#$&* course Mth 151 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `a** If there was one 11-year-old horse the sum of the remaining ages would have to be 122 - 11 = 111, which isn't divisible by 9. If there were two 11-year-old horses the sum of the remaining ages would have to be 122 - 2 * 11 = 100, which isn't divisible by 9. If there were three 11-year-old horses the sum of the remaining ages would have to be 122 - 3 * 11 = 89, which isn't divisible by 9. If there were four 11-year-old horses the sum of the remaining ages would have to be 122 - 4 * 11 = 78, which isn't divisible by 9. If there were five 11-year-old horses the sum of the remaining ages would have to be 122 - 5 * 11 = 67, which isn't divisible by 9. The pattern is 122 - 11 = 111, not divisible by 9 122 - 2 * 11 = 100, not divisible by 9 122 - 3 * 11 = 89, not divisible by 9 122 - 4 * 11 = 78, not divisible by 9 122 - 5 * 11 = 67, not divisible by 9 122 - 6 * 11 = 56, not divisible by 9 122 - 7 * 11 = 45, which is finally divisible by 9. Since 45 / 9 = 5, we have 5 horses age 9 and 7 horses age 11. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qQuery 1.3.32 (previously 1.3.10) divide clock into segments each with same total YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I do not understand this question and it does not match up to my book. confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): Still confused. Don’t understand the question. ------------------------------------------------ Self-critique Rating: 0 ********************************************* Question: `qQuery 1.3.48 (previously 1.3.30) Frog in well, 4 ft jump, 3 ft back. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: OK, this one does not match up to my book either, but I see that it is question 1.3.40 in my book ( I have the 11th edition). Now I can see that the previous question must have been 1.3.22. He would reach the top of the well on day 17. He gains 1 foot per day. At the end of day and night #1, he is at the 1 foot level. (0 + 4 - 3 = 1) At the end of day and night #1, he is at the 2 foot level. (1 + 4 - 3 = 2) At the end of day and night #16, he would be at the 16 foot level. On day 17, he would jump 4 feet which would get him out of the well. (16 + 4 = 20) He would not be there for the 3 foot fall that night. I drew a chart to help figure this one out. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** COMMON ERROR: 20 days CORRECTION: The frog reaches the 20-foot mark before 20 days. On the first day the frog jumps to 4 ft then slides back to 1 ft. On the second day the frog therefore jumps to 5 ft before sliding back to 2 ft. On the third day the frog jumps to 6 ft, on the fourth to 7 ft., etc. Continuing the pattern, on the 17th day jumps to 20 feet and hops away. The maximum height is always 3 feet more than the number of the day, and when max height is the top of the well the frog will go on its way. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qQuery 1.3.73 (previously 1.3.48) How many ways to pay 15 cents? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This one is 1.3.56 in my book. 1 dime, 1 nickle 1 dime, 5 pennies 3 nickles 2 nickles, 5 pennies 1 nickle, 10 pennies 15 pennies confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qQuery 1.3.68 (previously 1.3.52) Given 8 coins, how do you find the unbalanced one in 3 weighings YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This one is 1.3.58 in my book. First weighing: Put 4 coins on each side of the scale. Whichever side weighs less contains the fake coin. You know the heavy side contains all good coins so you can set them aside. Second weighing: Now take the remaining 4 coins and divide them. Put 2 coins on each side of the scale. Again, whichever side is lighter contains the fake. You have now narrowed it down to 2 coins. Third weighing: Put the last 2 coins on the scale, one on each side. Whichever side is lighter is the fake coin. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Divide the coins into two piles of 4. One pile will tip the balance. Divide that pile into piles of 2. One pile will tip the balance. Weigh the 2 remaining coins. You'll be able to see which coin is heavier. ** " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qQuery 1.3.68 (previously 1.3.52) Given 8 coins, how do you find the unbalanced one in 3 weighings YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This one is 1.3.58 in my book. First weighing: Put 4 coins on each side of the scale. Whichever side weighs less contains the fake coin. You know the heavy side contains all good coins so you can set them aside. Second weighing: Now take the remaining 4 coins and divide them. Put 2 coins on each side of the scale. Again, whichever side is lighter contains the fake. You have now narrowed it down to 2 coins. Third weighing: Put the last 2 coins on the scale, one on each side. Whichever side is lighter is the fake coin. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!