1. **Problem 1: Calculate total sales revenues for Kamukamu and Tweziimbe groups** 1. Given: - Milk price per litre = 700 - Maize price per kg = 1000 - Beans price per kg = 3500 - 1 bag maize or beans = 120 kg 2. Calculate total revenue for each group and sales period: - Kamukamu sales 1: 2520 litres milk, 35 bags maize, 10 bags beans - Milk revenue = $$2520 \times 700 = 1764000$$ - Maize revenue = $$35 \times 120 \times 1000 = 4200000$$ - Beans revenue = $$10 \times 120 \times 3500 = 4200000$$ - Total = $$1764000 + 4200000 + 4200000 = 10164000$$ - Tweziimbe sales 1: 2314 litres milk, 41 bags maize, 9 bags beans - Milk revenue = $$2314 \times 700 = 1619800$$ - Maize revenue = $$41 \times 120 \times 1000 = 4920000$$ - Beans revenue = $$9 \times 120 \times 3500 = 3780000$$ - Total = $$1619800 + 4920000 + 3780000 = 10349800$$ - Kamukamu sales 2: 3254 litres milk, 42 bags maize, 8 bags beans - Milk revenue = $$3254 \times 700 = 2277800$$ - Maize revenue = $$42 \times 120 \times 1000 = 5040000$$ - Beans revenue = $$8 \times 120 \times 3500 = 3360000$$ - Total = $$2277800 + 5040000 + 3360000 = 10697800$$ - Tweziimbe sales 2: 2719 litres milk, 32 bags maize, 11 bags beans - Milk revenue = $$2719 \times 700 = 1903300$$ - Maize revenue = $$32 \times 120 \times 1000 = 3840000$$ - Beans revenue = $$11 \times 120 \times 3500 = 4620000$$ - Total = $$1903300 + 3840000 + 4620000 = 10363300$$ 2. **Problem 2: Determine student counts fitting different sizes** Given: - Medium = Large = 100 students - Small = 76 - Small & Large = 50 - Medium & Large = 70 - Small & Medium = 60 - None = 4 - Total students = 140 - Some fit in all three sizes (x) Using inclusion-exclusion principle for three sets: Total in at least one size = Total - None = $$140 - 4 = 136$$ Formula: $$ ext{Small} + ext{Medium} + ext{Large} - (SM + SL + ML) + ext{All three} = ext{At least one}$$ Plug in values: $$76 + 100 + 100 - (60 + 50 + 70) + x = 136$$ $$276 - 180 + x = 136$$ $$96 + x = 136$$ $$x = 40$$ So, 40 students fit in all three sizes. 3. **Problem 3: Determine the number of students liking foodstuffs** Given: - Prices: posho 20,000, rice 30,000, cassava 10,000 (not needed for set problem) - Likes: Cassava (C) = 47, Posho (P) = 53, Rice (R) = 45 - Only posho = 23, only cassava = 10, only rice = 5 - All three = 15 Find students in combined groups: Using inclusion-exclusion: Calculate first the number who like at least one foodstuff: $$|P \cup C \cup R| = |P| + |C| + |R| - |P \cap C| - |P \cap R| - |C \cap R| + |P \cap C \cap R|$$ We know: - Only posho = 23 = $$|P| - |P \cap C| - |P \cap R| + |P \cap C \cap R|$$ - Only cassava = 10 - Only rice = 5 - All three = 15 Compute pairs intersections: Let: $$a = |P \cap C|, b = |P \cap R|, c = |C \cap R|$$ From only posho count: $$23 = 53 - a - b + 15 \, \Rightarrow \, a + b = 53 - 23 + 15 = 45$$ From only cassava count: $$10 = 47 - a - c + 15 \, \Rightarrow \, a + c = 47 - 10 + 15 = 52$$ From only rice count: $$5 = 45 - b - c + 15 \, \Rightarrow \, b + c = 45 - 5 + 15 = 55$$ Summing: $$(a + b) + (a + c) + (b + c) = 45 + 52 + 55 = 152$$ But this equals $$2(a + b + c)$$ so: $$2(a + b + c) = 152 \Rightarrow a + b + c = 76$$ Using sums: $$a + b = 45$$ $$a + c = 52$$ $$b + c = 55$$ Subtract pairs to find individuals: Subtract first from second: $$(a + c) - (a + b) = 52 - 45 \Rightarrow c - b = 7$$ Subtract first from third: $$(b + c) - (a + b) = 55 - 45 \Rightarrow c - a = 10$$ From $$c - b = 7$$ and $$c - a = 10$$, rearranged: $$c = b + 7$$ $$c = a + 10$$ Thus: $$b + 7 = a + 10 \Rightarrow b = a + 3$$ Recall $$a + b = 45\Rightarrow a + (a + 3) = 45 \Rightarrow 2a = 42 \Rightarrow a = 21$$ Then: $$b = 24$$ $$c = a + 10 = 31$$ Final counts: - Posho & Cassava = 21 - Posho & Rice = 24 - Cassava & Rice = 31 This fully describes the set intersections. **Final answers:** - Problem 1 totals: Kamukamu1=10164000, Tweziimbe1=10349800, Kamukamu2=10697800, Tweziimbe2=10363300 - Problem 2 all three fit: 40 - Problem 3 intersections: P\&C=21, P\&R=24, C\&R=31