| <<O>> Difference Topic RussianMathPart3 (r1.12 - 07 Jul 2005 - CatherineJohnson) |
RussianMathPart3Posted on Jun 14, 2005 @ 20:08 by CatherineJohnson | ||||||||
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| <<O>> Difference Topic RussianMathPart3 (r1.11 - 16 Jun 2005 - CatherineJohnson) |
RussianMathPart3Posted on Jun 14, 2005 @ 20:08 by CatherineJohnson | ||||||||
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| -- BarryGarelick - 16 Jun 2005 | ||||||||
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Barry--thank you. That's exactly right. I still have trouble expressing this aspect of math, and it's what constructivists seem never, ever, to address. I have in mind some future posts on this subject.... -- CatherineJohnson - 16 Jun 2005 | |||||||
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| <<O>> Difference Topic RussianMathPart3 (r1.10 - 16 Jun 2005 - BarryGarelick) |
RussianMathPart3Posted on Jun 14, 2005 @ 20:08 by CatherineJohnson | ||||||||
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| -- CatherineJohnson - 15 Jun 2005 | ||||||||
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With respect to Steve H's comment, it would appear that a discussion of proportion and ratios would provide the proper background to solve the problem. The comments with solutions used such approach. Constructivists would look at this page and delight in how many "different ways" people solved the problem. The constructivist's conversation would end there, without pointing out that all the different ways have the same thing in common: ratio. BG -- BarryGarelick - 16 Jun 2005 | |||||||
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| <<O>> Difference Topic RussianMathPart3 (r1.9 - 15 Jun 2005 - CatherineJohnson) |
RussianMathPart3Posted on Jun 14, 2005 @ 20:08 by CatherineJohnson | ||||||||
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| -- AnneDwyer - 15 Jun 2005 | ||||||||
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Whoa. OK, I'm gonna dive in here SOON. But first, dinner. -- CatherineJohnson - 15 Jun 2005 | |||||||
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| <<O>> Difference Topic RussianMathPart3 (r1.8 - 15 Jun 2005 - AnneDwyer) |
RussianMathPart3Posted on Jun 14, 2005 @ 20:08 by CatherineJohnson | ||||||||
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| -- SteveH - 15 Jun 2005 | ||||||||
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I looked at this problem as a ratio. Every time the chainring turns, the freewheel turns 2.2 times. Then it is obvious that you have to multiply by 5 so that the number ends in zero. The other option is to say that every time that chainring turns, the freewheel turns 11/5 times. Multiply by 5 and you get 5 turns for the chainring and 11 turns for the freewheel. -- AnneDwyer - 15 Jun 2005 | |||||||
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| <<O>> Difference Topic RussianMathPart3 (r1.7 - 15 Jun 2005 - SteveH) |
RussianMathPart3Posted on Jun 14, 2005 @ 20:08 by CatherineJohnson | ||||||||
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| -- JdFisher - 15 Jun 2005 | ||||||||
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An important question is whether the problem was posed with or without any preliminary discussion about how this type of problem can be solved. It would be nice if students could figure this out without any talk of LCM (or any other technique), as a constructivist would want, but not as the regular teaching methodology. Another question is whether the teacher reviews the proper or best methods afterwards and assigns the students (for individual work) other questions that fit this class. As I have said in the past, what is the use of constructivism if one child in the group "discovers" the solution and teaches it (perhaps poorly) to the other kids? -- SteveH - 15 Jun 2005 | |||||||
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| <<O>> Difference Topic RussianMathPart3 (r1.6 - 15 Jun 2005 - JdFisher) |
RussianMathPart3Posted on Jun 14, 2005 @ 20:08 by CatherineJohnson | ||||||||
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| -- CarolynJohnston - 15 Jun 2005 | ||||||||
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Seems like a relatively straightforward 'least common multiple' problem. Multiply the two numbers (44 and 20) and divide by their greatest common factor (4)--a great formula for LCM, by the way, that they don't give in modern textbooks. Divide by 44 to find the number of times the chainwheel must turn to return to its original position (5). Then divide by 20 to find the number of times the flywheel must turn to return to its original position (22). Many of these types of mechanisms are designed so that the numbers of 'teeth' in the contact wheels are relatively prime--i.e., have a GCF of 1--which reduces wear and tear. Also a handy mathematical tool--given the right data inputs--for determining when certain planets are aligned. Something like this was probably used to determine when the best time would be to launch a rover to Mars! -- JdFisher - 15 Jun 2005 | |||||||
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| <<O>> Difference Topic RussianMathPart3 (r1.5 - 15 Jun 2005 - CarolynJohnston) |
RussianMathPart3Posted on Jun 14, 2005 @ 20:08 by CatherineJohnson | ||||||||
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It doesn't say the teeth are the same size, but assume they are, because otherwise I think it's undoable. | |||||||
| Both gears are in their original positions whenever the number of teeth that have 'gone by' is a multiple of the number of teeth on the wheel. So the least number of teeth that can 'go by', leaving both gears in their original position, is their least common multiple. That number is 220. The chainring has 44 teeth, so as Carolyn says, the chainring rotates 5 times in that time. Greetings from hot and muggy DC. | ||||||||
| <<O>> Difference Topic RussianMathPart3 (r1.4 - 15 Jun 2005 - CarolynJohnston) |
RussianMathPart3Posted on Jun 14, 2005 @ 20:08 by CatherineJohnson | ||||||||
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| -- CarolynMorgan - 15 Jun 2005 | ||||||||
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It doesn't say the teeth are the same size, but assume they are, because otherwise I think it's undoable. Both gears are in their original positions whenever the number of teeth that have 'gone by' is a multiple of the number of teeth on the wheel. So the least number of teeth that can 'go by', leaving both gears in their original position, is their least common multiple. That number is 220. The chainring has 44 teeth, so as Carolyn says, the chainring rotates 5 times in that time. Greetings from hot and muggy DC. -- CarolynJohnston - 15 Jun 2005 | |||||||
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| <<O>> Difference Topic RussianMathPart3 (r1.3 - 15 Jun 2005 - CarolynMorgan) |
RussianMathPart3Posted on Jun 14, 2005 @ 20:08 by CatherineJohnson | ||||||||
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| -- CarolynMorgan - 15 Jun 2005 | ||||||||
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Oooops!! Make that "give" not "tive" !! -- CarolynMorgan - 15 Jun 2005 | |||||||
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| <<O>> Difference Topic RussianMathPart3 (r1.2 - 15 Jun 2005 - CarolynMorgan) |
RussianMathPart3Posted on Jun 14, 2005 @ 20:08 by CatherineJohnson | ||||||||
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Look here for syntax help.
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The freewheel returns to its original position every 20 teeth, correct? So you just need a multiple of 20, which will end in a zero. There are only 2 numbers that will tive you a product that ends in zero when you multiply times 44 (a 10 and a 5) Five x 44 equals 220, so the chainwheel must turn 5 times. Maybe the constructivist would like that since I just reasoned "my own method" through this problem!!!! :) -- CarolynMorgan - 15 Jun 2005 | |||||||
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| <<O>> Difference Topic RussianMathPart3 (r1.1 - 15 Jun 2005 - CatherineJohnson) |
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RussianMathPart3Posted on Jun 14, 2005 @ 20:08 by CatherineJohnsonI'm stumped. The chainring (attached to the pedals) on a one-speed bicycle has 44 teeth while the freewheel (on the back wheel) has 20 teeth. Determine the least number of turns the chainring must make in order for both the chainring and feewheel to return to their original positions. How many turns does the freewheel make during this time?This is a problem from Mathematics 6: an award winning textbook from Russia by Enn Nurk and Aksel Telgmaa, a 6th grade book. I've prime factored each number, since that seems to be what's called for. But now I'm stuck. Our Favorite Supplements RussianMath RussianMathPart2 RussianMathPart3 WhyILoveCarolyn ItTakesChops Mike McKeown comment IndusAcademy <!--
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Revision r1.1 - 15 Jun 2005 - 00:08 - CatherineJohnson Revision r1.12 - 07 Jul 2005 - 21:54 - CatherineJohnson |