by Shawn » Sat Apr 21, 2012 10:02 pm
On being "good", I used to compete. Good coaching, an enjoyment of the problems, and just a lot of practice is what I would say made me "good." I believe the statement applying equations to problems is somewhat flawed. If this was the case, then you would just memorize a list of formulas and memorize a matching problem set. I actually believe this is why I wasn't "very good", is because for quite a few type of problems this was my mindset. Instead of reasoning through the problems and understanding the solutions, I just memorized the solution and problem. This results in a small change not being easily added, because instead of understanding the problem you just know divide this multiply that done. I would say the contest is designed to make contestants use reasoning skills, to apply knowledge to novel problems. To be "good", you obviously need to know (not memorize) all the equations, but to excel you really need to understand why the solutions works.
Virtually all the problems you post I've seen and done before (there are only so many tests that have been written). Now I can't say that memorization can't be beneficial, because the tests are standardized and repeat. This is how I am able to remember how I solved the problem before, (or if I forgot I just apply reasoning and what I remember to it or I know what to Google to recall). I encourage you to try to think through the problem and come up with your own way of doing it. I would actually be happier and more willing to offer longer explanations if you can show me exactly what you don't get or explain your attempt at a solution. This way I can point out why it didn't work. I understand that at this point in time you haven't done all the "knowing" of the basic information you need to solve the problems, which is really a prerequisite to a lot of these problems. It is hard to do a free-fall problem without knowing acceleration due to gravity or some earth related problem without knowing the radius of the earth. The contest manual should be able to help you with these things.
On to problems:
7)
Dimensionless=without dimension (no unit). All this means is that he doesn't want ft/in or in/ft he wants in/in or ft/ft.
Ratio of diameters in equal units is all you need to do: (I choose to take everything to inches)
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46) Looking at your solution, you scaled mm{size}/$ squared times a bunch of stuff that cancels out (17 and 2.54) and then scaled based off of size again. This just doesn't make a lot of sense, unless you can explain the logic, I don't really understand.
Since we can't have partial pearls, you can't necessarily scale the number of pearls, which would just be multiply scaled cost by 7/10.
Find number of pearls on each:
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[unparseable or potentially dangerous latex formula] (43 no partials)
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[unparseable or potentially dangerous latex formula](61 no partials)
So we have our first scaling factor number of peals reduction of [unparseable or potentially dangerous latex formula]
Now scale off of size:
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This would represent the cost if the same number of pearls were used, but that is not the case earlier we said 43/61 of what was used before was used, so scale this down by that.
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Dollar problem so,
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I am a penny off, I would get it right, I probably rounded somewhere.
50)
First deduct volume of hemisphere.
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Next find the height of cone:
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Next soh-cah-toa to find angle
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Multiply by 2 (the triangle only covered 1/2) and convert to radians
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