Consider 7 sets called S1-S7 as follows:
S1=A (only) players; S2=B (only players); S3=C (only) players;
S4=A+B, not C; S5=A+C, not B; S6=B+C, not A; S7=all 3 sports.
Given conditions:
C1) Total number of sport A players=S1+S4+S5+S7=60;
C2) Total number of sport B players=S2+S4+S6+S7=80;
C3) Total number of sport C players=S3+S5+S6+S7=40.
C4) Total number of players=S1+S2+S3+S4+S5+S6+S7=90.
Total number of players who play only one sport=S1+S2+S3.
Total number of players who play more than one sport=90-(S1+S2+S3).
90-(S1+S2+S3)=2(S1+S2+S3), 3(S1+S2+S3)=90,
Deductions:
D1) S1+S2+S3=30
D2) S4+S5+S6+S7=60.
S1+S2+S3+2(S4+S5+S6)+3S7=60+80+40=180; 2(S4+S5+S6)+3S7=180-30=150, 2(60-S7)+3S7=150, 120-2S7+3S7=150⇒(D3) S7=30.
S4+S5+S7=60-S6; S4+S6+S7=60-S5; S5+S6+S7=60-S4;
S1+60-S6=60⇒(D4) S6=S1; S2+60-S5=80⇒(D5) S5=S2-20; S3+60-S4=40⇒(D6) S4=S3+20;
D7) S4+S5+S6=30.
We now have a fixed value for S7 (30 people play all 3 sports), and S4, S5, S6 are in terms of S1, S2, S3. And we know that S4+S5+S6=S1+S2+S3=30.
(1) From D6 and D7, S3+20+S5+S6=30, S3+S5+S6=10, therefore S3≤10, that is, S3 max cannot exceed 10. Let S3=10, then from D6, S4=30 and S5=S6=S1 (D4)=0.
Check out the given conditions, and applying D3: (C1) 0+30+0+30=60; S5=0⇒S2=20 (D5), so (C2) 20+30+0+30=80; (C3) 10+0+0+30=40; (D1) 0+20+10=30; (D2) 30+0+0+30=60; (C4) 0+20+10+30+0+0+30=90. All conditions check out, so maximum for C-only players is 10.
(2) S4 min=20, S3=0 (D6), S5 min=0, S2=20 (D5), S6≤10, and S1=S6 (D4). Let S6=S1=10, now check out the conditions:
(C1)=10+20+0+30=60; (C2) 20+20+10+30=80 (C3) 0+0+10+30=40; (C4) 10+20+0+20+0+10+30=90. All conditions check out, so 10 play A only when B and C players are maximum.