We can create 8 sets of students where the numbers in the sets are represented by the letters A to H:
A=quant only
B=verbal only
C=AR only
D=quant+verbal, but not AR
E=quant+AR, but not verbal
F=verbal+AR, but not quant
G=quant+verbal+AR
H=none of quant, verbal, AR
So using the given figures: D+G=30, F+G=35, E+G=45, D+E+F+3G=110, 3G=110-(D+E+F)
(1) A+D+E+G=100, A+E=100-30=70, A+D=100-45=55;
(2) B+D+F+G=50, B+D=50-35=15, B+F=50-30=20;
(3) C+E+F+G=60, C+F=60-45=15, C+E=60-35=25;
(4) A+B+C+D+E+F+G+H=200, 70+15+15+G+H=200, G+H=200-100=100;
(5) A+B+C+2(D+E+F)+3G=210=A+B+C+D+E+F+G+110, A+B+C+D+E+F+G=210-110=100, 100+H=200, H=100 (fixed)⇒G=0
G cannot exceed 30, because D+G=30. So let G=30 and test the conditions:
D=0, A=55 (from (1)), B=15 (from (2)), F=5, E=15, H=70, 55+15+C+0+15+5+30+70=200, C=10.
The conditions all check out OK, so max G=30 (students learning all 3 subjects).
When H is minimum, the number of students learning at least one subject is maximum. Since G≤30, and H=100-G, H≥70.
More to follow in due course...