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"Simple" Maths Problems - CLOSED!
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Fred W Send message Joined: 13 Jun 99 Posts: 2524 Credit: 11,954,210 RAC: 0 |
OK, Luke, was it the first solution or the one I offered after Fred W asked what he asked? Sarge, No rounding errors. The question does not state that the tank is totally enclosed (e.g. the cold water tank in my roof does not have a top). The wall of the tank needs a sheet that is pi*7*10=219.94. In order to have enough sheet to cut a circle with diameter 7 from the sheet then the sheet needs to be extended by 7 on its longest axis (least waste) so the sheet would have to be 28.994 * 10 to provide enough material for an open-top tank. i.e Area = 289.94 (to 2 d.p.) F. |
Sarge Send message Joined: 25 Aug 99 Posts: 12273 Credit: 8,569,109 RAC: 79 |
Agreed as to your approach. Still, if the tank is open at the top, yes, you have the correct result for the area of the lateral surface (approximately 219.91 cubic ... quickly erasing that and replacing it with square ... grrr ... whatever the units were), but ... . With only one circle, at the bottom, Pi*r^2 approximately 3.14 * (7/2)^2 = 3.14 * 3.5^2 = 3.14 * 12.25 = 38.465. Adding the two, we're at approximately 258.375, where as your answer was very close to mine, where I made the assumption of both the bottom and top enclosed by circles. Capitalize on this good fortune, one word can bring you round ... changes. |
Fred W Send message Joined: 13 Jun 99 Posts: 2524 Credit: 11,954,210 RAC: 0 |
Agreed as to your approach. Still, if the tank is open at the top, yes, you have the correct result for the area of the lateral surface (approximately 219.91 cubic ... quickly erasing that and replacing it with square ... grrr ... whatever the units were), but ... . I think we're at cross purposes a little. You are calculating the surface area of the tank. I am calculating the area of sheet metal that the components can be cut from. What you are saying is that the scrap left over after I have cut my one circle is almost enough area (though entirely the wrong shape) for a second circle. F. |
Sarge Send message Joined: 25 Aug 99 Posts: 12273 Credit: 8,569,109 RAC: 79 |
Ahhh, but it is metal and so perhaps it could be melted down and reshaped. :) Also, unless told otherwise, a mathematical problem assumes ideal situations, such as no material loss. Capitalize on this good fortune, one word can bring you round ... changes. |
Fred W Send message Joined: 13 Jun 99 Posts: 2524 Credit: 11,954,210 RAC: 0 |
Ahhh, but it is metal and so perhaps it could be melted down and reshaped. :) Yeah. But the question asks for the "amount of sheet metal". If I buy an amount of sheet metal I get charged for any offcuts too (and probably for the cutting). In my professional life this comes under the heading of "inadequate specification" ;) Judgement of Solomon would be "Shoot the messenger"? Guess we have to wait for Solomon - er... Luke. F. |
Sarge Send message Joined: 25 Aug 99 Posts: 12273 Credit: 8,569,109 RAC: 79 |
I still say, it's metal, melt it! Or, use Luke's lightsaber. ;) Capitalize on this good fortune, one word can bring you round ... changes. |
Sarge Send message Joined: 25 Aug 99 Posts: 12273 Credit: 8,569,109 RAC: 79 |
Solve for "Planet X." :) Capitalize on this good fortune, one word can bring you round ... changes. |
Luke Send message Joined: 31 Dec 06 Posts: 2546 Credit: 817,560 RAC: 0 |
There are two winners here... Fred wins 1 Point for the Volume Answer and.... Sarge wins 1 Point for the first correct Surface Area answer... Correct Answer: Q25 Volume: 384.9 Cubic Units Working: 2Àrh + 2Àr^2 = 2À(35 + 49/4) = 189/2 À ≈ 296.9 Q25 Surface: 296.9 Square Units Working: Àr^2h = À(49/4)(10) = 245/2 À ≈ 384.9 And The Updated Standings: 1. Mr. Kevvy - 7 Points 2. William Rothamel - 5 1/2 Points 3. WinterKnight - 5 Points 4. Sarge - 4 Points 5. Dominique - 3 1/2 Points 6. Fred W - 1 1/2 Points 7. John McLeod VII - 1 Point 8. Scary Capitalist - 1/2 Point 9. TBD...... Sarge and Fred quickly make a big impact to the Standings Table... doing we have a few new leaders blossoming?.... Also, My lightsaber is all powerful.... It has advanced mathematics technology.. And its Blue.... Q26, Q27 coming soon.. May the Force be with you.... Luke. - Luke. |
Luke Send message Joined: 31 Dec 06 Posts: 2546 Credit: 817,560 RAC: 0 |
Here are 3 more questions for everyone....... Q26: There is a free gift in my breakfast cereal. The manufacturers say that the gift comes in four different colors, and encourage one to collect all four (thus eating lots of cereal). Assuming there is an equal chance of getting any one of the colors, what is the expected number of boxes I must consume to get all four? Q27:Given a $100,000 mortgage, a payment period of 30 years, monthly payments (the first due at the end of the first month), and an interest rate of 7.5% compounded monthly, what will be the monthly payments? Q28: Thirteen pirates put their treasure in a safe. They decide that the safe should be able to be opened if any majority of pirates agree but not be able to be opened if any minority agree. The pirates don't trust each other so they consult a locksmith. The locksmith puts a specific number of locks on the safe such that every lock must be opened to open the safe. Then he distributes keys to the pirates such that every pirate has some but not all of the keys. Any given lock can have multiple keys but any given key can only open one lock. What is the least number of locks required? Best Regards, Luke. - Luke. |
William Rothamel Send message Joined: 25 Oct 06 Posts: 3756 Credit: 1,999,735 RAC: 4 |
The mortgage problem solution is $843.86 per month with the rate divided by 12 and amortization set up for 360 periods. |
William Rothamel Send message Joined: 25 Oct 06 Posts: 3756 Credit: 1,999,735 RAC: 4 |
Thoughts on the pirate problem: 13 pirates so a majority is seven the locksmith must make a lock for each distinct group of 6. So you can select 6 from 13 the following number of ways: (13x12x11x10x9x8) / (1x2x3x4x5x6) = 1716 locks !! I must be nuts or a little rusty on combinatorics but it seems that this is the only way that any 7 but not any 6 could open the chest. Bill |
Sarge Send message Joined: 25 Aug 99 Posts: 12273 Credit: 8,569,109 RAC: 79 |
The mortgage problem solution is $843.86 per month with the rate divided by 12 and amortization set up for 360 periods. r = 0.075 m = 12 t = 30 i = r / m = -.00625 n = m * t = 360 Periodic payments = 100000 / ((1 - (1 + i)^(-1 * n)) / i) = 699.21 Capitalize on this good fortune, one word can bring you round ... changes. |
William Rothamel Send message Joined: 25 Oct 06 Posts: 3756 Credit: 1,999,735 RAC: 4 |
Another try--my math is a bit sloppy: Thats why I like to use a program or EXCEL. sum of (1+.075/12)^N for n=1 to 360 times the 100,000 all divided by 360 to make the same monthly payments and the answer is 699.21 Sarge is right my answer was for a 15 year loan and was wrong anyway. Regards, Bill |
Sarge Send message Joined: 25 Aug 99 Posts: 12273 Credit: 8,569,109 RAC: 79 |
Another try--my math is a bit sloppy: Thats why I like to use a program or EXCEL. Bill, I just taught the material recently, so I figured I better take a shot at it or else I'd have a heck of time scoring the final exam! I think you had the right approach. On the pirates one, no clue yet. On the number of cereal boxes, I am trying to turn myself from what originally came to mind because I don't think I was applying the correct concept at first. Capitalize on this good fortune, one word can bring you round ... changes. |
William Rothamel Send message Joined: 25 Oct 06 Posts: 3756 Credit: 1,999,735 RAC: 4 |
Sarge--I need to give this more rigor instead of the intuitive stuff I like which is usually wrong. On the problem of the 4 colors. I think about the following. The chance (probability) of having an acceptable draw at trial 1 is 1.00 And that having been done the chance of having a successful draw on trial 2 is 3/4 or .75 and that having been done the chance of having a successful draw on the third trial is .5 since only half of the color choices would be successful. On the fourth trial the probability of success is .25 . Therefore .75 x .5 x .25 =.0975 Taking the inverse of this is 1/.0975 =10.25 so I say that the expected number of trials would be 11. Of course I seem to remember something about conditional probability that is missing here in my analysis. I don't even know if Bayes theory needs to be applied here --I think not Anyhow: My answer is 11 trials |
Sarge Send message Joined: 25 Aug 99 Posts: 12273 Credit: 8,569,109 RAC: 79 |
I am going to have to think about that one more, Bill. I just read your approach quickly. On the second draw, would you not have to consider both the situation of picking the same thing you did the first time versus picking one of the 3 remaining colors? Capitalize on this good fortune, one word can bring you round ... changes. |
William Rothamel Send message Joined: 25 Oct 06 Posts: 3756 Credit: 1,999,735 RAC: 4 |
Another try--my math is a bit sloppy: Thats why I like to use a program or EXCEL. Guess what I am going to be teaching statistics here in a grad school starting this summer--How do you spell Chebyschev ??? The only good thing is that as I relearn each concept i can figure out what the difficult ideas are and steer the students to grasping these particular ideas. |
Sarge Send message Joined: 25 Aug 99 Posts: 12273 Credit: 8,569,109 RAC: 79 |
I have had many experiences where I get a refresher, or am offered up challenges, that serve me well a few months later when teaching. I've seen two other spellings of Chebyshev, I believe. http://en.wikipedia.org/wiki/Pafnuty_Chebyshev His name is transliterated variously as Chebychev, Chebyshov, Tchebycheff or Tschebyscheff (French and German transcriptions) among many others. One of the issues is the use of a different alphabet. Also, there are slight differences in what syllables a child learns to pronounce, around the world. But, I do not know if this is the case for Russian vs. English speakers. Capitalize on this good fortune, one word can bring you round ... changes. |
Sarge Send message Joined: 25 Aug 99 Posts: 12273 Credit: 8,569,109 RAC: 79 |
P.S. - don't completely throw out the intuitive stuff. :) Capitalize on this good fortune, one word can bring you round ... changes. |
Luke Send message Joined: 31 Dec 06 Posts: 2546 Credit: 817,560 RAC: 0 |
Congratulations to Bill & Sarge. Bill was right on Question 28: Answer is 1716 locks. 1 Point Sarge was right on Question 27: Answer is $699.21. 1 Point Well Done both of you!!! And The Updated Standings: 1. Mr. Kevvy - 7 Points 2. William Rothamel - 6 1/2 Points 3. WinterKnight - 5 Points 4. Sarge - 5 Points 5. Dominique - 3 1/2 Points 6. Fred W - 1 1/2 Points 7. John McLeod VII - 1 Point 8. Scary Capitalist - 1/2 Point 9. TBD...... Bill is just 1/2 a point away from taking the title of Best Problem Solver.... Sarge is making a big impact on the table as well. Q26 is still here: Q26: There is a free gift in my breakfast cereal. The manufacturers say that the gift comes in four different colors, and encourage one to collect all four (thus eating lots of cereal). Assuming there is an equal chance of getting any one of the colors, what is the expected number of boxes I must consume to get all four? - Luke. |
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