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Question

IMO it should be 60. Can you please share the answer?

fluke · Answer

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"

subhajeet · Answer

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"

I am unable to understand this part. Can someone please explain me this.

fluke · Answer

I am unable to understand this part. Can someone please explain me this.

From the earlier part, we found that people who had 
only Snacks=40
only Juice=10

Our motive is to maximize the number of people who had food. So, why not usher the same number for people who had food as well. Thus, those people whom we thought had just snacks Or just juice actually had food as well.

So,
only Snacks=40 is changed to:
Snacks+food=40

only Juice=10 is changed to:
Juice+food=10

And we know: 10 had only food(given)

So,
40+10+10=60.

Spidy001 · Answer

this can be easily visualized with a venn diagram.

Given J & S = 40, S = 80 => S but not J = 40
 as we are trying to maximize F , all the above 40 could have had F as well. 
 i.e max(S & F ) = 40

All three = 0.
F only = 10
everybody has at-least one. => neither = 0.

=> J but not S = 100 - 80-10 = 10

as we are trying to maximize F, all the above 10 could have had F as well.

i.e max( J & F) =10

=> F max = F only + max(J & F) + max(S & F)
 10+10+40 = 60

Answer is E.

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