sparsh.21 wrote: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

a. 65
b. 55
c. 45
d. 35
e. 25

getting the value of x is easy = 55

for the value of y i.e. lowest

lets say out of 55 nobody had cell phone

80-55 = 25

55-25 = 30

out of 30 nobody had DVD,

75-55 = 20

30-20 = 10

x - y = 55 - 10 = 45

I am not too sure about this. Kindly confirm the OA?

Yeah (C) should be right
a stands for all 3

i m+cm+dm+a=55
ii d+cd+dm+a=75
iii c+cd+cm+a=80

=> i+ii+iii
I c+d+m+(cd+cm+dm)2+3a=210
II c+d+m+cd+cm+dm+a=100

=> I-II
cd+cm+dm+2a=110

max 55 (I and II are ok!)
min 10 (c+d+m = 0)

IMO (C)

yes.... the OA is C
but i am still not able to understand how to get the value of y...
Can anybody please elaborate a little more ?

thanks

ok, lets have a look at these equotations:

I c+d+m+cd+cm+dm+a=100

II cd+cm+dm+2a=110

both need to be true.
We have no information about the value of cd,cm,dm (e.g. cd is cellphone AND dvd player; VENN DIAGRAMM!). So we get the max for a with cd=cm=dm=0 (sum of them is minimal).
=> 2a=110 <=> a=55

Let`s have a look for y now.
To find our min for a we know that the sum of cd,cm,dm should be maximal (a = (110-(cd+cm+dm))/2)
What is the max of this sum?
It is easy to find! Just look at I, the max should be reached when c,m,d all equal 0 ( cd+cm+dm=100-a-(c+m+d) ).


So: c=d=m=0
=>
(1) cd + cm + dm + a = 110
(2) cd + cm + dm + 2a = 100

(1)-(2) a=10