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, xy is
a. 65
b. 55
c. 45
d. 35
e. 25
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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, xy 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
8055 = 25
5525 = 30
out of 30 nobody had DVD,
7555 = 20
3020 = 10
x  y = 55  10 = 45
I am not too sure about this. Kindly confirm the OA?
No rest for the Wicked....
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=100a(c+m+d) ).
So: c=d=m=0
=>
(1) cd + cm + dm + a = 110
(2) cd + cm + dm + 2a = 100
(1)(2) a=10
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=100a(c+m+d) ).
So: c=d=m=0
=>
(1) cd + cm + dm + a = 110
(2) cd + cm + dm + 2a = 100
(1)(2) a=10