English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

A rancher buys 100 live animals for $100. Chickens cost $.50 each, goats cost $3.50 each, and cows cost $10 each. How many of each did the rancher buy? Is more than one combination possible?

2007-08-27 19:05:39 · 12 answers · asked by Mz_Monroe 3 in Science & Mathematics Mathematics

12 answers

You have 2 equations, 3 variables: C for number of chickens, G for goats, W for cows. You have an equation for the body count and one for the money:

C + G + W = 100
.5C + 3.5 G + 10W = 100

multiply 2nd equation by 2 to get rid of fractions:

C + G + W = 100
C + 7G + 20W = 200

subtract 1st from 2nd:

6G + 19W = 100, and solve it for G:

G = (100 - 19W)/6

while that has infinitely many solutions, we need G to be a whole number, so there's only 1 solution:
19(4) = 76, 100-76 = 24, 24/6 = 4 goats.

so then it's 4 cows, 92 chickens, and only the 1 solution.

2007-08-27 19:25:47 · answer #1 · answered by Philo 7 · 0 0

40 chickens .50 x 40= $20
20 Goats 3.50 x 20= $70
1 Cow $10

$100.00

2007-08-27 19:17:05 · answer #2 · answered by lvillejj 4 · 0 1

If x be the number of chickens, y the number of goats and z the number of cows, we can write:

0.5x + 3.5y + 10z = 100

Also x + y + z = 100 (number of animals)

We have three variables and only two equations. So, we cannot solve for definite values of x, y and z.

Yes, multiple solutions are possible to satisy that one equation.

2007-08-27 19:21:05 · answer #3 · answered by Swamy 7 · 0 0

I won't solve it for you but I'll get ou started. Set up an equation:

C= chickens
G= goats
W= cows

100= .5C + 3.5G + 10W

the 100 is what he spends
the .5, 3.5 and 10 represent the cost of each noted animal

to get rid of the .5 and 3.5 and make this problem easier you can mult. everything by 10

1000= 5C + 35G + 100W

reduce the problem

200= C + 7G + 20W

now you need to slove for each variable C, G and W, which you should know how to do and yes it takes some effort.

2007-08-27 19:16:37 · answer #4 · answered by hefillsmylifewithgoodthings 2 · 0 1

lots of possibilities
for example 4 cows 40
10 goats 35
leaves25 dollars to buy 50 chickens

9 cows 90
1 goat 3.50 makes 93.50 and leave 6.50 for 13 chickees

2007-08-27 19:17:46 · answer #5 · answered by Anonymous · 0 1

let no. of chicken ,goats and cows be a , b and c then :
a+b+c=100
and
0.5a+3.5b+10c=100
there are three variables and two equations so no unique solution will be written .
there will be more than one combination .

2007-08-27 19:19:18 · answer #6 · answered by Anonymous · 0 0

Four cows, four goats, and ninety-two chickens is the only combination possible.

2007-08-27 19:20:42 · answer #7 · answered by Anonymous · 0 1

Just say he bought a horse for $100 and call it even.

2007-08-27 20:44:56 · answer #8 · answered by Anonymous · 0 0

Paul Ding is correct

2007-08-27 19:24:25 · answer #9 · answered by David G 3 · 0 0

There is nooooooo way I am doing a math problem if I don't have to. Get a tutor, or ask your teacher.

2007-08-27 19:12:51 · answer #10 · answered by maddy 3 · 0 1

fedest.com, questions and answers