In a village of 100 households, 80 have at least one DVD player, 90 have at least one cell phone, and 65 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 option: A) 65 B) 55 C) 45 D) 30 E) 20
Am I missing something here??? it seems straightforward......   The obvious maximum that have all 3 is 65, 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    100-80= 
<gadget id="calculator"> 100-80</gadget>
<output>20</output>
20 don't have DVD   and  100-65= 
<gadget id="calculator"> 100-65</gadget>
<output>35</output>
35 don't have MP3   SO a total of   10+20+35= 
<gadget id="calculator">  10+20+35</gadget>
<output>65</output>
65 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  100-65= 
<gadget id="calculator"> 100-65</gadget>
<output>35</output>
35.   65-35= 
<gadget id="calculator">65-35</gadget>
<output>30</output>
30   30
<result>D</result>
In a village of 100 households, 80 have at least one DVD player, 90 have at least one cell phone, and 65 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: