math word problem
Cheryl
farmer has $100.00 and is to buy exactly 100 animals. Sheep are
$10.00 each, Goats are $3.50 each and chickens are 0.50 cents each.
Keep coming up with 99 animals. Any ideas???
Thanks.
--
submissions: post to k12.ed.math or e-mail to k12...@k12groups.org
private e-mail to the k12.ed.math moderator: kem-mo...@k12groups.org
newsgroup website: http://www.thinkspot.net/k12math/
newsgroup charter: http://www.thinkspot.net/k12math/charter.html
Joseph Sroka
<Cher...@hotmail.com> wrote:
> have tried and tried but cannot come up with the answer for this. A
> farmer has $100.00 and is to buy exactly 100 animals. Sheep are
> $10.00 each, Goats are $3.50 each and chickens are 0.50 cents each.
> Keep coming up with 99 animals. Any ideas???
>
> Thanks.
What combination of 99 animals did you come up with? Check the total cost
of that combination. Are you saying that the total cost of your 99 animal
combo is $100?
It's late, so I "cheated" and used MS Excel to solve the problem.
Are you expected to solve this problem by hand? By calculator? By using
a spreadsheet program?
--- Joe
--
------------------------------------
Delete the second "o" to email me.
Chergarj
>farmer has $100.00 and is to buy exactly 100 animals. Sheep are
>$10.00 each, Goats are $3.50 each and chickens are 0.50 cents each.
>Keep coming up with 99 animals. Any ideas???
s=sheep; g=goat; c=chicken
Price per unit as given in quoted problem description.
How many animals? s+g+c=100.
How much money as dollars? 10s+3.5g+0.5c=100
Set Up and work with matrix:
[1 1 1 100]
[10 3.5 0.5 100]
.... row operations can give you this matrix:
[1 1 1 100]
[19 6 0 100]
but this seems like a dead end in that you still have three unknowns and two
equations.
What if you let s=k, and that k is a constant (although you do not yet know
what its value is.) Then, using original equations, you can make:
g + c = 100 - k
7g + c = 200 - 20k
Subtract the first of these equations from the second and find that the 'c'
drops out:
6g=100 - 20k - (100 - k)
OR, g=(100-19k)/6
Using s and g, based on the constant k, a generalized solution is
s=k; g=(100-19k)/6; c=(500+13k)/6 (I combined steps and used original animals
equation to obtain c).
Now, to obtain a specific solution, you want some whole number k for which s,
g, and c are also whole numbers, and which conform to the given original
information. I find that k=4 gives a good result:
s=4, g=4, c=92
G C
Darrell
news:6438j09slb1csnt9i...@4ax.com...
> have tried and tried but cannot come up with the answer for this. A
> farmer has $100.00 and is to buy exactly 100 animals. Sheep are
> $10.00 each, Goats are $3.50 each and chickens are 0.50 cents each.
> Keep coming up with 99 animals. Any ideas???
3 years ago I addressed a similar problem on another NG, specifically
whether or not any trial and error is necessary to obtain the final
solution, so I dug up my response and included it below. Although it
certainly should not take much trial and error if one so chose to go that
route, it is not necessary. Although the given information can be modeled
by two equations in three unknowns (at face value indicating some trial and
error may be necessary) remember that all variables must be integers between
0 and 100. The procedure outlined below describes how to use this fact
(integers between 0 and 100) to analytically solve the problem. The types
of animals and prices may vary slightly from your problem, but I believe you
will find the general procedure will work out very much the same when
applied to your problem.
****************************************
response to similar problem posted 9-01
with $10 cows, $5 pigs, and $.50 sheep
****************************************
That's one way of approaching it, but here's another approach that
addresses your original question---can it be done without any trial
and error. The answer is yes!
As we know, just because a system has infinitely many solutions does
not mean it is satisfied by any set of values. I know there are
restrictions on the variables in this particular problem, but I'm
speaking in more general terms for the moment. For ex. consider the
equation x+y=5. There are infinitely many solutions, but not just
*any* replacements for x and y will satisfy the equation. But, if we
choose a value for x (say t) then y is fixed to be 5-t. We say
(t,5-t) is a parametric solution to x+y=5 (a solution given in terms
of a parameter.) Once you have a parametric solution, you can then
use the restrictions of the problem to narrow down the possible
"choices" for the parameter that will satisfy the restrictions of the
problem. In the case of this problem, you can narrow it down to only
*one* possible value.
Using row reduction to isolate c and p, the system:
10c + 5p + (1/2)s = 100
c + p + s = 100
is equivalent to the system:
c - (9/10)s = -80
p + (19/10)s = 180
For the parameter, I'll let s=10t (just to clear fractions):
c = -80 + 9t
p = 180 - 19t
So the parametric solution is:
(c, p, s) = (-80+9t,180-19t,10t).
From that, you can narrow t down to only the possibility(ies) that
satisfies(y) the restrictions on (c,p,s) that each must be integers
between 0 and 100.
Spoiler follows:
Now start narrowing down the possible values for t. Each of these
must be between 0 and 100:
0 <= -80 + 9t <= 100
80 <= 9t <= 180
(80/9) <= t <= 20
0 <= 180-19t <= 100
-180 <= -19t <= -80
(180/19) >= t >= (80/19)
0 <= 10t <= 100
0 <= t <= 10
Considering the top inequality, 80/9 is 8.8.. and since s=10t (and s
must be an integer) then t must be an integer between 9 and 20.
Considering the second inequality, (180/19)~9.47 and (80/19)~4.21 so t
must be an integer between 5 and 9. No need to even consider the last
inequality as obvious as it is------we just narrowed it down to only
one possible value for t and that value is 9. Therefore, the solution
is:
(-80+9t,180-19t,10t)
(1, 9, 90) = (c, p, s)
1 cow, 9 pigs, 90 sheep
Baa,
Darrell
Robert Morewood
: I have tried and tried but cannot come up with the answer for this.
: A farmer has $100.00 and is to buy exactly 100 animals. Sheep are
: $10.00 each, Goats are $3.50 each and chickens are 0.50 cents each.
^^^^ ----------
There is ONLY ** 1 ** way to get 99 animals (for $100 dollars).
?? Either you only tried it ONCE, or
?? you've been using the same numbers
?? and expecting different results!?!
?? (classic definition of insanity!)
?? or your arithmetic is really bad.
Perhaps you should have SHOWN what you were trying.
Are you an arithmetic student or an beginning algebra
student or an advanced algebra student?
Since I refuse to do your homework for you, I will do
the problem that you claim to have already solved:
> If sheep cost $10, goats cost $3.50, and chickens cost $0.50
> and you got 99 animals for $100, how many of each did you get?
An ARITHMETIC student should use trial and error.
Pick three numbers that add up to 100
- Maybe 1 and 2 and 97 ?
and see what 1 sheep, 2 goats, and 97 chickens
would cost: $10x1 + $3.50x2 + $0.50x97 = $65.50
That is not enough. Try three different numbers:
- Maybe 2 and 1 and 97 ?
$10x2 + $3.50x1 + 0.50x97 = $72
That is closer. Keep trying till you find the answer.
Persistance and good arithmetic are required.
Excellent arithmetic practice for elementary students.
Any ALGEBRA student should immediately write:
" Let S = Number of Sheep, "
" G = Number of Goats, and "
" C = Number of Chickens. "
and then translate "you got 99 animals"
into algebra: "S + G + C = 99"
The other translation is a little trickier:
" If sheep cost $10, goats cost $3.50, and "
" chickens cost $0.50, you got animals for $100 "
becomes: " 10 x S + 3.50 x G + 0.50 x C = 100 "
So you have to find three POSTIVE WHOLE numbers
which satisfy that "System of Linear Equations":
10S + 3.50G + 0.50C = 100
S + G + C = 99
Beginning algebra students might now go back to
trial and error, plugging in different numbers.
More experienced students might try substitution,
or subtraction, or matrices. I like subtraction.
Double the first equation and subtract the second:
20S + 7G + 1C = 200
-{ S + G + C = 99 }
----------------------
19S + 6G = 101
Most algebra students would NOW go back to
trial and error to solve this simplier equation.
Really advanced students would use Modular Arithmetic
to finish. Looking at the equation "Mod 6" (disgarding
all multiplies of six) leaves just: S = 5 Mod 6
[I disgarded 18S+6G from one side and 96 from the other.]
That means that S is 5 more than some multiple of 6.
( S = 5 or 11 or 17 or ... )
But since 10S is at most 100, S can only be 5.
Working back: 19x5 + 6G = 101 so G = (101-19x5)/6 = 1
and: 5 + 1 + C = 99 so C = 99-5-1 = 93.
Checking S=5, G=1, and C=93:
10S+3.50G+0.50C = 100 -> 10x5+3.50x1+0.50x93 = 100 Check.
S + G + C = 99 -> 5 + 1 + 93 = 99 Check.
So if you got 99 animals for $100 (and you used postive
whole numbers and didn't make arithmetic mistakes)
you MUST have used 5 Sheep, 1 Goat, and 93 Chickens.
Now YOU try to get 100 animals for $100.
Robert
|)|\/| || Burnaby South Secondary School
|\| |ore...@olc.ubc.ca || Beautiful British Columbia
Mathematics & Computer Science || (Canada)
P.S. You really didn't mean 0.50 cents (that's half a penny!)
You must write either 0.50 dollars OR 50 cents.
(With half a penny instead of 50 cents, there is no solution.)
Bill Dubuque
>
> I have tried and tried but cannot come up with the answer for this.
> A farmer has $100 and is to buy exactly 100 animals. Sheep are
> $10 each, Goats are $3.50 each and chickens are 0.50 cents each.
1) 20s + 7g + c = 200
2) s + g + c = 100
------------------
3) 19s + 6g = 100 via 1) - 2)
3) => s = 4 (mod 6). By 1) s <= 200/20 so s = 4 (s=10 => g=c=0 =><=)
3) => g = 4 (mod 19). By 1) g <= 120/7 so g = 4 (using s = 4)
=> c = 92 by 2) and/or 1)
--Bill Dubuque
Valery
2) s + g + c = 100
------------------
3) 19s + 6g = 100 via 1) - 2)
s<= 100/19; i.e. s <= 4;
19s must be even, if 6g and 100 - even; i.e. s=2 or s=4;
5) s=4; => 6g=24; g=4; (see 4) ) and c=92;
6) s=2; 6g=100-38=62; g=62/6 -it is not natural number;
Reply (s,g,c)=(4,4,92); Valery, Moscow.
------------------------------------------------------
"Bill Dubuque" <w...@nestle.csail.mit.edu> ???????/???????? ? ????????
?????????: news:dnack0p4pl0mj3940...@4ax.com...