PS

Juice=J
Snacks=S
Food=F

Let's first consider just in terms of J Or S:
\(n(J \hspace{3} \cup \hspace{3} S)= n(J)+n(S)-n(J \hspace{3} \cap \hspace{3} S)\)
\(n(J \hspace{3} \cup \hspace{3} S)=90\) {:Because 10 is taken by only food}
\(n(S)=80\)
\(n(J \hspace{3} \cap \hspace{3} S)=40\)

\(90=n(J)+80-40\)
\(n(J)=50\)

Now, let's consider "Only Juice" & "Only Snacks":

\(n(only \hspace{3} J)=n(J)-n(J \hspace{3} \cap \hspace{3} S)=50-40=10\)
\(n(only \hspace{3} S)=n(S)-n(J \hspace{3} \cap \hspace{3} S)=80-40=40\)

Now, we can fit in the % of people who had (food&Snacks) & who had (food&juice). Note: We can't fit food consumers with \(n(J \hspace{3} \cap \hspace{3} S)\)

So,
Maximum number who had \(n(F \hspace{3} \cap \hspace{3} S)=40\)
Maximum number who had \(n(F \hspace{3} \cap \hspace{3} J)=10\)
People who had only food=10

Combined:
Maximum people who could have had food=40+10+10=60

Ans: "E"