magical cook wrote:There are 68 children in the cafeteria of a school and all of the children have something for lunch. Thirty-four of the children brought lunches from home, 23 of the children bought a drink from the cafeteria beverage machine, and 32 of the children bought fruit in the cafeteria. If 18 children did at least 2 of these things, how many children did exactly two of these things?
(A) 3
(B) 6
(C) 9
(D) 13
(E) 15
This is an overlapping sets question.
Let's start by simplifying. If 18 children did at least two of these things, all children did at least one (everyone had something to eat), and we have 68 children, then 50 children did exactly one of these things (68-18).
If we add up all the "things", we get 34 + 23 + 32 = 89. If 50 kids did exactly one each, then we have 89 - 50 = 39 "things" done by the remaining 18 kids.
The easiest way to proceed from this point is to backsolve - i.e. plug in the choices. When we backsolve, we generally want to start with (b) or (d). My spidey sense tells me that I want a bigger answer, so I'm going to start with (d).
(d) 13
If 13 kids did exactly 2, then that would account for 26 of the items, leaving us with 5 more kids for 13 (39 - 26) items. The kids in this group who don't do exactly 2 items are doing 3 items. Does 5 * 3 = 13? Nope, eliminate (d).
We ended up with too many things done (13*2 + 5*3 = 41, not 39). To decrease the number of things done, we want more kids doing 2 things and fewer doing 3 things. Since the only answer bigger than (d) is (e), (e) must be correct.
On test day you'd choose (e) without checking it, but for the sake of completeness:
(e) 15
If 15 kids do exactly 2 things, that's 30 out of the 39 accounted for. We have 3 kids left doing 3 things each, so that's 9 more. 30 + 9 = 39.. perfect, (e) is correct.
