course Mth 151 ΒەV}Lɓassignment #008
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21:12:59 1.3.6 9 and 11 yr old hosses; sum of ages 122. How many 9- and 11-year-old horses are there?
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RESPONSE --> There are 5 9-year old horses and 7 11-year old horses confidence assessment: 3
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21:13:10 ** 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. **
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RESPONSE --> OK self critique assessment: 3
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21:15:03 Query 1.3.10 divide clock into segments each with same total
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RESPONSE --> I would draw one line from below 11 to below 2, and a second line from above 8 to above 5. So the three segements would be: 11, 12, 1, and 2 in the first segment; 10, 9, 3, and 4 in the second segment; and 8, 7, 6, and 5 in the third segment. The total of each segment in 26. confidence assessment: 3
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21:15:13 ** 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. **
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RESPONSE --> OK self critique assessment: 3
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21:25:20 Query 1.3.18 M-F 32 acorns each day, half of all acorns eaten, 35 acorns left after Friday
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RESPONSE --> They started with 128 acorns confidence assessment: 2
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21:25:45 ** 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. **
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RESPONSE --> That's the method I used self critique assessment: OK
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21:27:02 Query 1.3.30 Frog in well, 4 ft jump, 3 ft back.
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RESPONSE --> It will take 20 days, because the frog progresses one foot each day confidence assessment: 3
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21:27:39 ** 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. **
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RESPONSE --> I didn't take into consideration the last day where the frog would not fall back self critique assessment: 2
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21:30:57 Query 1.3.48 How many ways to pay 15 cents?
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RESPONSE --> 6 ways: 15 pennies; 10 pennies, 1 nickel; 5 pennies, 2 nickels; 5 pennies, 1 dime; 1 dime, 1 nickel; 3 nickels confidence assessment: 3
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21:31:07 ** 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 **
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RESPONSE --> OK self critique assessment: 3
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21:33:18 Query 1.3.52 Given 8 coins, how do you find the unbalanced one in 3 weighings
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RESPONSE --> Divide the 8 coins into two groups of 4 - that will tell you which group of 4 has the fake coin. Then take that group of 4 and divide it into two groups of 2, and you will know which group of 2 has the fake coin. Then weigh the two coins from that group against each other. confidence assessment: 3
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21:33:25 ** 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. **
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RESPONSE --> OK self critique assessment: 3
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