 are dimes, find 4 nickels that add up to 40 cents. 1c4) If 3 are dimes, find 3 nickels that add up to 30 cents. 1c5) If 4 are dimes, find 2 nickels that add up to 20 cents. 1c5) If 5 are dimes, find 1 nickel that adds up to 10 cents. 1c5) If 6 are dimes, find 0 nickels that add up to 0 cents. Here, (1c5) works out, and the others are so obviously wrong that I can cross them off without having to work very hard. But if none of these worked out, I would know that the answer had to be down one of the other paths. Now, the important thing here isn't to answer the question. No one cares about the answer. What's interesting here is that I could systematically change each problem into a set of smaller problems, until I get one so small I can solve it in my head. It's kind of like if you've lost your watch, and you're looking for it in your house. You could start looking around randomly, or you could formulate a search plan. For example, It's got to be in the basement, or on the first floor, or on the second floor: 1) basement 2) first floor 3) second floor If it's on the second floor, it's got to be in one of the rooms there: 1) basement 2) first floor 3) second floor 3a) My room 3b) Pat's room 3c) Mom and Dad's room 3d) The bathroom If it's in my room, it's got to be in one of a few places: 1) basement 2) first floor 3) second floor 3a) My room 3a1) In or on the dresser 3a2) In the closet 3a3) On the floor, behind or under some furniture 3a4) In the bed . . 3b) Pat's room 3c) Mom and Dad's room 3d) The bathroom The nice thing about this is that, if the watch is somewhere, you WILL find it (assuming you've been careful about constructing your list); and you won't ever look in the same place more than once. A lot of math problems are basically the same way. You start with the original problem, and say: Well, the answer has to be down one of these paths. And then you WRITE DOWN THE PATHS! That's the crucial part, that people tend to forget. Because when you write down what you're doing, it helps you (1) figure out the next thing to do, and (2) avoid doing the same things over and over again. And then for each path you've started, YOU DO THE SAME THING AGAIN, i.e., you look at the paths that extend the ones you've started. And again, you write it all down. Note that sometimes a problem like this won't be to find ONE way to get the right amount of money, but to find ALL ways to get it. (In the case of searching the house, maybe you've lost two earrings, or five baseball cards, or whatever, and you need to find all of them.) Also, sometimes a problem is to show that something CAN'T be true. (In the case of searching the house, maybe a friend thinks he might have left something valuable at your house, and you have to show that this isn't the case.) In either of these cases, you have to keep going until you've examined all of the possibilities. So a list isn't just convenient, it's indispensable. Does this make sense? Try doing this with your second problem, and let me know how that goes, okay? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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