course Mth 151 ???Y?????????assignment #008?w?????}??yO?????Liberal Arts Mathematics I
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22:00:49 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 --> This is not a question from the book, but I'll still try to answer it. 122-9-9-9-9-9=77 77-11-11-11-11-11-11-11=0 5 9-year-old horses 7 11-year-old horses
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22:02:09 ** 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 --> I did the opposite. I used 9 first and found the first number which is divisible by 11, which is 77.
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22:04:30 Query 1.3.10 divide clock into segments each with same total
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RESPONSE --> I worked and worked on this. The farthest I got was adding the numbers together (1+2+3...+12 = 78) and then dividing that by 3 to get the same totals for each region, which was 26. I even got so far as to find that 11+12+1+2 = 26 and 5+6+7+8 = 26, but the other numbers didn't work out.
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22:06:23 ** 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 --> Oh, I didn't think about horizontal lines. Now that I draw it out, I can see it. Well, I got some of the problem.
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22:15:22 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 --> 128 acorns. Start with Saturday and the 35 acorns. 35 x 2 - 32 = 38 (Friday) (do the opposite of what was actually done so that you can move backwards) 38 x 2 - 32 = 44 (Thursday) 44 x 2 - 32 = 56 (Wednesday) 56 x 2 - 32 = 80 (Tuesday) 80 x 2 - 32 = 128 (Monday) They started out with 128 and then check it with adding 32 and dividing it by 2, repeat until Saturday.
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22:16:02 ** 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 --> ok
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22:21:39 ** 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 hit the enter response on accident. I was going to say 20 days. 4-3=1. The frog will climb 1 ft a day. I got that wrong. After I drew it on paper rather than trying to do it in my head I see the correct answer.
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22:23:49 Query 1.3.48 How many ways to pay 15 cents?
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RESPONSE --> 6 ways 15 p 10 p + 1n 5 p + 2n 5 p + 1d 3n 1n + 1d
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22:24:33 ** 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
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22:33:29 Query 1.3.52 Given 8 coins, how do you find the unbalanced one in 3 weighings
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RESPONSE --> Divide the coins into 3 groups. 3 coins, 3 coins and 2 coins. Weigh the groups of coins. If the groups of 3 coins are the same, then divide the group by 3 to determine the true weight of each coin. After you weigh the group of 2 it should be 2 x one true coin. If the weight is not correct then you know it is wrong. If the coin happens to be in one of the groups of 3 then the difference will be seem in the scales. Divide the group of 2 by 2 and get the correct weight per coin. Then multiply that single coin number by 3 to get the correct weight. If the coin is in the group of 2 in the final weighing that will be easy, just weigh them separately. If the fake coin is in the group of 3 then weigh only 2 of the coins. You will either find the smaller coin or the 2 that you weigh will be the correct weight and the one that you didn't weigh is the fake.
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22:34:34 ** 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 --> That seems much easier than what I just spent 20 minutes typing out.
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