In a village of 100 households, 75 have at least one DVD player, 85 have at least one cell phone, and 70 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:	Choices:
A) 65
B) 55
C) 40
D) 35
E) 25
Am I missing something here??? it seems straightforward......   The obvious maximum that have all 3 is 70, 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:    100-85= 
<gadget id="calculator"> 100-85</gadget>
<output>15</output>
15 don't have Cell    100-75= 
<gadget id="calculator"> 100-75</gadget>
<output>25</output>
25 don't have DVD   and  100-70= 
<gadget id="calculator"> 100-70</gadget>
<output>30</output>
30 don't have MP3   SO a total of   15+25+30= 
<gadget id="calculator">  15+25+30</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.   70-30= 
<gadget id="calculator">70-30</gadget>
<output>40</output>
40   40
<result>C</result>
In a village of 100 households, 75 have at least one DVD player, 85 have at least one cell phone, and 70 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: