sharmashivng wrote:
Hi
yashikaaggarwalJust having a little trouble with this one. Need your help.
In a village of 100 households, 75 have at least one DVD player, 80 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:
Now,
We are given that we need to 'maximize' the number of households that have all three of these devices.
It will be 100, right? It is given that
55 have at least one MP3 player. It is very much possible that 100 folks have all the 3 devices.
As for minimum, I am little stuck. It should be 55.
So basically, 100-55= 45.
Is my thinking correct? I didn't use Venn diagrams, equations.
Yeah you are correct with both part, but let's understand the logic
the total is 100
and if we total all DVD users(Only 1, 2 or all 3) + Cell phone users(Only 1, 2 or all 3) + MP3 users(Only 1, 2 or all 3)
that equals to 75+80+55 = 210 which is 110 more than 100
so the largest interaction zone can be 110
BUT,
we can never go more than the original total, and
If we want to maximize those who like all 3, we have to maximize the value in the intersection. So, we have to minimize the value of the union. that can hold upto 100 value.
So, Maximum is 100 Now, Minimum.
If we want to minimize those who like all 3, we have to minimize the value in the intersection. So, we have to maximize the value of the union. the minimum value among all DVD users, Cell phone users, and MP3 users (75,80,55 is 55)
which can be a part of all 3 venn Interaction period at the lowest level
so the minimum value is 55 Difference between maximum and minimum is 100-55 = 45 _________________