#$&* course Mth 151 Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
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Given Solution: `a** If there was one 11-year-old horse the sum of the remaining ages would have to be 122 - 11 = 111, which isn't divisible by 9. If there were two 11-year-old horses the sum of the remaining ages would have to be 122 - 2 * 11 = 100, which isn't divisible by 9. If there were three 11-year-old horses the sum of the remaining ages would have to be 122 - 3 * 11 = 89, which isn't divisible by 9. If there were four 11-year-old horses the sum of the remaining ages would have to be 122 - 4 * 11 = 78, which isn't divisible by 9. If there were five 11-year-old horses the sum of the remaining ages would have to be 122 - 5 * 11 = 67, which isn't divisible by 9. The pattern is 122 - 11 = 111, not divisible by 9 122 - 2 * 11 = 100, not divisible by 9 122 - 3 * 11 = 89, not divisible by 9 122 - 4 * 11 = 78, not divisible by 9 122 - 5 * 11 = 67, not divisible by 9 122 - 6 * 11 = 56, not divisible by 9 122 - 7 * 11 = 45, which is finally divisible by 9. Since 45 / 9 = 5, we have 5 horses age 9 and 7 horses age 11. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): That is what I got, but I did it a little different, but the same. I made a chart with 9 year olds and 11 year olds. Then what I did to start off the number that I needed I took the 122/9 and got 13; then took the 122/11 and got 11. Then I took the 9 and times it by 12 one under less and went down until I got 9 * 1. I did the same thing with the 11’s then I add the two columns together until I got a combination that add up to 122. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery 1.3.10 divide clock into segments each with same total YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I am a little confused on what to do in this problem, but here goes nothing. I would say that 12 will go to 6, 1 will go to 7, 2 will go to 8, 3 will go to 9, 4 will go to 10, 5 will go to 11. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The total of all numbers on the clock is 78. So the numbers in the three sections have to each add up to 1/3 * 78 = 26. This works if we can divide the clock into sections including 11, 12, 1, 2; 3, 4, 9, 10; 5, 6, 7, 8. The numbers in each section add up to 26. To divide the clock into such sections the lines would be horizontal, the first from just beneath 11 to just beneath 2 and the second from just above 5 to just above 8. Horizontal lines are the trick. You might have to draw this to see how it works. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I drew it on the book and I understand how you got what you did. The thing I didn’t get was how I was supposed to add all the numbers up and then divide them. I know this is going to sound strange, I understand and then I don’t understand. Self-critique Rating: ********************************************* Question: `qQuery 1.3.18 M-F 32 acorns each day, half of all acorns eaten, 35 acorns left after Friday YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I believe that they would have started with 4 acorns on Monday. Because if you do a chart that looks like this: ACORNS DALIE ATE (1/2) M 32 16 T 32 + 16 = 48 24 W 32 + 24 = 56 28 TH 32 + 28 = 60 30 F 32 + 30 = 62 31 If you have a total of 35 left and by the math you should have 31, you must have had 4 acorns beforehand. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** You have to work this one backwards. If they were left with 35 on Friday they had 70 at the beginning (after bringing in the 32) on Friday, so they had 70 - 32 = 38 at the end on Thursday. So after bringing in the 32 they had 2 * 38 = 76 at the beginning of Thursday, which means they had 76 - 32 = 44 before the 32 were added. So they had 44 Wednesday night ... etc. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I so didn’t get this one. I am confused on working the backward way. Where did you get the 70 at?????? I also don’t understand when you had 2 * 38, why would you times it by 2??????? Self-critique Rating: ********************************************* Question: `qQuery 1.3.30 Frog in well, 4 ft jump, 3 ft back. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: After 17th days you would only have 3 feet left. So on the 18th day you would get out. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** COMMON ERROR: 20 days CORRECTION: The frog reaches the 20-foot mark before 20 days. On the first day the frog jumps to 4 ft then slides back to 1 ft. On the second day the frog therefore jumps to 5 ft before sliding back to 2 ft. On the third day the frog jumps to 6 ft, on the fourth to 7 ft., etc. Continuing the pattern, on the 17th day jumps to 20 feet and hops away. The maximum height is always 3 feet more than the number of the day, and when max height is the top of the well the frog will go on its way. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I got it right and wrong; I put on the 17 day as zero on my paper, but then said oh he slipped 3 feet and leaving 3 feet. I guess I had a blonde moment on that one. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery 1.3.48 How many ways to pay 15 cents? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I believe there is 6 different ways: 1. 15 pennies 5. 3 nickels 2. 10 pennies & 1 nickel 6. 1 nickel & 1 dime 3. 5 pennies & 2 nickels 4. 5 pennies & 1 dime confidence rating #$&*:: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** To illustrate one possible reasoning process, you can reason this out in such a way as to be completely sure as follows: The number of pennies must be 0, 5, 10 or 15. If you don't use any pennies you have to use a dime and a nickle. If you use exactly 5 pennies then the other 10 cents comes from either a dime or two nickles. If you use exactly 10 pennies you have to use a nickle. Or you can use 15 pennies. Listing these ways: 1 dime, 1 nickel 1 dime, 5 pennies 2 nickels, 5 pennies 3 nickels 15 pennies 1 nickel 10 pennies ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Got it!!!!!!!! Self-critique Rating: ********************************************* Question: `qQuery 1.3.52 Given 8 coins, how do you find the unbalanced one in 3 weighings YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I don’t understand the question or at least I don’t understand how I would do it. I guess you could weight one and see how it feels in your hand then compare the others by themselves and hope you get it right on the third weighing. confidence rating #$&*:: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Divide the coins into two piles of 4. One pile will tip the balance. Divide that pile into piles of 2. One pile will tip the balance. Weigh the 2 remaining coins. You'll be able to see which coin is heavier. ** ****That solution is so easy that it was hard!!!!!!****** Another blonde moment. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qQuery 1.3.48 How many ways to pay 15 cents? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I believe there is 6 different ways: 1. 15 pennies 5. 3 nickels 2. 10 pennies & 1 nickel 6. 1 nickel & 1 dime 3. 5 pennies & 2 nickels 4. 5 pennies & 1 dime confidence rating #$&*:: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** To illustrate one possible reasoning process, you can reason this out in such a way as to be completely sure as follows: The number of pennies must be 0, 5, 10 or 15. If you don't use any pennies you have to use a dime and a nickle. If you use exactly 5 pennies then the other 10 cents comes from either a dime or two nickles. If you use exactly 10 pennies you have to use a nickle. Or you can use 15 pennies. Listing these ways: 1 dime, 1 nickel 1 dime, 5 pennies 2 nickels, 5 pennies 3 nickels 15 pennies 1 nickel 10 pennies ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Got it!!!!!!!! Self-critique Rating: ********************************************* Question: `qQuery 1.3.52 Given 8 coins, how do you find the unbalanced one in 3 weighings YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I don’t understand the question or at least I don’t understand how I would do it. I guess you could weight one and see how it feels in your hand then compare the others by themselves and hope you get it right on the third weighing. confidence rating #$&*:: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Divide the coins into two piles of 4. One pile will tip the balance. Divide that pile into piles of 2. One pile will tip the balance. Weigh the 2 remaining coins. You'll be able to see which coin is heavier. ** ****That solution is so easy that it was hard!!!!!!****** Another blonde moment. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&! ********************************************* Question: `qQuery 1.3.48 How many ways to pay 15 cents? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I believe there is 6 different ways: 1. 15 pennies 5. 3 nickels 2. 10 pennies & 1 nickel 6. 1 nickel & 1 dime 3. 5 pennies & 2 nickels 4. 5 pennies & 1 dime confidence rating #$&*:: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** To illustrate one possible reasoning process, you can reason this out in such a way as to be completely sure as follows: The number of pennies must be 0, 5, 10 or 15. If you don't use any pennies you have to use a dime and a nickle. If you use exactly 5 pennies then the other 10 cents comes from either a dime or two nickles. If you use exactly 10 pennies you have to use a nickle. Or you can use 15 pennies. Listing these ways: 1 dime, 1 nickel 1 dime, 5 pennies 2 nickels, 5 pennies 3 nickels 15 pennies 1 nickel 10 pennies ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Got it!!!!!!!! Self-critique Rating: ********************************************* Question: `qQuery 1.3.52 Given 8 coins, how do you find the unbalanced one in 3 weighings YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I don’t understand the question or at least I don’t understand how I would do it. I guess you could weight one and see how it feels in your hand then compare the others by themselves and hope you get it right on the third weighing. confidence rating #$&*:: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Divide the coins into two piles of 4. One pile will tip the balance. Divide that pile into piles of 2. One pile will tip the balance. Weigh the 2 remaining coins. You'll be able to see which coin is heavier. ** ****That solution is so easy that it was hard!!!!!!****** Another blonde moment. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!#*&!