Set Theory: Level 2

Solution (C): Union (C or T) = Total – Neither = \( 100 – 20 = 80 \). \( 80 = 60 + 50 – \text{Both} \). \( 80 = 110 – \text{Both} \implies \text{Both} = 30 \).

Solution (B): Union = \( 80 + 70 – 60 = 90\% \). Failed Both = \( 100\% – 90\% = 10\% \).

Solution (A): Only Soccer = \( 40 – 15 = 25 \). Only Tennis = \( 30 – 15 = 15 \). Exactly one = \( 25 + 15 = 40 \).

Solution (D): Total = (Only Golf) + (Only Tennis) + (Both). Wait, easier way: Total = (Golf) + (Only Tennis). (Because “Golf” includes “Both”). \( 50 = 35 + \text{Only Tennis} \). \( 50 = 35 + 10 \). Wait, this adds up to 45, not 50. Ah, the question implies sets are Golf and Tennis. Total = Only Golf + Only Tennis + Both. We know Total = 50. Only Tennis = 10. So (Only Golf + Both) = 40. But (Only Golf + Both) is exactly the set “Golf”. So “Golf” = 40. But the problem states “35 play golf”. This implies 50 – 35 = 15 people do NOT play golf. These 15 must play Only Tennis. But problem says 10 play Only Tennis. There is a discrepancy. Unless “Everyone plays at least one” is false. If 5 play neither, then Total Players = 45. Golf = 35. Only Tennis = 10. \(35 + 10 = 45\). This works. So 5 play neither. The question asks “how many play only golf?”. Golf = Only Golf + Both. We know Total Players (45) = Only Golf + Only Tennis + Both. \(45 = \text{Only Golf} + 10 + \text{Both}\). \(\text{Only Golf} + \text{Both} = 35\). We need to find Only Golf. Actually, if Total=50 and everyone plays at least one, then \(50 = \text{Golf} + \text{Only Tennis}\). \(50 = 35 + 10 = 45\). Contradiction. Let me re-read carefully. Maybe I misread the standard problem type. Let’s assume “35 play golf” means “Only Golf”. If Only Golf = 35, Only Tennis = 10. Total = 45. Still not 50. Let’s assume standard logic: Total = 50. Golf = 35. Only Tennis = 10. This means \( 50 – 35 = 15 \) people are not in the Golf circle. These 15 people MUST be “Only Tennis”. But the problem says Only Tennis = 10. This means 5 people are unaccounted for. They must play **neither**. So, Neither = 5. Now, to find “Only Golf”: We know Tennis = Only Tennis + Both. We don’t know Both. We know Golf = Only Golf + Both = 35. We know Total = Only Golf + Only Tennis + Both + Neither. \( 50 = \text{Only Golf} + 10 + \text{Both} + 5 \). \( 35 = \text{Only Golf} + \text{Both} \). This matches the Golf count. We have 1 equation with 2 unknowns. We cannot solve for Only Golf specifically without knowing “Both” or “Tennis Total”. **Correction:** Re-evaluating the question design. If the question asks “How many play only golf?”, and the options are numeric, there must be a unique solution. Let’s try: Total = 50. Only Tennis = 10. Golf = 35. Maybe “10 play only tennis” implies “Tennis = 10”? No. Maybe “35 play golf” means “Only Golf”? If so, answer is 35 (not an option). Let’s assume the question meant: **Total = 45**. Then \(45 = \text{Golf} + \text{Only Tennis} = 35 + 10\). So Only Golf = \( \text{Golf} – \text{Both} \). Actually, if Total = Golf + Only Tennis, then Only Golf = Total – Tennis? No. Only Golf = Total – (Only Tennis + Both + Neither). Let’s look at the structure: Group 1: Golf (includes Both). Group 2: Only Tennis (excludes Both). These two groups are disjoint and cover everyone (if Neither=0). So Golf + Only Tennis = Total. If the numbers given were 40 Golf and 10 Only Tennis, then Total = 50. Then Only Golf = Golf – Both. We still can’t find “Only Golf” without “Both”. **Let’s try Option D (40)**. If Only Golf = 40. Then Both = -5. Impossible. **Let’s try Option E (30)**. If Only Golf = 30. Then Both = 5. Tennis = 15. Total = 30+10+5 = 45. (Close to 50). Let’s Assume: The question meant **”Total 50, 35 Golf, 40 Tennis”**. Then \(50 = 35 + 40 – B \implies B = 25\). Only Golf = \(35-25=10\). Let’s go with a standard problem from the image style (Set difference). Question: “Of 50, 35 Golf, 10 Only Tennis”. If “Only Tennis” = 10, then “Tennis but not Golf” = 10. People who don’t play Golf = 50 – 35 = 15. These 15 are (Only Tennis) + (Neither). \(15 = 10 + \text{Neither} \implies \text{Neither} = 5\). This doesn’t help find “Only Golf”. **Alternative interpretation:** “10 play only tennis” means the rest (40) play Golf or Neither. If everyone plays at least one, then the rest (40) play Golf. Since 35 play Golf, this is a contradiction. **Conclusion:** The numbers 50, 35, 10 are inconsistent for a “at least one” scenario. I will replace this with a valid mathematical question.

Solution (D): Total Active = \( 50 – 10 = 40 \). \( 40 = 30 + 25 – \text{Both} \). \( 40 = 55 – \text{Both} \implies \text{Both} = 15 \). Only Cricket = Total Cricket – Both = \( 30 – 15 = 15 \).

Solution (B): Union (at least one) = \( 100\% – 30\% = 70\% \). \( 70 = 60 + 50 – \text{Both} \). \( 70 = 110 – \text{Both} \implies \text{Both} = 40\% \). Only Cola = Total Cola – Both = \( 60 – 40 = 20\% \).

Solution (C): Ratio 3:1 means \(3+1=4\) parts total. 1 part = \(40/4 = 10\). Passed = 3 parts = 30.

Solution (E): Union contains all unique elements: {2, 4, 6, 8, 10, 3, 9, 12}. Count is 8. (Check formula: \(5 + 4 – 1 (\text{common is 6}) = 8\)).

Solution (D): Union = \( A + B + C – (AB+BC+AC) + ABC \). \( 45 + 35 + 40 – (15 + 10 + 12) + 5 \). \( 120 – 37 + 5 = 88 \). None = \( 100 – 88 = 12 \).

Solution (A): \( 50 = 35 + 25 – \text{Both} \implies \text{Both} = 10 \). Only Cat = Total Cat – Both = \( 25 – 10 = 15 \).

Solution (B): \( 45 = 20 + 30 – \text{Intersection} \). \( 45 = 50 – \text{Int} \implies \text{Int} = 5 \).