In a hotel, 60% had vegetarian lunch while 30% had non-vegetarian lunch and 15% had both type of lunch. If 96 people were present, how many did not eat either type of lunch ? Pick
A) 21
B) 22
C) 23
D) 24
E) 25
n(A)=(60/100 ∗96)=288/5   n(B)=(30/100 ∗96)=144/   n(A∩B)=(15/100 ∗96)=72/5   People who have either or both lunchn(A∪B)=288/5 +144/5 −72/5 =360/5= 
<gadget id="calculator">360/5</gadget>
<output>72</output>
72   So People who do no have either lunch were= 96 -72= 
<gadget id="calculator">96 -72</gadget>
<output>24</output>
24 ANSWER : 24
<result>D</result>
In a hotel, 60% had vegetarian lunch while 30% had non-vegetarian lunch and 15% had both type of lunch. If 96 people were present, how many did not eat either type of lunch ?