i have a stupid math problem?

Question

my friend just called me with a stupid questions the bad part is i have a BS in computer science....

ok this is what it is

chicken are 10cents

sheep are 2dollars

and pigs are 5dollars

now a 100 aminals need to equal $100

i just need some help with how to write it out as a problem...

thank you

lol i'm not in school, i'm 26, my friend just called me becasue they saw it online and could not figure it out easy, i know i can sit here and plug in numbers and figure it out easy and i already know the layout of the problem... just trying to see if there is a more simple way of doing it

Answer 1

Oh no! Someone with a BS in computer science should TOTALLY be able to do this!!! :(

Let x = # of chickens, y = # of sheep, z = # of pigs

Now you have 2 equations:

x + y + z = 100
.10x + 2y + 5z = 100

Now solve (but since you have 2 equations and 3 unknowns, you will have multiple answers not just one solution)

HTH! :-)

Answer 2

The statement of the problem yields two equations:

10c + 200s + 500p = 100*100 (coefficients are cents, thus $100 is 100*100 cents)

The second equation is
c + s + p = 100

You have two equations and three unknowns. The trick here is that you don't have fractional animals. So you need c, s, and p to be positive integers.

This is really a number theory problem but I'd recommend eliminating one variable. Take 10 times equation 2 and subract it from equation 1. You'll eliminate c and have a relationship between s and p. Then search for integers that solve it.

Write back if you want more help.

Answer 3

Let
c - number of chickens
s - number of sheep
p - number of pigs

Given
c + s + p = 100 ... (1)

And
0.1c + 2s + 5p = 100 ... (2)

Take (2) x 10 - (1),

c - c + 20s - s + 50p - p = 1000 - 100

19s + 49p = 900

s = (900 - 49p)/19 ... (3)

For integer value solutions,
the factor (900 - 49p) needs to be a multiple of 19
therefore, (900 - 49p) = 19n (for some n = 1, 2, ... )

Solving, we get
p = (900 - 19n)/49

if n = 19, p = 11

Substituting p = 11 into (3),

We get s = (900 - 49(11))/19

s = 19

Substituting p = 11 and s = 19 into (1),

we get

c + (19) + (11) = 100

c = 70

So, one solution is

==============
c = 70 (chickens)
s = 19 (sheep)
p = 11 (pigs)
==============

Answer 4

70 chickens, 19 sheep, and 11 pigs

Yes, you can use everyone's formula's, but remember that you need an integer number of each animal (unless they're going to the butcher).

Because the chicken's are a dime, you'll need a multiple of 10 of them. You'll quickly find you need 60, 70, or 80 of them.

Using other's formulas, you'll find that using 60 or 80 chicken's leaves you with a fraction of other animals.

Using 70 chickens, 19 sheep, and 11 pigs gets you $100 and 100 animals.

Answer 5

There are three unknowns (the number of chicken, sheep and pigs).

But you can only have two equations. One relating to the total cost, the other to the total of animals purchased.

Two equations with three unknowns means that there can be no non-zero solutions.

Answer 6

The best you could do is 0.1c + 2s + 5p =100 and c+s+p=100, but you need one more equation to complete the system.

Answer 7

this isn't really the (x,y) method however nearly the correct answer.. so, rebecca thinks that during yaer five (t5) she is going to want 225 seats. so t5=225. within the typical system tn=a+(n-one million)d, tn=225, n=five, d=15 (average change; the quantity she raises it via every yr). a is the first actual time period, so that you ought to discover that. tn=a+(n-one million)d (225)=a+((five)-one million)(15) 225=a+60 a=225-60 a=a hundred sixty five tn=a+(n-one million)d tn=a hundred sixty five+(n-one million)(15) tn=a hundred sixty five+15n-15 tn=15n+one hundred fifty

Answer 8

let ( x )= 1
100 animals = (1)x = $100.00
It seems to me that something is missing from the problem. I need more information.

Answer 9

You can buy 20 pigs
or
1000 chickens
or
50 sheeps

But you can't have an equation. There is too many unknown variables

Answer 10

it can't be solved since you have 3 unknowns but only 2 equations