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.