Posted: 8/26/2008 8:12:49 AM EDT
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I am taking a math history class, and it has been about 11 years since I have had to use any math at all. I can do it if I have the book for a few examples, but this is not the case with last night's HW. 2. I have a reed. I know not its dimension. I broke off from it 1 cubit and walked 60 times along its length. I restored to it what I have broken off, then walked 30 times along is length. The area is 375 square cubits. What was the original length of the reed? I think that those who enjoy math might have the opportunity for some fun problems over the next couple of months. 
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Holy crap! Check out problem 3. I found a stone, but I did not weigh it: after I added 1/7 of its wueght and hen 1/11 of this new weight, I weighed the total at 1 mina. What was the original weight of the stone? Scribe's answer: The original weight of the stone was 2/3 mina, 8 sheqels and 22 1/2 se. [If 1 mina=60 shequels and 1 sheqel- 180 se, verify the correctness of the scribe's answer.] |
I understand that, but I think we are forced to make an assumption that he is making a rectangle, and that width is not a factor. |
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Thanks. I actually know what the answers are supposed to be due to the extra sheet that he gave us. I just need to figure out how to get the answer. |
I think this is what they mean: length = x (60(x-1)) * 30x = 375 1800x^2 - 1800x - 375 = 0 Solve for x using quadratic equation: x=1.1770032003863302 or -0.1770032003863301. Since length can't be negative, length = 1.1770032003863302 |
What in the world kind of crack head class is this? Sorry, I can do this decade's math not that... sorry I can be of no help! |
Thanks so much. I was able to use what you did there to find my mistake. This is coming back very quickly, but not quickly enough. I am also thinking that I played too much pool while drinking during math class the first time around, although I still got a B. 
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"Teach middle school". Hmmm.... You can count to 10, right? 
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I got to 27 the other day before I had to start over. |
That's not the way problems are posed. Your job as a teacher will be to give questions that measure performance and learning, not guessing and assumptions for which there is no basis. In this problem, there is not even the slightest indication or implication that the reeds describe a rectangle, only that the person walks. |
Again, true. This is a math history class though, and these problems are supposedly similar to some writings on ancient Babylonian tablets. They were a lot better at math than I. |
Changing the units does not change the math. It's a verbal trick designed to trip you up. People form mental relationships with units. For example, if I say that I am 5'5" tall, you already have a picture in your head of my height approximate to yours and that, as a male, I'm a little shorter than average. However, if I say that I am 16 1/4 hands high, that doesn't create the same mental relationship. If I say that I have $11.37 in my pocket, you understand what that looks like, how many coins and bills I must have at a minimum and probably what I can buy with it as well. However, if I say that I have $360.00 Taiwan dollars in my pocket, you have no idea what that looks like or what it's worth. Initially, it might sound like that's a lot of money. In reality, it's the same thing as $11.37USD. If you are to become proficient in math, you need to understand algebra and number manipulation independent of the units. Of course, units are important but the core skills should be practiced over a wide variety of systems so that you learn to not let units trip you up. Incidentally, the above problem is inherrently unsolvable since there are excessive unknowns. You get something like the following: x = weight of stone y = weight of "this new weight" 1/7 x + 1/11 y = 1mina The scribe says: x = 2/3 mina + 8/60 mina + 22.5/(60*180) mina So without knowing what y ("this new weight") is, you cannot verify that the scribe is correct. It's just simple algebra with different units in order to trip you up. Once you get it in your head that the units don't matter, then it all will make sense. |
I would have to agree. That your teacher gave you a B is irrelevant. The real question is this. Did you learn the material? I'm afraid the answer is , no. Neither of those problems is particularly difficult. They are wrapped up in a lot of archaic, difficult language and arcane units of measure trying to baffle you. I would say that the writer of those word rpoblems probably got a B in English grammar and math, as well, as they are not well written from either a mathematical or grammatical perspective. |
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That does sum it up pretty weel. However, I am tripped up by the wording of many of the questions. Number 7 really confuses me. I can't even set the problem up to work it backwards with the give answer. 7, I found a stone, but did not weigh it. After I weighed out its weight and added 2 sheqels, I then added one third of one seventh of this amount multiplied by 24. The weight was then 1 mana what was the original weight? (1 mana is 60 sheqels.) Answer:4.33 sheqels |
You have made a mistake. The formula is this: (1+1/7)X * (1+1/11) = 1 mina BTW, the scribe is correct in his weight. |
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If someone could point me in the right direction with this one it would help me to all but finish up the sheet except for #7 posted above. I have two fields of grain, From the first field, I get 2/3 a bushel of grain/unit area. From the second field, I get 1/2 bushel of grain/unit area The yield of the first exceeds the second by 50 bushels. The total area of the two fields together is 300 square units. What is the area of each field? |
If you can't translate a description of a problem from english into mathematical symbols, you can't very well solve, can you? And I agree the problem in the OP requires a logical leap not supported by the given information. |
Set the area of the first field as x. The area of the other field is then 300-x the yeild of the first feild is 2/3 x. the yeild from the second is 1/2 * (300-x) since the yeild of the second field is 50 less than the first: 2/3 x - 50 = 1/2 * (300-x) |
Correct. There is no indication that the walker made a 90 degree turn somewhere in there, and is trying to measure a rectangle. |
Thank you so much. Once I know how to do it, it comes easy, but it has been 10-11 years since I have even thought about any of this stuff. |
call the stone's weight x add two sheqels: x + 2 add one third of one seventh of this amount multiplied by 24 This is the same as adding (x + 2) * 1/3 * 1/7 *24 to the original amount, (x + 2) you now have (x + 2) * (1 +1/3 * 1/7 * 24) or (x+2) * (1 + 24/21) or (x+2) * 45/21 This weight is equal to one mana, which is 60 sheqels, so (x + 2) * 45/21 = 60 sheqels Just try to break the problems into small peices and follow the chain of logic. And don't forget to check my adding; I suck at that  |
Yep, the problem, as posed, doesn't make sense. I assumed that the walker made a 90 degree turn but, even making that assumption, you have to make another assumption that the area mentioned in the question refers to the area bounded by a rectangle of which the path walked was two of the four sides. Maybe this is a bad translation of babylonian?
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That's why I am here. I have taught elementary school for two years, and I am a great teacher. I only need one class to finish off my math cert for middle school though. As I have stated it has been 10-11 years since I have used any of this. The first problem honestly took me an hour this morning. Looking at some of the solutions posted, I have just incorrectly interpreted what the questions state for a few of them. Others have only taken me a couple of minutes. I am confident that by the end of the semester I will be good to go. Also, thanks for not being a dick about it. 
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Recommended reading for teachers: The Aviation Instructor's Handbook Just ignore the Aviation part. |
That looks pretty interesting. I think I will give it a read. |
I got it now. Yes, it is solvable. In my defense, I didn't follow the link. I used what killingmachine had posted, which is somewhat different. "its weight" and "this new weight" are the same thing. The word "new" leads one to believe that we are talking about two different stones. |
I can follow everything except for where you get the 1 in this problem. You have the right answer, but I just can's see where that 1 comes from when I read the problem. When I read it I get this. (x+2) + (X+2)*1/3*1/7*24 =60 |
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I just wanted to say thank you again to all of those that helped. I looked pretty darn good today when I walked into class. I could not believe that most kids only got 1-2 problems done. The only one I am missing is #7 still just because I don't understand what the problem is trying to tell me. I will be a great math teacher yet!
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Sorry, I factored out (x + 2) without telling you .(x + 2) + (x + 2) *1/3 * 1/7 * 24 = (x + 2)*(1 + 1/3*1/7*24) So you in fact were doing it correctly, you should still have gotten the correct answer. |
It's kinda hard to help with these questions when you make a shitload of typographical errors. The first problem you posted doesn't even make sense. Re-read the problems, and then post them exactly as stated in the book. |
Answer is right here in red: one cubit. For the original length is before it was broken. Besides, how would the person who did the experiment know the area if he did not know the length of the broken part. |
you win . 
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Anyways, to answer the quoted question: the scribe's answer is correct. This problem is extremely easy to solve. First convert the scribe's answer to "se": 2/3 mina = 40 sheqels + 8 sheqels = 48 sheqels 48 sheqels = 8640 se + 22.5 se = 8662.5 se So the scribe says that the stone originally weighed 8662.5 se, and we know the final weight of the stone is 1 mina, or 10800 se. 8662.5 * 1/7 = 1237.5 8662.5 + 1237.5 = 9900 9900 * 1/11 = 900 9900 + 900 = 10800 |
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That is a good way to do it. If you post much more in this thread you will have doubled your post count. Glad to have provided a thread that you wanted to contribute to.
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Sorry, I didn't realize the problem you were solving was #7. I was referring to the similarly worded problem #3. |
No worries, I think you can solve number 7 without algebra anway. |