math13QA

#$&*

course Mth 151

2/15/2012 5:05PM

008.  `Query 8 

 

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Question:  `q (previously 1.3.6)  There are a number of  9 and 11 yr old horses in the barn and the sum of their ages is 122.  How many 9- and 11-year-old horses are there?

 

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Your solution: 

First I started adding up number of 11 year old horses and subtracting that from 122 until I got a number 9 could divide into. I finally got that if there are 7 11 year old horse there would 45 years left out of the 122 that I could divide 9 into. So 9 divided by 45 equals 5, so there are five 9 year old horses and seven 11 year old horses.

 

 

confidence rating #$&*: 3

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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.  **

 

 

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Self-critique (if necessary): OK

 

 

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Self-critique Rating: OK

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Question:  `qQuery 1.3.32 (previously 1.3.10) divide clock into segments each with same total

 

 

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Your solution: 

First I added up all the numbers on the clock to get total, which is 78. Then I divided that by 3 to get the number each region should add up to, which is 26. After a lot of trial and error I figured out if you put a line from the top of 10 to the top of 3 and another one from the top of 8 and the top of 5, each section will have 26. The top section will have 11, 12, 1, 2 which added together equals 26, the middle section will have 10, 3, 9, 4 which added together equals 26, and the bottom section has 8, 7, 6, 5 which added together equals 26.

 

 

confidence rating #$&*: 3

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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.  **

 

 

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Self-critique (if necessary): OK

 

 

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Self-critique Rating: OK

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Question:  `qQuery  1.3.48 (previously 1.3.30)  Frog in well, 4 ft jump, 3 ft back.

 

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Your solution: 

It will take the frog 20 days to get to the top of the 20 foot well because he is only totaling 1 foot a day.

 

 

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.  **

 

 

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Self-critique (if necessary):

I see why the frog would reach the top before the 20th day.

 

 

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Self-critique Rating: 3

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Question:  `qQuery  1.3.73 (previously 1.3.48)  How many ways to pay 15 cents?

 

 

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Your solution: 

15 pennies

10 pennies and a nickel

5 pennies and a dime

5 pennies and 2 nickels

a nickel and a dime

3 nickels

 

 

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

 

**

 

 

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Self-critique (if necessary): OK

 

 

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Self-critique Rating: OK

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Question:  `qQuery  1.3.68 (previously 1.3.52)  Given 8 coins, how do you find the unbalanced one in 3 weighings

 

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Your solution: 

First I would weigh 4 on each side and see which side went up, because the side that went up would have the fake coin in it. I would take 4 coins from the side that went up and put 2 in each side and do the same as before. Then I would take the 2 coins from the side that went up and put 1 in each side and the side that goes up that time will be the one with 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. **

"

end document

Self-critique (if necessary):

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Self-critique rating:

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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: 

First I would weigh 4 on each side and see which side went up, because the side that went up would have the fake coin in it. I would take 4 coins from the side that went up and put 2 in each side and do the same as before. Then I would take the 2 coins from the side that went up and put 1 in each side and the side that goes up that time will be the one with 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. **

"

end document

Self-critique (if necessary):

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Self-critique rating:

#*&!

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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: 

First I would weigh 4 on each side and see which side went up, because the side that went up would have the fake coin in it. I would take 4 coins from the side that went up and put 2 in each side and do the same as before. Then I would take the 2 coins from the side that went up and put 1 in each side and the side that goes up that time will be the one with 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. **

"

end document

Self-critique (if necessary):

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Self-critique rating:

#*&!#*&!

&#Very good work. Let me know if you have questions. &#