At a picnic, 78% children ate sandwiches, 82% ate cake, 61% ate pizza

Question

At a picnic, 78% children ate sandwiches, 82% ate cake, 61% ate pizza and some children ate fruit. If 30% children ate all four things, what is the maximum number of children that could have eaten only fruit?

(A) 12%
(B) 18%
(C) 24%
(D) 30%
(E) 32%

(A) 12%
(B) 18%
(C) 24%
(D) 30%
(E) 32%

KarishmaB · Accepted Answer

Looks like this question turned out to be harder than I imagined. I made it for a few of my students to help them internalise the concept of maximum/minimum in sets. Here is how we can solve it logically:

We need to maximize “only fruit” which means that we need to maximise that part of the fruit circle that does not overlap with anything else. So, we should try to pack everything else together such that we have maximum children left for "only fruit". 
We know how to pack everything together – we just put them in one circle after another in a venn diagram. 
So, the 78% sandwiches circle is put inside the 82% cake circle and the 61% pizza circle can be put inside the 78% sandwiches circle. We have one constraint – 30% children ate all four things. So, 30% children of the pizza circle would have had fruit but not the other 31%.
The 18% children outside the cake circle can now have only Fruit.

Answer (B)

Note that there are many other ways that will give the same answer but since we need to maximise, we can twist the situation to make it easy for us to visualise.

chetan2u · Answer

We can solve it by overlapping sets for 4 items. But since these are not tested on GMAT, let us try to solve it logically.

Let f be the number of children eating only fruits.
 
 If 82% are having cake, then surely children eating ONLY fruits  ,  f\leq{100-82} . Thus all options greater than 18 can be eliminated.

Only A and B are left.
 We are looking at the maximum value of f, so if I did not know how to proceed further or I am short of time, I would mark B and move ahead.  Reason for this is that there are no restrictions on number of children who eat exactly two items or on number of children who eat only one item.

Let us concentrate on only children eating sandwiches, cakes and pizza. Let it be x.
We have to minimize the value of x to get maximum value of  f .

So, now 
(i) 78-30% or 48% children ate sandwiches alone or one more item with it.
(ii) 82-30% or 52% ate cake alone or one more item with it.
(iii) 61-30 or 31% ate pizza alone or one more item with it.
 If we maximize children who ate 2 fruits, the value of x will be minimum. 
 
Let children who ate both pizza and sandwich and not cake be y. 
Then  48-y ate both pizza and cake  and not sandwich, and  31-y ate both cake and sandwich , and not pizza.
So  48-y+31-y\leq{52} 
 79-2y\leq{52}  
71-2y should be odd, so max possible value = 79-2y=51.....y=14

Thus 
(i) children who ate both pizza and sandwich, and not cake =y=14
(ii) children who ate both pizza and cake, and not sandwich= 48-y=34
(iii) children who ate both sandwich and cake, and not pizza= 31-y=17
(iv) Children who ate only cake = 52-34-17=1

So, 82% had all the three items, and remaining 18% is the maximum value for f.

B

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At a picnic, 78% children ate sandwiches, 82% ate cake, 61% ate pizza

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