1. **Problem statement:** We have 110 members playing at least one of football, basketball, volleyball. Given: - 20 play football and basketball only - 15 play football and volleyball only - 26 play basketball and volleyball only - $x$ play all three games - $2x$ play only one game We want to find how many play basketball altogether. 2. **Formula and rules:** Using the principle of inclusion-exclusion for three sets: $$|F \cup B \cup V| = |F| + |B| + |V| - |F \cap B| - |F \cap V| - |B \cap V| + |F \cap B \cap V|$$ Also, total players = 110. 3. **Define variables:** Let: - $a$ = number who play only football - $b$ = number who play only basketball - $c$ = number who play only volleyball - Given $a + b + c = 2x$ (only one game players) 4. **Express total players:** Total = only one game + exactly two games + all three games $$110 = (a + b + c) + (20 + 15 + 26) + x = 2x + 61 + x = 3x + 61$$ 5. **Solve for $x$:** $$3x + 61 = 110 \implies 3x = 49 \implies x = \frac{49}{3} \approx 16.33$$ Since number of players must be integer, assume $x=16$ (closest integer). 6. **Calculate only one game players:** $$a + b + c = 2x = 32$$ 7. **Find total basketball players:** Basketball players include: - Only basketball: $b$ - Football and basketball only: 20 - Basketball and volleyball only: 26 - All three: $x=16$ So total basketball players: $$b + 20 + 26 + 16 = b + 62$$ 8. **Find $b$ using total only one game players:** We know $a + b + c = 32$, but no further info to separate $a,b,c$. Assuming equal distribution or no further info, we cannot find exact $b$. **However, since question asks for basketball players altogether, and only basketball players is $b$, we can express answer as $b + 62$ where $b$ is unknown.** --- **Second problem:** 1. **Problem statement:** 290 readers read at least one of Daily Times (D), Guardian (G), Punch (P). Given: - $|D|=181$, $|G|=142$, $|P|=117$ - $|D \cap G|=75$, $|D \cap P|=60$, $|G \cap P|=54$ - Find: i. Number reading all three papers ii. Number reading exactly two papers iii. Number reading exactly one paper iv. Number reading Guardian alone 2. **Formula:** Using inclusion-exclusion: $$|D \cup G \cup P| = |D| + |G| + |P| - |D \cap G| - |D \cap P| - |G \cap P| + |D \cap G \cap P|$$ Since all read at least one paper, total = 290. 3. **Calculate $|D \cap G \cap P|$:** $$290 = 181 + 142 + 117 - 75 - 60 - 54 + x$$ $$290 = 440 - 189 + x$$ $$290 = 251 + x \implies x = 39$$ 4. **Calculate exactly two papers:** Exactly two = sum of pairwise intersections minus thrice the triple intersection: $$= (75 + 60 + 54) - 3 \times 39 = 189 - 117 = 72$$ 5. **Calculate exactly one paper:** Exactly one = total - exactly two - all three $$= 290 - 72 - 39 = 179$$ 6. **Calculate Guardian alone:** $$|G| - |D \cap G| - |G \cap P| + |D \cap G \cap P| = 142 - 75 - 54 + 39 = 52$$ --- **Final answers:** - Basketball players altogether = $b + 62$ where $b$ is unknown - Number reading all three papers = 39 - Number reading exactly two papers = 72 - Number reading exactly one paper = 179 - Number reading Guardian alone = 52