jainrahul1985 wrote:In a village of hundred households , 75 have at least one DVD player, 80 have at least one cell phone ,and 55 have at least one MP3 player. Every village has at least one of these three devices.If X and Y are respectively the greatest and the lowest possible number of households that have all the three of these devices, X-Y is :
A ) 65
B ) 55
C ) 45
D) 35
E) 25
OA C . Easy way to solve this question ?
Minimum:
To find the minimum, add 75+80+55=210, and then subtract twice the total number of households: 210-2*100=10, so minimum is 10. Edit: note that if you get a negative number with this approach, the minimum is zero.
Maximum:
Add 75+80+55=210, subtract the total, and divide this number by 2, ignoring the remainder: 210-100=110. 110/2=55. This number, or the least of the three individual groups, will be the maximum, whichever is less. In this case they are both 55, so the maximum is 55.
55-10=45.
Note that the maximum in situations like this, where they specify that each household has at least one thing, will not always just be the lowest of the three values. For example, if we had had DVD-60, Cell phone-65, mp3-55, then if we put 55 as the number of households that have all 3, then we could put 10 households as having just a cell phone and 5 households as having just a dvd player, but that would only give us a total of 70 households, and we don't have any more space in any of the circles to get the total number up to 100. If they hadn't said every household has at least one thing, we could just say that there could be 30 households that don't own any of them and be done. But if every household has to have at least one of these things, we have a problem. In this case, the maximum number of households that own all three would be 40, by the method outlined above.
