In a village of 100 households, 85 have at least one DVD player, 90 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:	Choose the correct choice from the following choices:
A) 65
B) 55
C) 45
D) 35
E) 25
Am I missing something here??? it seems straightforward......   The obvious maximum that have all 3 is 55, because you are limited by the SMALLEST number.   The minimum is simply the sum of the max of each people who DONT have the product, so:    100-90= 
<gadget id="calculator"> 100-90</gadget>
<output>10</output>
10 don't have Cell    100-85= 
<gadget id="calculator"> 100-85</gadget>
<output>15</output>
15 don't have DVD   and  100-55= 
<gadget id="calculator"> 100-55</gadget>
<output>45</output>
45 don't have MP3   SO a total of   10+15+45= 
<gadget id="calculator">  10+15+45</gadget>
<output>70</output>
70 combined who might NOT have SOME combination of the 3 products. So subtract that from 100, to give you the minimum of the people who COULD have all 3 and you get  100-70= 
<gadget id="calculator"> 100-70</gadget>
<output>30</output>
30.   55-30= 
<gadget id="calculator">55-30</gadget>
<output>25</output>
25   25
<result>E</result>
In a village of 100 households, 85 have at least one DVD player, 90 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: