corn distribution word problem-please help!

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evergreens

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Corn Distribution word problem

100 bushels of corn are to be divided among 100 men, women and children. Men get 3 bushels each, women get 2 bushels each, children get 1/2 bushel each. If there is at least one person from each category (men, women, children), what are the possible numbers of people in each category? Find three solutions.

Men Women Children Total People
1. 100
2. 100
3. 100
 
Hello, evergreens!

Corn Distribution word problem

100 bushels of corn are to be divided among 100 men, women and children.
Men get 3 bushels each, women get 2 bushels each, children get 1/2 bushel each.
If there is at least one person from each category (men, women, children),
what are the possible numbers of people in each category?
Find three solutions.
Click to expand...

Let \(\displaystyle M\) = number of men
Let \(\displaystyle W\) = number of women
Let \(\displaystyle C\) = number of children.

Since there are 100 people: \(\displaystyle \;M\,+\;W\,+\;C\;=\;100\)

. . . . . Distribution of corn: \(\displaystyle \:3M\,+\,2W\,+\,\frac{1}{2}C\;=\;100\)


Multiply the second equation by 2: \(\displaystyle \;\;6M\,+\,4W\,+\,C\:=\;200\)

. . . . . Subtract the first equation: \(\displaystyle \;-M \,- \;W\,-\;C\;=\,-100\)

And we get: \(\displaystyle \:5M\,+\,3W\;=\;100\;\;\Rightarrow\;\;W\:=\:\frac{100\,-\,5M}{3}\)

Since \(\displaystyle W\) is a positive integer, \(\displaystyle M \:=\:2,\,5,\,8,\,11,\,14,\,17\)

There are six solutions.

Code: 
    Men    Women    Children    Total
1.   2      30         68        100
2.   5      25         70        100
3.   8      20         72        100
4.  11      15         74        100
5.  14      10         76        100
6.  17       5         78        100
 
How come you can't have 1 man, 1 woman, and 190 children? The problem says at least one from every catagory :?:
 
RTQ ... the total of men, women, and children has to equal 100.
 
evergreens said:
100 bushels of corn are to be divided among 100 men, women and children. Men get 3 bushels each, women get 2 bushels each, children get 1/2 bushel each. If there is at least one person from each category (men, women, children), what are the possible numbers of people in each category? Find three solutions.
Click to expand...
1--Let M, W, and C be the respective numbers of people.
2--Then, M + W + C= 100
3--Also, 3M + 2W + C/2 = 100
4--Multiplying (3) by 2 and subtracting (2) fro the result results in 5M + 3W = 100
5--Dividing through by the lowest coefficient yields 1M + 2M/3 + W = 33 + 1/3
6--(2M - 1)/3 must be an integer as does (4M - 2)/3
7--Divding again yields 1M + M/3 - 2/3
8--(M - 2)/3 must be an integer k making M = 3k +== 2
9--Substitutuing back into (3) yields W = 30 -- 5k
10--k.....0.....1.....2.....3.....4.....5
.....M.....2.....5.....8....11....14...17
.....W...30...25....20...15....10....5
.....C...68...70....72...74....76...78 or 7 solutions in all, as soroban has already pointed out.

1Man, 1Women and 190 Children does not equal 100 people as the problem requires.

The problem states that there must be "at least" 1 of each, not that there must be, or can be 1 of each,meaning that zero of any of the 3 cannot be a solution. For k = 6, we would have 20 Men, 80 Children and no women which cannot be a solution.
 
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