Korean Job Discussion Forums

[Qinella]

In the spirit of stultifying, perilous word problems we all love so much, I present to you one I solved tonight. Feel free to post your answer here, and the logic you used!







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Roosters: $5 each
Hens: $1 each
Chicks: $0.05 each

You have $100. You must buy exactly 100 chickens, at least one of each kind, spending exactly $100.

How many of each do you buy?











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Do us proud, Davers!

[cosmo]

100 hens, duh.

[Qinella]

Gotta buy at least one of each.

30 views and no answers yet.. where's blynch when you need him???!!!

[cosmo]

I thought it was a trick question.

Since its a real math problem, I'll let someone else do it.

[cosmo]

A farmer has 100 dollars and needs to buy exactly

100 animals. He can buy Cows which cost $10 each, Pigs

which cost $3each, and Chickens which cost $0.50 each.

He must buy exactly 100 animals. He must buy at least one

of each of the three types of animals. He must spend all

of the $100. How many of each type does he buy?

Solution:

Let c = # of cows, p = # of pigs, k = # of chickens

The constraints are:

c>=1, p>=1, k>=1; c,p,k are integers.

c + p + k = 100

10.00c + 3.00p + 0.50k = 100

Since cows cost $10, the solution must contain a limited

number of them, So look at their possibilities first.

Using the process of elimination, we cannot purchase more

than 5 cows because 6,7,8,9,or 10 cows in the solution will

not leave enough money to buy the remaining 94,93,92,91, or

90 amimals needed:

10 cows = $100 leaves $0 dollars for 90 more animals.

9 cows = $90 leaves $10 dollars for 91 more animals.

(could buy at most 20 chickens)

8 cows = $80 leaves $20 dollars for 92 more animals.

(could buy at most 40 chickens)

7 cows = $70 leaves $30 dollars for 93 more animals.

(could buy at most 60 chickens)

6 cows = $60 leaves $40 dollars for 94 more animals.

(could buy at most 80 chickens)

5 cows = $50 leaves $50 dollars for 95 more animals.

(could buy 100 chickens)

So, the highest feasible value for c is 5.

Substituting this into the two equations, we get

5 + p + k = 100

50 + 3.00p + 0.50k = 100

which rearranges to:

p + k = 95

3.00p + 0.50k = 50

and solves by either substitution or elimination to

p =1, k = 94, giving a solution of:

5 cows = $50.00

1 pig = $ 3.00

94 chickens = $47.00

--------------------------

100 animals $100.00

If we similarly try c = 4,3,2,and 1, the resulting two

equations in p and k do not possess integer values for

both p and k, so this is the only possible real-life

solution.

[kprrok]

The way the original post is written, it doesn't matter. 100 chickens (the ones you must) buy equals $5, then use the remaining $95 to buy whatever animals you want.

KPRROK

[Qinella]

Cosmo: Was that a copy and paste job? Be honest~

kprrok: I like the logic you used for that. You initially understood that whatever combination of chicks and hens purchased must equal a number that leaves a remainder divisible by 5, right?

[cosmo]

Yes, and the worst part is that it was worth ONLY 5 points on an exam.

[Qinella]

[dogshed]

I have 100 hens for $100.
I need one of each.
I trade one hen for 20 chicks and now I have 19 extra.
Every time I trade a hen for 20 chicks I have 19 extra animals.
If I trade 5 hens for 1 rooster I get 4 less animals.
After playing around a bit I get that the smallest number divisible
by 19 and 4 is 19 x 4 which is 76.
I start over.
I get 19 roosters for 19 x 5 = $95 (76 less)
I get $4 worth of chicks = 80 chicks. (76 more)
I have 99 birds and I spent $99.
I add one hen.
All it took was a nap.
I have no idea how to do it with algebra.

[kprrok]

[dogshed]

[cbclark4]

I tried this 5X+X+X/20=100

But then I forgot how to work those numbers.

So I just opened a spreadsheet and shot numbers at it.


19 Rooster
1 Hen
80 Chicks

So what should my formula have been?

x+y+z=A

5x+y+z/20=A

Where A is 100

I think?

cbc

[mithridates]

80 chicks = $4
1 hen = $1
19 roosters = $95

Edit: Good, somebody else got it. I promise I didn't look.

[bosintang]