At Matilda's birthday party, the ratio of people who ate ice cream to people who ate cake was $3: 2$. People who ate both ice cream and cake were included in both categories. If 120 people were at the party, what is the maximum number of people who could have eaten both ice cream and cake?
(A) 24
(B) 30
(C) 48
(D) 80
(E) 72
Solution 1

Since we want the maximum number of people, we start with the largest of the given choices and see if we can make it work.

Is it possible that 80 people could have eaten both ice cream and cake?

If so, there would be at least 80 people who ate cake and at least $\frac{3}{2}(80)=120$ people who ate ice cream. (Note that there can be overlap between these two sets of people.)

Is this possible?

It is if we say that exactly 120 people ate ice cream and exactly 80 people ate cake, with all of these 80 eating ice cream as well.

So an overlap of 80 is possible, which means that the answer is (D).

## Solution 2

Suppose that the there were $x$ people who ate cake only, $y$ people who ate ice cream only, $b$ people who ate both cake and ice cream, and $n$ people who ate neither.

We know that $x+y+b+n=120$ and so $x+y+b=120-n$.

We also know that the total number who ate cake was $x+b$ and the total number who ate ice cream was $y+b$, so $\frac{y+b}{x+b}=\frac{3}{2}$ or $2(y+b)=3(x+b)$ or $2 y=3 x+b$ or $y=\frac{3}{2} x+\frac{1}{2} b$.

Therefore, $\frac{3}{2} x+\frac{1}{2} b+x+b=120-n$ and so $\frac{5}{2} x+\frac{3}{2} b=120-n$.

Multiplying both sides by 2 , we obtain $5 x+3 b=240-2 n$.

Since $x, b$ and $n$ are non-negative, then the left side is at most 240 and so $b$ can be at most $\frac{1}{3}(240)=80$, if $y=n=0$.

We saw in Solution 1 that $b=80$ is possible.

ANSWER: (D)