In a village of 100 households, 75 have at least one DVD player, 80 have at least one cell phone, and 55 have at least one MP3 player. If x and y are respectively the greatest and lowest possible number of households that have all three of these devices, x – y is:

(A) 65

(B) 55

(C) 45

(D) 35

(E) 25

You only have 100 households so there has to be overlap considering that the numbers, 75, 80, and 55 sum to 210. Let’s look at the DVD Player and Cell Phone groups first because they are the biggest and will create the most obvious constraints. They sum to 155. Considering that there are only 100 households it must be that there are at least 55 overlaps, people who have both DVD and Cell. You could have the MP3 group completely contained in that 55. In that case, 55 people have three of the devices. That’s the max that have all three of the devices.

Now let’s try to minimize the people who have DVD, Cell Phone, and MP3 (I call these triples). With that in mind we want to make the MP3 people as independent as possible. There are 25 people who DON’T have a DVD (100 – 75) and 20 people who DON’T have a Cell Phone (100 – 80). That’s 45 people total. Now, some of the 25 no DVD *could be *a part of the 20 Cell Phone but they don’t have to be. The 20 and 25 could be independent and in this case we want those groups as unique as possible to maximize the number of people who are in ONLY MP3 player. Those 45 people *could *be ONLY MP3 player. So, because there are then ten leftover MP3 Player people (55 – 45) those 10 must be in all three groups. 55 (max in all three) – 10 (min in all three) = 45. **Choose C.**

You could also use the three group overlapping sets formulas:

Group A + Group B + Group C + None – Doubles – 2(Triples) = Total

210 – d – 2t = 100

-d – 2t = -110

d + 2t = 110

So now we know that that sum of d + 2t is 110. So the max t could be is 55 (2*55 = 110). A quick common sense check verifies that it is possible that all of the MP3 people are contained in the overlap between the DVD and Phone people. Now let’s try to maximize the doubles. Can they be 110? Nope. You only have 100 households total. Can they be 100? Nope. Then you’d have 5 triples to balance out the equation. Again your max d + t is 100. Can they be 90? Then t is 10. That could work 90 + 10 = 100.

You can also visualize with Venn diagrams. For the most part I don’t like Venns for three group (or two group) sets but for this one it can actually work relatively well. I’ll circle back with a Venn Diagram solution.