you have to buy 100 animals. roosters cost 5.00, hens cost 1.00 and chicks cost .05cents. how many of each animal can I buy?
4 answers
You left out the amount of money you have to spend.
sorry 100.00
You 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.
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.
please what is the answer