not enough information.
How many different animals, how much do those animals cost?
Look at this post, you question is very similar.
From the information you've given, the only thing i can come up with is that the animals cost one dollar each since it's 100 dollars for 100 animals
As stated, your problem does not provide sufficient information.
Consider one of the typical ways this type of problem is stated:
You have $100 dollars and you need to buy exactly 100 animals. Each chicken cost .50 cents, each pig cost $3.00 and each cow cost $10.00. How many of each animal can you buy. Remember you have $100 and need exactly 100 animals.
This is one of the oldest of this type of problem. I will show you three ways to derive the solution, two by simple observation or logic and the other the analytical method. The other stipulation that must be made with the problem is that he bought at least one of everything. I'll give it to you in terms of x, y, and z items.
Restating, you are to buy three items, say x, y, and z (all integers) and the sum of these items is 100. You have $100 to spend on the three items with their individual costs being $10.00, $3.00, or $.50, respectively. You are to buy at least one of each item.
With all of this information, you can derive the following relationships:
(1)--x + y + z = 100 (1)
(2)--10x + 3y + .5z = 100 (2)
3---Multiplying (1) by .5 yields .5x + .5y + .5z = 50 (3)
4---Subtracting (3) from (2) yields 9.5x + 2.5y = 50 (4)
5---Dividing (4) through by 2.5 yields 3.8x + y = 20 (5)
6---By observation, the only value of x that will make 3.8x an integer is 5
7---Therefore 3.8(5) + y = 20 making y = 1
8---Thus, z = 94.
Check: 5 + 1 + 94 = 100 and 10(5) + 3(1) + .5(94) = 50 + 3 + 47 = 100. Okay.
Another way of viewing it is to say that the average cost of each item is $1.00. Each x's cost differs from the average by +$9.00, each y's cost by + $2.00, and each z's cost by - $.50. Therefore, for each x he must buy 18 z's and for each y he must buy 4 z's. Thus, since 5(1 + 18) + (1 + 4) = 100, he must buy 5 x's, 1 y, and 94 z's.
(3)---Multiplying (2) by 2 yields 19x + 5y = 100 (3)
(4)---Dividing (3) through by 5 yields 3x + 4x/5 + y = 20 (4)
(5)---Rearranging, 4x/5 = 20 - 3x - y
(6)---By definition, 4x/5 must be an integer but we want a unit coefficient for x.
(7)---Multilply 4x/5 by 4 and divide by 5 again yielding 16x/5 = 3x + x/5.
(8)---x/5 must also be an integer so let x/5 = k making x = 5k
(9)---Substituting back into (3) we get 95k + 5y = 100 or y = 20 - 19k
(10)---In order to have at least one of each, clearly k can only be 1 making
(11)---k 0 1 2
x 0 5 10
y 1 1 neg neg = negative
z 99 94 neg
(12)---Therefore the only valid solution is x = 5, y = 1, and x = 94.
If you are truly interested in the historical and/or further aspects of recreational mathematics, I strongly reccommend you send for the Dover Publications catalog of Math and Science books. They publish one of the widest varieties of math books I have run across, many of them, reprints of classics in the field. Their address is 31 East 2nd Street, Mineola, NY 11501, Tel. No. 516-294-7000. The linear indeterminate problem methodology is best described in the book titled, Number Theory and Its History by Oystein Ore.
There might very well be many other books available in your school, or local, library or through your teachers. These, will probably be able to give you additional information on the specific methods for solving Linear Diophantine problems. If you are interested in exploring other problems of this type, with 2, 3, and 4 unknowns, I offer you some problems below to pursue on your own. Some of them were used in the examples above. I would first advise you to get some references from Dover or your library and learn the procedure unless you have been able to learn it from the above examples. The methods allow you to solve many problems that, at first glance, appear unsolvable.
Linear Indeterminate Algebra or Linear Diophantine Problems
1----A man buys 100 diamonds for $10,000. Some cost $50 apiece, some $300 apiece and some $1000
apiece. Assuming he bought at least one of each, how many of each kind did he buy?
2----A farmer must buy 100 head of animals with $100. Calves cost $10 each, lambs cost $3 each, and pigs
cost 50 cents each. If the farmer buys at least one of each kind of animal, how many of each kind does he buy?
3----Your boss has asked you to purchase three different types of ballpoint pens. The first costs 50 cents,
the second $5.50, and the third $9.50. He has given you $100 and told you to purchase 100 pens in any
combination as long as you spend exactly $100 for 100 pens. Just one solution is possible. Can you find it?
4----A merchant has three items on sale-a radio for $50.00, a clock for $30.00, and a flashlight for $1.00. At
the end of the day, she has sold a total of 100 of the three sale items and has taken in exactly $1000.00 on the
total sales. How many of each item did she sell?
5----When 100 bushels of grain are distributed among 100 persons so that each man receives three bushels,
each woman two bushels, and each child half a bushel, how many men, womenm and children are there?
6----One duck may be bought for $5.00, one chicken for $1.00, and 20 starlings for $1.00. You have $100
and are asked to buy 100 birds. How many of each kind will you buy?
7----At an inn, a party of 20 persons pay a bill for $20. The party consists of men, women, and maidens,
each man paying $3, each woman $2, and each maiden 50 cents. How was the party composed?
8--(590)--What is the multiplier by which two hundred and twenty-one being multiplied and sixty-five added to the
product, the sum divided by one hundred and ninety-five becomes exhausted?
9----A school has a theater with a seating capacity of 100. It wishes to admit 100 people in such a proportion
that will enable him to take in $1.00 with prices as follows: men 5 cents, women 2 cents, children 10 for one cent.
How many of each must be admitted?
10----A party of 41 persons, men, women, and children, take part in a meal at the inn. The bill is for $40 and
each man pays $4, each woman $3, and every child 1/3 of a dollar. How many men, women, and children were
I dont know
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