In a hotel, 60% had vegetarian lunch while 30% had non-vegetarian lunch and 15% had both types of lunch. If 96 people were present, how many did not eat either type of lunch ?

$$\eqalign{ & n\left( A \right) = \left( {\frac{{60}}{{100}} \times 96} \right) = \frac{{288}}{5} \cr & n\left( B \right) = \left( {\frac{{30}}{{100}} \times 96} \right) = \frac{{144}}{5} \cr & n\left( {A \cap B} \right) = \left( {\frac{{15}}{{100}} \times 96} \right) = \frac{{72}}{5} \cr & \therefore n\left( {A \cup B} \right): \cr & = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right) \cr & = \frac{{288}}{5} + \frac{{144}}{5} - \frac{{72}}{5} \cr & = \frac{{360}}{5} \cr & = 72 \cr} $$
So, people who had either or both types of lunch = 72
Hence, people who had neither type of lunch = (96 - 72) = 24