course mth 151 ?D??????????assignment #008???????y|C????
......!!!!!!!!...................................
22:20:51 1.3.6 9 and 11 yr old hosses; sum of ages 122. How many 9- and 11-year-old horses are there?
......!!!!!!!!...................................
RESPONSE --> Five nine-year olds and seven eleven-year olds.
.................................................
......!!!!!!!!...................................
22:21:16 ** 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. **
......!!!!!!!!...................................
RESPONSE --> Got it.
.................................................
......!!!!!!!!...................................
22:30:05 Query 1.3.10 divide clock into segments each with same total
......!!!!!!!!...................................
RESPONSE --> 6 segments
.................................................
......!!!!!!!!...................................
22:31:44 ** 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. **
......!!!!!!!!...................................
RESPONSE --> i was confused by what was wanted because i could not find that question in my book as well as the question before, 6 in my book as asking for a age of a bus driver and number 10 was asking how old chris is, maybe i'm not looking at this right but i'm pretty sure i am.
.................................................
......!!!!!!!!...................................
22:47:08 Query 1.3.18 M-F 32 acorns each day, half of all acorns eaten, 35 acorns left after Friday
......!!!!!!!!...................................
RESPONSE --> 128 acorns
.................................................
......!!!!!!!!...................................
22:49:13 ** 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. **
......!!!!!!!!...................................
RESPONSE --> the question asks what they had on monday morning not wednesday and i got 128 acorns using same method, by working backwards. I took 35 and times it by 2 and subtracted by 32, i did this 6 times.
.................................................
......!!!!!!!!...................................
22:53:20 Query 1.3.30 Frog in well, 4 ft jump, 3 ft back.
......!!!!!!!!...................................
RESPONSE --> It will take him 20 days because he is basically only jumping 1 foot a night.
.................................................
......!!!!!!!!...................................
22:54:34 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
22:56:39 Query 1.3.48 How many ways to pay 15 cents?
......!!!!!!!!...................................
RESPONSE --> There are 6 ways to use pennies, dimes, and nickels to get 15 cents.
.................................................
......!!!!!!!!...................................
22:56:51 ** 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 **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
23:17:44 Query 1.3.52 Given 8 coins, how do you find the unbalanced one in 3 weighings
......!!!!!!!!...................................
RESPONSE --> you begin by placeing 4 coins on one side of the balance scale and place the other 4 on the other side. The fake will be on the side that is lighest. Then take the four heavy coins and place them aside. Then divide the coins in half place 2 on one side and 2 on the other. Again the lighest will be the fake. Then take the 2 and divide them by placing one on one side and one on the other. The lightest coin will be the fake.
.................................................
......!!!!!!!!...................................
23:17:57 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
"