In a village, $60\%$ of families have a cow, $30\%$ of families have a buffalo and each of the $15\%$ of families has both a cow and a buffalo. In all, there are $960$ families in the village. How many families do not have a cow or a buffalo?

Let $C$ be the set of families having a cow and $B$ be the set of families having a buffalo.
Given:

$$n(C)=60\%$$

$$n(B)=30\%$$

$$n(C\cap B)=15\%$$

The percentage of families having either a cow or a buffalo or both is:

$$n(C\cup B)=n(C)+n(B)-n(C\cap B)$$

$$n(C\cup B)=60\%+30\%-15\%=75\%$$

The percentage of families that do not have either a cow or a buffalo is:

$$100\%-n(C\cup B)=100\%-75\%=25\%$$

Total number of families = $960$.
Number of families without a cow or a buffalo:

$$\text{Number of families}=25\%\text{ of }960=\frac{25}{100}\times 960=\frac{1}{4}\times 960=240$$