In a village each of 60% of the family has a cow , each of the 30% of the family has a buffalo and each of the 15% of the families has both cow and buffalo. if in all there are 96 families in the village. how many families do not have a cow or buffalo?
Answers
Answer:
Step-by-step explanation:
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
= 96100
× 25
= 24
Answer:24
Step-by-step explanation: Here,
Let the families be denoted by U, the families who have cow and who have buffalo by C and B.
Total families=96
n(U)= 100percent
n(C)= 60 percent
n(B)= 30 percent
n(C intersection B)= 15 percent
n(CUB)' =n(U)-n(CUB)
So, we should find n(CUB) first,
n(CUB)= n(C) +n(B) - n(C intersection B)
=60 percent + 30 percent -15 percent
= 75 percent
Now,
As in percent, n(CUB)' = 75 percent and the no. Of total families is 96
n(CUB)' =75 percent of 96
= 75/100 ×96
= 24