# Math

There is not enough information to completely solve the problem as given.

Let M represent customers who ordered margaritas, Q customers who ordered quesadillas, and Ch customers who ordered chili.

Let a be people only in M, b people in `M nn Ch` , c people who ordered all 3,...

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There is not enough information to completely solve the problem as given.

Let M represent customers who ordered margaritas, Q customers who ordered quesadillas, and Ch customers who ordered chili.

Let a be people only in M, b people in `M nn Ch` , c people who ordered all 3, d people in Ch only, e people in `Ch nn Q` , f people in Q only, and g people in `M nn Q` .

Draw a Venn diagram with 3 circles labeled M,Q, and Ch.

We have the following:

a+b+c+d+e+f+g=100-23=77

g=13 (13 people ordered M and Q, but not Ch.)

c+e+f+g=41 => c+e+f=28 (Q has 41 people total.)

d+e+f=29 (29 people ordered Ch and Q but not M)

b+c=26 (26 people ordered Ch and M)

b+c+e+g=46 => b+c+e=33 (46 people ordered at least 2)

Since b+c=26 e=7.

Now a+b+c+d+e+f+g=77; we know g=13 and e=7 so
a+b+c+d+7+f+13=77 => a+b+c+d+f=57
But b+c=26 and d+f=22 (d+e+f=29 and e=7) so a=9.

So far we have a=9,e=7,g=13.

Our remaining equations are:

c+f=21
d+f=22
b+c=26 which is 3 equations and 4 unknowns. There cannot be a unique solution for these equations.

Suppose c=13; then b=13, d=14, and f=8. This solution satisfies all constraints:

c+e+f+g=13+7+8+13=41 so 41 ordered Q.

d+e+f=29 so 29 ordered Ch and Q.

b+c=26 so 26 ordered Ch and M .

g=13 so 13 ordered M and Q but not Ch

b+c+e+g=13+13+7+13=46 so 46 ordered at least 2

a+b+c+d+e+f+g=9+13+13+14+7+8+13=77 so 23 ordered none.

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Another solution would be a=9,b=14,c=12,d=13,e=7,f=9,g=13

You will see that the constraints are again satisfied.

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