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July 29, 2015

July 29, 2015

Posted by **carol** on Thursday, April 23, 2009 at 6:20pm.

- math -
**Ms. Sue**, Thursday, April 23, 2009 at 6:28pmYou left out the amount of money you have to spend.

- math -
**carol**, Thursday, April 23, 2009 at 6:31pmsorry 100.00

- math -
**TchrWill**, Thursday, April 23, 2009 at 6:35pmYou left out the cost of the purchases, typically $100.

Here is a similar version of this oldie.

If you had a $100.00 to spend and need to buy a 100 animals, and cows cost

$10.00, pigs $3.00, chichens .50 cents each,.how many of each can you buy?

Let C, P, and F be the numbers of cows, pigs, and fowl.

1--C + P + F = 100

2--10C + 3P + .5F = 100 or 100C + 30P + 5F = 1000

3--Multiplying (1) by 5 and subtracting from (2) yields 19C + 5P = 100

4--Dividing through by 5 gives P + 3C + 4C/5 = 20

5--4C/5 must be an integer as must be C/5

6--Let C/5 = k making C = 5k

7--Substituting (6) back into (3) gives 95k + 5P = 100 making P = 20 - 19k

8--k can only be 1 making C = 5, P = 1, and F = 94

Check: 10(5) + 3(1) + .5(94) = $100

Alternatively

Let C, P, and F be the numbers of cows, pigs, and fowl.

1--C + P + F = 100

2--10C + 3P + .5F = 100 or 100C + 30P + 5F = 1000

3--Multiplying (1) by 5 and subtracting from (2) yields 19C + 5P = 100

4--Solving for P, P = 20 - 19C/5

5--19C/5 must be an integer meaning that C must be evenly divisible by 5.

6--Thus, C must be 5 making P = 1, C = 5, and F = 94.

I'll let you work it out with your numbers.