1. Let m = the number of men (an integer)

2. Let w = the number of women (an integer)

3. Let c = the number of children (an integer)

4. Number of bushels = 3m+2w+.5c = 100

5. Number of people = m+w+c=100

6. 0<m<100

7. 0<w<100

8. 0<c<100

If we assume that there are 10 children, then we can solve for two equations and two unknowns:

1. 3m+2w+5 = 100

2. m+w+10=100

We can solve this by subtracting two times the second equation from the first equation:

1. 3m+2w+5 = 100

2. 2m+2w+20=200 (this is two times the second equation shown above)

3. m          - 15 = -100 (this is the result of subtracting two times the second equation from the first equation)

4. m = -85

5. w=100-10- (-85) = 175

The problem with this result is that the values for m and w are outside the allowable range, so clearly the assumption that there are 10 children is wrong. Let’s try increasing the number of children to 80 and see how that impacts the result from m:

1. 3m+2w+40 = 100

2. 2m+2w+160=200

3. m          -125 = -100

4. m                 =  25

5. w = 100-80-25=-5

Here, the result for m is within the range, but the value for w is outside the range. Let’s try reducing the number of children to 70:

1. 3m+2w+35 = 100

2. 2m+2w+140=200

3. m          -105 = -100

4. m                    =    5

5. w = 100-70-5= 25

Now we have a solution that makes sense (m=5, w=25, c=70). However, there may be other answers that work, so let’s try increasing the number of students to 72 and see what happens:

1. 3m+2w+36 = 100

2. 2m+2w+142=200

3. m          -106 = -100

4. m                    =    6

5. w = 100-72-6= 22

This solution also makes sense (m=6, w=22, c=72). Clearly, there can be several correct solutions to this question.