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 Post subject: GMAT Overlapping SetsPosted: Mon Aug 02, 2010 7:18 am

Joined: Sun May 30, 2010 3:15 am
Posts: 424
In a 100-unit student dormitory building, 75 dorm rooms have at least one DVD player, 80 have at least one cell phone, and 55 have at least one MP3 player. Every dorm room has at least one of these three devices. If x and y are respectively the greatest and lowest possible number of dorms that have all three of these devices, x – y is:

A. 65
B. 55
C. 45
D. 35
E. 25

x, the greatest possible number of dorms that have all three devices, is simply the number of dorms that have at least one MP3 player. After all, it is possible that these 55 dorms (that have an MP3 player) also have a DVD player and a cell phone. Thus, x = 55.

Finding the value of y takes a little more thought:
If we add up 80 and 75 we get 155. There are only 100 dorms, so at absolute minimum, there is an overlap of 55. 155 – 100 = 55 dorms must have both a DVD player and a cell phone.
We also know that 100 – 55 = 45 dorms do not have an MP3 player, and it is possible that all of these dorms is among the 55 dorms that have both a DVD player and a cell phone. This means that, at minimum, 10 dorms (55 – 45) have an MP3 player, DVD player and cell phone.

Therefore y = 10 and x – y = 45, choice (C)
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Kindly can you provide much more detail explanation how the minimum value y is 10.

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 Post subject: Re: math (test 5, question 23): sets, venn diagram.Posted: Mon Aug 02, 2010 9:03 am

Joined: Sun May 30, 2010 2:23 am
Posts: 498
Let us use Venn diagram to analyze this question. In most general way it will look:

Let us consider only DVD players and cell phones.

We denote number of dorm rooms that have both a DVD player and a cell phone by z. Then the number of dorm rooms that have a DVD player but don't have a cell phone is 75 – z. The number of dorm rooms that have a cell phone but don't have a DVD player is 80 – z.

So the number of dorm rooms that have at least DVD player or a cell phone is (75 – z) + (80 – z) + z = 155 – z
We know that this number doesn't exceed 100, so 155 – z ≤ 100 or 55 ≤ z.

Now, let us consider the number of dorm rooms that have both: a DVD player and cell phone, together with the number of dorm rooms that have an MP3 player.

Let us denote the number of dorm rooms that have all: a DVD player, a cell phone and an MP3 player by y.

Then the number of dorm rooms that have either both: a DVD player and a cell phone or have an MP3 player is not less than
(55 – y) + (55 – y) + y = 110 – y
We know that this can not exceed 100, so 110 – y ≤ 100
or 10 ≤ y.

So y is not less than 10. And it can be 10 if Venn diagram is following:

Therefore y = 10 is the lowest possible number of dorms that have all three of these devices.

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