In a group of 150 students, 52 like tea, 48 like juice and 62 like coffee. If each student in the group likes at least one among tea, juice and coffee, then the maximum number of students that like more than one drink is:

Given total no of students 150. 
52 (liked tea) = only tea + (tea, coffee) + (tea, juice) + (tea, coffee, juice) .
48 (like juice) = only juice + (juice, tea) + (juice, coffee) + (coffee, juice, tea)
62 (like coffee) = only coffee + (coffee, juice) + (coffee, tea) + (coffee, juice, tea).

adding all three we get :
162 = Exactly 1 drink + 2(Exactly 2 drinks) + 3(Exactly all 3 drinks) . 
but given total 150 =  Exactly 1 drink + Exactly 2 drinks + Exactly all 3 drinks .
Substituting equation 2 in equation 1: 
162 = 150 + Exactly 2 drinks + 2(Exactly all 3 drinks) .

12 = {Exactly 2 drinks + Exactly 3 drinks} + {Exactly 3 drinks} .
12 = {More than 1 drink} + {Exactly 3 drinks}.

Therefore, the maximum of More than 1 drink occurs at exactly 3 drinks is 0. Hence, maximum of More than 1 drink is 12.