#$&* course Mth 151 2/24 12:35 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `a** If there was one 11-year-old horse the sum of the remaining ages would have to be 122 - 11 = 111, which isn't divisible by 9. If there were two 11-year-old horses the sum of the remaining ages would have to be 122 - 2 * 11 = 100, which isn't divisible by 9. If there were three 11-year-old horses the sum of the remaining ages would have to be 122 - 3 * 11 = 89, which isn't divisible by 9. If there were four 11-year-old horses the sum of the remaining ages would have to be 122 - 4 * 11 = 78, which isn't divisible by 9. If there were five 11-year-old horses the sum of the remaining ages would have to be 122 - 5 * 11 = 67, which isn't divisible by 9. The pattern is 122 - 11 = 111, not divisible by 9 122 - 2 * 11 = 100, not divisible by 9 122 - 3 * 11 = 89, not divisible by 9 122 - 4 * 11 = 78, not divisible by 9 122 - 5 * 11 = 67, not divisible by 9 122 - 6 * 11 = 56, not divisible by 9 122 - 7 * 11 = 45, which is finally divisible by 9. Since 45 / 9 = 5, we have 5 horses age 9 and 7 horses age 11. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery 1.3.32 (previously 1.3.10) divide clock into segments each with same total YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This question is not in the book. I do not know how may segments to divide it into with the information probided if I add the total of the numbers on the clock face: 1+2+3+ ... +12=78 I only know to divide this into three segements by looking at the given solutions 1/3 * 78 = 26. Then by trial and error I got the solution of 11, 12, 1, 2=26 3, 4, 9, 10=26 5, 6, 7, 8=26 confidence rating #$&*: ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 understand how to do this when all the information is provided ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery 1.3.48 (previously 1.3.30) Frog in well, 4 ft jump, 3 ft back. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1.3.40 day 1 starting at the bottom 4ft -3ft = stopping at the 1 ft point day 2 starting at the 1 ft point then 1ft + 4feet up - 3ft down = 2 ft point day 3 starting at the 2 ft point then 2ft + 4feet up -3 ft down = 3 ft point so on and so forth each day the frog jumps 3 feet higher than the number of the day, so then on the 17th day the frog would have jumped to 20feet and out of the well confidence rating #$&*: ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** COMMON ERROR: 20 days CORRECTION: The frog reaches the 20-foot mark before 20 days. On the first day the frog jumps to 4 ft then slides back to 1 ft. On the second day the frog therefore jumps to 5 ft before sliding back to 2 ft. On the third day the frog jumps to 6 ft, on the fourth to 7 ft., etc. Continuing the pattern, on the 17th day jumps to 20 feet and hops away. The maximum height is always 3 feet more than the number of the day, and when max height is the top of the well the frog will go on its way. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery 1.3.73 (previously 1.3.48) How many ways to pay 15 cents? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1.3.56 1 dime, 1 nickel 1 dime, 5 pennies 3 nickels 2 nickels, 5 pennies 1 nickel, 10 pennies 15 pennies 6 different ways confidence rating #$&*: ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ 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: 1.3.58 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. confidence rating #$&*: I had a hard time with this one. I would not have been able to answer without looking at the given solution. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Divide the coins into two piles of 4. One pile will tip the balance. Divide that pile into piles of 2. One pile will tip the balance. Weigh the 2 remaining coins. You'll be able to see which coin is heavier. ** " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qQuery 1.3.68 (previously 1.3.52) Given 8 coins, how do you find the unbalanced one in 3 weighings YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1.3.58 Divide the coins into two piles of 4. One pile will tip the balance. Divide that pile into piles of 2. One pile will tip the balance. Weigh the 2 remaining coins. confidence rating #$&*: I had a hard time with this one. I would not have been able to answer without looking at the given solution. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: #*&!