Question

Dave's sister baked $3$ dozen pies of which half contained chocolate, two thirds contained marshmallows, three-fourths contained cayenne, and one-sixths contained salted soy nuts. What is the largest possible number of pies that had none of these ingredients?

Answer 1

Number of pies = 3×12 = 36

Pies contained chocolate = $\frac{1}{2}$×36 = 18
Pies contained marshmallows = $\frac{2}{3}$×36 = 24
Pies contained cayenne = $\frac{3}{4}$×36 = 27
Pies contained soy nut = $\frac{1}{6}$×36 = 6

Pies that don't contain any of these 36-(18+24+27+6) = 25 pies