[Guest]

I am not sure if this is considered a puzzle or not, but it is to me. So here goes:

A farmer goes to market and buys 100 animals at a total cost of $1,000. If cows cost $50 each, sheep cost $10 each, and rabbits cost 50 cents each, how many of each kind does he buy? Please kindly show the STEPs that lead you to the final solution. In addition, please note that the solution of he buys only 100 sheep is unacceptable.

[Guest]

I am not sure if this is considered a puzzle or not, but it is to me. So here goes:

A farmer goes to market and buys 100 animals at a total cost of $1,000. If cows cost $50 each, sheep cost $10 each, and rabbits cost 50 cents each, how many of each kind does he buy? Please kindly show the STEPs that lead you to the final solution. In addition, please note that the solution of he buys only 100 sheep is unacceptable.

he bought 100 sheep. 1000/100 = 10

[Guest]

Given 3 variables (Cows, Sheep, Rabbits) and only 2 equations:

C+S+R=100

50C+10S+0.5R=1,000

There are virtually an infinite number of solutions. We need to have at least one more equation.

Edited April 29, 2011 by jagdmc

[Guest]

you need 80 Rabbits, 19 cows and 1 sheep.

you know that rabbits must be purchased in lots of 20, so you only have 20,40,60,80 rabbits available. With these options you have two equations and two unknowns which you can solve for in terms of cows and sheep. You try each of the 20/40/60/80 and solve for integer numbers of cows/sheep. Turns out the results are above.

[Guest]

he bought 100 sheep. 1000/100 = 10

In addition, please note that the solution of he buys only 100 sheep is unacceptable.

Actually, mordimar, I would have taken your answer as correct, but given this last constraint...

[Guest]

c=number of cows

s=number of sheep

r=number of rabbits

c+s+r=100

50c+10s+0.5r=1000

c+s+r=100=>c=100-s-r=>1000=50(100-s-r)+10s+0.5r=5000-50s-50r+10s+0.5r=5000-40s-49.5r

so far:

1000=5000-40s-49.5r=>4000=40s+49.5r=>100=s+(99/80)r so r=80 and s=1 could work, this means that c=19 by the first equation. This also fits the second equation so it is a solution.

19 cows, 1 sheep and 80 rabbits.

[curr3nt]

you need 80 Rabbits, 19 cows and 1 sheep.

you know that rabbits must be purchased in lots of 20, so you only have 20,40,60,80 rabbits available. With these options you have two equations and two unknowns which you can solve for in terms of cows and sheep. You try each of the 20/40/60/80 and solve for integer numbers of cows/sheep. Turns out the results are above.

Answer is correct but you didn't show why rabbits must be in lots of 20.

@Nutty - Not sure why you listed this here AND in homework help.

Edited April 29, 2011 by curr3nt

[Guest]

Answer is correct but you didn't show why rabbits must be in lots of 20.

@Nutty - Not sure why you listed this here AND in homework help.

Divide all prices by 10$, you will have each rabbit being a nickle and the others being in whole $$$$, Therefore, you need lots of 20 to get that cost up to full $$$$.

[Guest]

@Nutty - Not sure why you listed this here AND in homework help.

Oh I'm sure, you're sure. It surely was obvious to me just from reading the OP (he basically said show your work like we math teachers like to say )

[Guest]

James22 and MattB, you TWO have shown the way to the solution sought correctly. Therefore, James22 and MattB are at this moment declared as the "Winners". =) To all others, thank you for participating. There are more than one way to approach this problem, and I have seen two different approaches so far. I have seen two other different approaches, and so there are 4 different approaches to get to the final solution.

[Guest]

I think there are a multiple of solution to this but I going to show you the first one I found while I was trying to figure out the puzzle for myself however when I looked at others people solutions they were all the same so please correct me if I'm wrong.

20 Rabbits, 9 Sheep, 18 Cows

Since the rabbits are 50 cents each you need at least buy at least 2 rabbits to spend a whole dollar but because sheep and cows cost a double digit number of dollars you will need to buy enough rabbits so that the cost of rabbits is also a double digit number. This means you'll need to buy at least 20 rabbits or 40, 60, 80, 100 ...

I started off with 20 rabbits and worked from there.

First- 20 Rabbits = $10Second- I decided to bring the total cost so far ($10) up to a three digit number because if I wanted to buy more than one cow then the cost of cows would be a three digit number. I needded to add at least $90 dollars for the total cost to be a three digit number which meant I would need to buy at least 9 sheep. Once again I decided to go with the first possible option so...

9 Sheep = $90 (total cost so far $100)

Thirdly- I needed to spend $900 on the cows so the total price would be $1000. All this took was a simple equation- 900/5 which gave me the answer 18 so.....

18 Cows = $900 (total cost $1000)

I have a feeling I've probably miss read the puzzle or missed something out so please correct me and show me where I wen't wrong. Thx xx xx

Edited April 29, 2011 by Riddle_Lover

[Guest]

Riddle_Love, I think you misread the problem. The problem says that the farmer bought a total of 100 different animals with a total of 1,000 dollars.

Your solution of 20 rabbits, 9 sheep, and 18 cows only add up to 47 animals which is not even close to the given total of 100 animals.

Edited April 29, 2011 by NuttyNumbers

[Guest]

@Nutty - Not sure why you listed this here AND in homework help.

Curr3nt, I have always respected you for your kindness and mathematical abilities. However, I can reassure you that this puzzle and the equations problem in the homework forum are completely different. I think you can prove my statement above because the solution to this puzzle has to be positive numbers while the solution to the equations problem is not necessarily exclusively positive numbers.

[Guest]

Oh I'm sure, you're sure. It surely was obvious to me just from reading the OP (he basically said show your work like we math teachers like to say )

I don't exactly know what your meaning is, so I don't know how to react to it at all. I have seen different approaches to this problem and I am just curious how many more are out there. Do not let Curr3nt's comments mislead you. Please read post #13 and if you would like you can check out the homework forum to find out the truth yourself.

[Guest]

The equation may be written as 50c + 10s + r/2 = 1000 such that c is the number of cows, s is the number of sheep and r is the number of rabbits. As the number of each animal should be an integer, we can multiply each side of the equation by 2 to remove the fraction:

100c + 20s + r = 2000.

As seen by inspection, and as both the number of cows, sheep and 2000 are multiples of 20, the number of rabbits must be a multiple of 20.

There are not an infinite number of solutions but 1070 solutions (less one, as 0 cows 100 sheep and 0 rabbits has been deemed unacceptable). If at least one of each type of animal were purchased the number of solutions drops to 931 solutions.

For the minimal number of livestock purchased, such that c >= 1, s >= 1 and r >=1, the solution is: Cows = 19, Sheep = 4, Rabbits = 20 (Total livestock = 43 animals).

For the minimal number of livestock purchased, such that r >= s >= c >= 1, the solution is:

Cows = 16, Sheep = 19, Rabbits = 20 (Total livestock = 55 animals).

For the maximal number of livestock purchased, such that c >= 1, s >= 1, and r >= 1, the solution is: Cows = 1, Sheep = 1, Rabbits = 1880 (Total livestock = 1882 animals).

[Guest]

Sorry. While I had to step away from the computer, my daughter had submitted the incomplete post.

The equation may be written as 50c + 10s + r/2 = 1000 such that c is the number of cows, s is the number of sheep and r is the number of rabbits. As the number of each animal should be an integer, we can multiply each side of the equation by 2 to remove the fraction:

100c + 20s + r = 2000.

As seen by inspection, and as both the number of cows, sheep and 2000 are multiples of 20, the number of rabbits must be a multiple of 20.

There are not an infinite number of solutions but 1070 solutions (less one, as 0 cows 100 sheep and 0 rabbits has been deemed unacceptable). If at least one of each type of animal were purchased the number of solutions drops to 931 solutions.

For the minimal number of livestock purchased, such that c >= 1, s >= 1 and r >=1, the solution is: Cows = 19, Sheep = 4, Rabbits = 20 (Total livestock = 43 animals).

For the minimal number of livestock purchased, such that r >= s >= c >= 1, the solution is:

Cows = 16, Sheep = 19, Rabbits = 20 (Total livestock = 55 animals).

For the maximal number of livestock purchased, such that c >= 1, s >= 1, and r >= 1, the solution is: Cows = 1, Sheep = 1, Rabbits = 1880 (Total livestock = 1882 animals).

As there were 100 animals purchased, we have the second equation: c + s + r = 100.

100c + 20s + r = 2000

-1*(c + s + r) = -1*(100)

=========================

99c + 19s = 1900

99c = 1900 - 19s

99c = 19*(100 - s)

c = (19/99)*(100 - s)

As one can see, in order for c to be a non-negative integer, (100 - s) must be a multiple of 99 yet s can not be greater than 1, thus s must be equal to 1.

c = (19/99)*(100 - 1) = (19/99)*(99) = 19

And c, therefore equals 19.

Plugging the values back into the original equation:

50*(19)+ 10*(1) + r/2 = 1000.

950 + 10 + r/2 = 1000

960 + r/2 [ -960] = 1000 [ -960]

r/2 [*2] = 40 [*2]

r = 80

Thus, the number of livestock purchased was

Cows = 19, Sheep = 1, Rabbits = 80.

Edited April 30, 2011 by Dej Mar

[Guest]

Dej Mar, your final answer is correct. I enjoy reading the way you procedurally solve this problem. Thanks for participating. Just reading your steps alone gives me a brain workout already.

[Guest]

Thx Nutty for correcting me. I knew I must have misread the puzzle because if the puzzle only asked you to find a way to spend the $1000 then it'd be incredibly easy. Now that I realise that you need to buy 100 animals with a total cost of $1000 I agree with James22 and MattB. I'm not going to show working but I'll tell you that the way I figured it out was the way I showed in my last reply, then going through all the possible ways of spending $1000 until I found one that bought 100 animals. It's a really way of going around it but that's how I done it.xx

ANSWER - 80 Rabbits, 19 Cows and 1 sheepa

WORKING - Not showing for same reasons as written above

Thx alot for correcting me xx