Directions for question :
In a group of 300 families in a colony of Dharavi, each family likes at least one dish among Pao Bhaji, Vada Pao and Misal Pao. 160 families like Pao Bhaji and 140 families like Vada Pao.
Question:
If the number of families who like Misal Pao is 200 and the maximum possible number of families like only Vada Pao, what can be the maximum possible number of families who like both Pao Bhaji and Misal Pao, but not Vada Pao ?
80
100
120
160
Let PB, VP and MP in the Venn diagram stand for families who like Pao Bhaji, Vada Pao and Misal Pao respectively.
Since each family likes at least one dish among Pao Bhaji, Vada Pao and Misal Pao, therefore n = 0
It is given that :
(1) … a + (x + y) + r = 160
(2) … b + (x + z) + r = 140
(3) … (a +b + c) + (x + y + z) + r = 300
(4) … c + (y + z) + r = 200
(5) … b has the maximum possible value
Subtracting (4) from (3), we get
(6) … (a + b) + x = 300 – 200 = 100
Since the value of b in (6) is maximum,
Hence the value of a and x must be minimum in this case.
aMin can be 0
xMin can also be 0
Hence, the maximum possible number of families who like only Vada Pao
= maximum possible value of b
= 100 – (0 + 0) [Subtracting (aMin + xMin) from (6)]
= 100
As per the given criteria, the present scenario is as given below :
Hence,
(7) … z + r = 140 – (100 + 0) = 40 [From (2)]
and
(8) … y + r = 160 – (0 + 0) = 160 [From (1)]
For y to be maximum (which is what has been asked from us), r has to be the minimum.
rMin can be considered to be 0
Hence, maximum value of z can be 40 – 0 = 40 [From (7)]
and
maximum value of y can be 160 – 0 = 160 [From (8)]
and
value of c has to be 200 – (160 + 40 + 0) = 0 [From (4)]
Hence, the maximum possible number of families who like both Pao Bhaji and Misal Pao, but not Vada Pao = 160