In a village of 100 households, 70 have at least one DVD player, 80 have at least one cell phone, and 60 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 options: A) 65 B) 50 C) 45 D) 35 E) 25
Am I missing something here??? it seems straightforward......   The obvious maximum that have all 3 is 60, 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:    100-80= 
<gadget id="calculator"> 100-80</gadget>
<output>20</output>
20 don't have Cell    100-70= 
<gadget id="calculator"> 100-70</gadget>
<output>30</output>
30 don't have DVD   and  100-60= 
<gadget id="calculator"> 100-60</gadget>
<output>40</output>
40 don't have MP3   SO a total of   20+30+40= 
<gadget id="calculator">  20+30+40</gadget>
<output>90</output>
90 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  100-90= 
<gadget id="calculator"> 100-90</gadget>
<output>10</output>
10.   60-10= 
<gadget id="calculator">60-10</gadget>
<output>50</output>
50   50
<result>B</result>
In a village of 100 households, 70 have at least one DVD player, 80 have at least one cell phone, and 60 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: