households

[GmatKiss]

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

[Stuart@KaplanGMAT]

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

Hello!

First, a key strategy on min/max questions: whenever a question asks you to minimize one thing, maximize others; whenever a question asks you to maximize one thing, minimize others.

Next, let's break this question down into two parts. First, the max number of households that has all three; second, the min number.

The max number is easy - since the smallest number is 55 (MP3s), there's a maximum of 55 households that could have all 3 items.

So, our final answer will be 55 - y. Eliminate (A).

Next, let's focus on the minimum value. If we want to minimize the number of households that have all 3, and we know that there's at least some overlap (since 75+80+55>100), we want to maximize the number of households that have exactly 2 of the 3.

This would be easiest to do with a Venn diagram to help us picture what's going on; sadly, computer-generated diagrams are not my forte! So, we'll close our eyes and visualize (go to your happy place!):

If we have add all 3 components together, we get 210 total objects owned by the households. So, we have 110 extras to distribute.

Our 3 overlapping groups will be:

DVD/Cell
Cell/MP3
DVD/MP3

Keeping our individual numbers in mind, the max we can set our 3 groups to is:

DVD/Cell: 50
Cell/MP3: 30
DVD/MP3: 25

However, that only adds up to 105, so we need at least some people in our triple group.

Since y cannot be 0, eliminate B(55).

Now we got really close (105/110 were used up), so we don't need many in the all-3 group; a smart test taker would just choose (C)45 and move on with their lives.

If, on the other hand, you're reviewing the question and want to prove the answer, we'd test (C) as follows:

For C to be correct, y=10, i.e. there are 10 people with all 3 devices. That means that we have 90 households left with a total of 65 DVDs, 70 Cell phones and 45 MP3s. 65+70+45=180, so we have 180-90=90 extra devices.

We could set up our double groups as:

DVD/Cell: 45
Cell/MP3: 25
DVD/MP3: 20

45+25+20=90... perfect, we've used up all of our extras. Choose C!

[GmatKiss]

Hi Stuart,

Could you please explain the below step,

Keeping our individual numbers in mind, the max we can set our 3 groups to is:

DVD/Cell: 50
Cell/MP3: 30
DVD/MP3: 25

How did you arrive at these numbers ?

TIA,
GK

[Stuart@KaplanGMAT]

Hi Stuart,

Could you please explain the below step,

Keeping our individual numbers in mind, the max we can set our 3 groups to is:

DVD/Cell: 50
Cell/MP3: 30
DVD/MP3: 25

How did you arrive at these numbers ?

TIA,
GK

Hi!

I played with the numbers a bit to maximize the duplicates, keeping in mind that I wanted to use as many of the DVDs, cell phones and MP3 players as possible twice.

I started with the MP3s, since that was the smallest group. Then I noted that there were 5 more Cell phones than DVDs, so I gave 5 more MP3s to the Cell phone people than to the DVD folk. After that I just maximized the number of duplicates based on how many DVDs/Cell phones were left.