Algebra word problem?

Question

A farmer gives her daughter $100.00 and instructs her to go the the village and buy some sheep, some goats and some chickens. The sheep cost $3.00 each; the goats cost $2.00 each and the chickens cost $0.50 cents each. She goes to the village and spends the entire $100.00 and comes home with exactly 100 animals. She bought at least 1 sheep, at least 1 goat, and at least 1 chicken.

1). How many of each animal did she buy?
2). If there is more than 1 solution, how does one find all the solutions?

Charles

Answer 1

catarthur got halfway there when he derived:
5S + 3G =100.
Look at 3G = 100 - 5S. S being an integer, 100 -5S is a multiple of 5, therefore, if G is an integer, it must be a multiple of 5 also.
Since there is at least one goat, G is at least 5, and to keep the number of sheep positive, G must be less than 33.
So there are six solutions:
G = 5, 10, 15, 20, 25, or 30.
S = (100 - 3G)/5
C = 100 - S - G.

Answer 2

This can be written as:
Find the integers S, G and C such that
100 = 3S + 2G + 0.5C
S+G+C = 100

first let's reduce this by substituting for one variable
C= 100-S-G

therefore
100 = 3S +2G +50 -0.5S - 0.5G
50 = 2.5S +1.5G
100 = 5S +3G

Now we can try varying S between 1 and 19
S = 1, G = 95/3 not an integer
S = 2, G = 30, C = 68
S = 3, G = 85/3 not an integer
S = 4, G = 80/3 not an integer
S = 5, G = 25,C = 70
S = 6, G = 70/3 not an integer
S = 7, G = 65/3 not an integer
S = 8, G = 20, C = 72
S = 9, G = 55/3 not an integer
S = 10, G = 50/3 not an integer
S = 11, G = 15, C = 74
S = 12, G = 40/3 not an integer
S = 13, G = 35/3 not an integer
S = 14, G = 10, C = 76
S = 15, G = 25/3 not an integer
S = 16, G = 20/3 not an integer
S = 17, G = 5, C = 78
S = 18, G = 10/3 not an integer
S = 19, G = 5/3 not an integer

Answer 3

As you may gather from the answers given before, there are quite a number of ways in which this problem may be solved. The solution I came up with follows:

Let x, y, z, be the number of sheep, goats, and chicken, respectively, bought by the farmer's daughter. You may use s, g, and c if you like. You're given two equations, namely:

x +y +z = 100 (eq. I)
3x + 2y + ½ z = 100 (eq. II)

Both equations have 3 unknowns; thus, you cannot find x, y, z by solving a regular 3-equation, 3-unknown (3 × 3) system of simultaneous equations. An additional constraint can be imposed, namely, that all unknowns are positive integers, other than zero. You may proceed as follows:

Multiply second equation by 2:

6x +4y + z = 200

Subtract first eq. from this:

6x + 4y + z = 200
x + y + z = 100
————————
5x + 3y + 0z = 100

Solve for y:

y = (100 − 5x)/3 (eq. III)

As 100 = 99 +1, and 5x = 6x − x, you may write:

y = (99 + 1 − 6x + x)/3,

which reduces to:

y = 33 − 2x + (1 + x)/3.

Clearly, 33 is a positive integer; 2x as well, in the assumption that x is also a positive integer. But then, if y is a positive integer as well, we're forced to admit that the ratio (1 + x)/3 is just another positive integer. Let n be that last positive integer; i.e., let n = (1 + x)/3. Therefore, solving for x,

x = 3n − 1 (eq. IV)

We can now obtain a similar expression for y by subtituting the result above in eq. III:

y = [100 − 5(3n − 1)] / 3 = 105/3 − 5n = 35 − 5n (eq. V)

We have now "y" in terms of "n", also. As for "z", just solve for "z" in eq. I:

z = 100 − 3n + 1 − 35 + 5n = 66 + 2n (eq. VI).

All unknowns are now in terms of "n", which we know is a (nonzero) positive integer. As all equations derived above for x, y, z, involve only positive integers, we may be sure the unknowns are positive integers as well.

Now, in order for x to be ≥ 1, according to eq. IV, n ≥ 1. If, for instance, n = 0, or negative, x would turn out negative, which isn't acceptable. Likewise, in order for y to be ≥ 1, n ≤ 6. Thus, we've: 1 ≤ n ≤ 6; this is the valid range of "n" for our present purpose. This means there are 6 sets of (valid) solutions. These are, in tabular form:

n x y z
――――――
1 2 30 68
2 5 25 70
3 8 20 72
4 11 15 74
5 14 10 76
6 17 5 78

Answer 4

OK... so first we note: x=number of sheep, y=number of goats, z=number of chicken.
x*3+y*2+z*0.5=100 and x+y+z=100;
if x+y+z=100, then z=100-x-y;
We replace that in the first equation and we have:
3*x+2*y+0.5(100-x-y)=100 =>2.5*x+1.5*y=50 => 5*x+3*y=100;
Now, what we need to do is replace x for any natural number from 1 on, because we know that there is at least one sheep, and all the numbers are natural.
Something like this:
x=1 =>3*y=95 (not possible, y will not be natural)
x=2 =>3*y=90 =>y=30 =>z=100-30-2=68.
And so on...

Answer 5

let the number of sheep bought be A , goats be B and chikens be C
that is
sheep------A
goats------B and
chikken------C

so if you multiply the number of sheep to the there prices will give you $100 and add the number of sheep it gives you 100

A + B + C = 100 ------------eqn 1

A*3 + B*2 + C*0.5 = 100---eqn 2

there is one information missing that will link and of the two
to get the third formula

Answer 6

If the scholars present day grade is a ninety% then we’re of direction assuming it’s out of one hundred% ninety/one hundred Now via fact the scholars grade better 20% from the final grading era then you definately upload that to the previous one hundred% supplying you with ninety/ a hundred and twenty = .seventy 5 and because grades are in possibilities you multiply .seventy 5 by applying one hundred supplying you with seventy 5% via fact the scholars previous grade