In a village, each of the 60% of families has a cow; each of the 30% of families has a buffalo and each of the 15% of families has both a cow and buffalo. In all there are 96 families in the village. How many families do not have a cow or a buffalo ?

Correct Answer: 24
15% of families have cow and buffalo.
Families only have cow = 60 - 15 = 45%
Families only have buffalo = 30 - 15 = 15%
Required families which do not have a cow or a buffalo
= 100 - (Families only have cow + Families only have buffalo + Families have cow and buffalo)
= 100 - (45 + 15 + 15)
= 25%
According to the question,
Required number
= $$\frac{96}{100}$$ × 25
= 24