1. **QUESTION ONE: Organize sales data and calculate total sales for both groups** - Kamukamu last month: 2520 litres milk, 35 bags maize, 10 bags beans - Kamukamu this month: 3254 litres milk, 42 bags maize, 8 bags beans - Tweziimbe last month: 2314 litres milk, 41 bags maize, 9 bags beans - Tweziimbe this month: 2719 litres milk, 32 bags maize, 11 bags beans - Price per litre milk = 700 - Price per kg maize = 1000 - Price per kg beans = 3500 - Each bag of maize or beans = 120 kg Calculate total sales for each period: 1. Calculate total quantities sold (kg or litres) per group for each product: Kamukamu last month maize = $35 \times 120 = 4200$ kg Kamukamu last month beans = $10 \times 120 = 1200$ kg Tweziimbe last month maize = $41 \times 120 = 4920$ kg Tweziimbe last month beans = $9 \times 120 = 1080$ kg Kamukamu this month maize = $42 \times 120 = 5040$ kg Kamukamu this month beans = $8 \times 120 = 960$ kg Tweziimbe this month maize = $32 \times 120 = 3840$ kg Tweziimbe this month beans = $11 \times 120 = 1320$ kg 2. Calculate sales money for each group per period: Last month Kamukamu sales: Milk: $2520 \times 700 = 1,764,000$ Maize: $4200 \times 1000 = 4,200,000$ Beans: $1200 \times 3500 = 4,200,000$ Total Kamukamu last month = $1,764,000 + 4,200,000 + 4,200,000 = 10,164,000$ Last month Tweziimbe sales: Milk: $2314 \times 700 = 1,619,800$ Maize: $4920 \times 1000 = 4,920,000$ Beans: $1080 \times 3500 = 3,780,000$ Total Tweziimbe last month = $1,619,800 + 4,920,000 + 3,780,000 = 10,319,800$ This month Kamukamu sales: Milk: $3254 \times 700 = 2,277,800$ Maize: $5040 \times 1000 = 5,040,000$ Beans: $960 \times 3500 = 3,360,000$ Total Kamukamu this month = $2,277,800 + 5,040,000 + 3,360,000 = 10,677,800$ This month Tweziimbe sales: Milk: $2719 \times 700 = 1,903,300$ Maize: $3840 \times 1000 = 3,840,000$ Beans: $1320 \times 3500 = 4,620,000$ Total Tweziimbe this month = $1,903,300 + 3,840,000 + 4,620,000 = 10,363,300$ 3. Calculate total sales for both groups for each month: Last month total sales = $10,164,000 + 10,319,800 = 20,483,800$ This month total sales = $10,677,800 + 10,363,300 = 21,041,100$ 2. **QUESTION TWO: Evaluate whether at least 70% fit in two sizes** Given: - Students who fit medium = 100 - Students who fit large = 100 - Students who fit small = 76 - Students fit small & large = 50 - Students fit medium & large = 70 - Students fit small & medium = 60 - Students fit none = 4 - Class total = 140 - Some fit all three sizes 1. Find number fit in at least two sizes. 2. Use Inclusion-Exclusion for triple intersections: Let $x$ = students who fit all three sizes. Number fit in exactly two sizes = (sum of pairs) - 3x Pairs sum = $50 + 70 + 60 = 180$ 3. Total fit in none = 4, so total fit in at least one size = $140 - 4 = 136$ 4. Total in exactly one size = (sum individuals) - (sum pairs) + x Sum individuals = $76 + 100 + 100 = 276$ From inclusion-exclusion: $$136 = 276 - 180 + x \Rightarrow x = 40$$ 5. Calculate number fitting at least two sizes: $$= \text{pairs sum} - 3x + x = 180 - 3\times40 + 40 = 180 - 120 + 40 = 100$$ 6. Percentage with at least two sizes: $$\frac{100}{140} \times 100 = 71.43\%$$ Since 71.43% > 70%, we will buy the suggestion. It satisfies the captain's condition. 3. **QUESTION THREE (a): Arrange foodstuffs bought and find which school spent most** Prices: Posho = 20,000 Rice = 30,000 Cassava = 10,000 Week 1 quantities (bags): - Nalinya: Posho=2, Rice=1, Potatoes=3(not priced, ignore) - Ndejje: Posho=2, Rice=2 - Sazza: Posho=1, Rice=1, Cassava=2 Week 2 quantities: - Nalinya: Posho=3, Cassava=2 - Ndejje: Posho=1, Rice=2, Cassava=1 - Sazza: Posho=3, Cassava=1 Arrays (rows: schools, columns: foodstuffs posho, rice, cassava): Week 1: $$\begin{bmatrix}2 & 1 & 0 \\ 2 & 2 & 0 \\ 1 & 1 & 2\end{bmatrix}$$ Week 2: $$\begin{bmatrix}3 & 0 & 2 \\ 1 & 2 & 1 \\ 3 & 0 & 1\end{bmatrix}$$ Price vector: $$\begin{bmatrix}20000 \\ 30000 \\ 10000\end{bmatrix}$$ Calculate total spent per school: Week 1 total per school: - Nalinya: $2 \times 20000 + 1 \times 30000 + 0 = 70000$ - Ndejje: $2 \times 20000 + 2 \times 30000 + 0 = 100000$ - Sazza: $1 \times 20000 + 1 \times 30000 + 2 \times 10000 = 70000$ Week 2 total per school: - Nalinya: $3 \times 20000 + 0 + 2 \times 10000 = 80000$ - Ndejje: $1 \times 20000 + 2 \times 30000 + 1 \times 10000 = 90000$ - Sazza: $3 \times 20000 + 0 + 1 \times 10000 = 70000$ Total for two weeks: - Nalinya: $70000 + 80000 = 150000$ - Ndejje: $100000 + 90000 = 190000$ - Sazza: $70000 + 70000 = 140000$ Ndejje S.S spent the most. 3. **QUESTION THREE (b): Survey data analysis** Given: - Liked cassava (C) = 47 - Liked posho (P) = 53 - Liked rice (R) = 45 - Liked posho only = 23 - Liked cassava only = 10 - Liked rice only = 5 - Liked all three = 15 - All students liked at least one (i) Arrange in Venn diagram categories: Calculate exactly two sizes: Posho and cassava only = $P \cap C - P \cap R \cap C - P \cap C \cap R??$ Total of sets: $$|P \cup C \cup R| = ?$$ First find number liking exactly two foods: Total in P = 53 Only P = 23 All three = 15 So: Two food only involving P: $$\text{two only} = |P| - |P \text{ only}| - |P \cap C \cap R| = 53 - 23 - 15 = 15$$ Similarly for C: Two only for C: $$47 - 10 - 15 = 22$$ Two only for R: $$45 - 5 - 15 = 25$$ (ii) Number of students preferring at least two foods = sum exactly two + all three = $15 + 22 + 25 + 15 = 77$ (iii) Total students: Sum all single only + exactly two + all three Only P + only C + only R = 23 + 10 + 5 = 38 Exactly two = 15 + 22 + 25 = 62 All three = 15 Total = $38 + 62 + 15 = 115$ (iv) Probability student prefers rice: $$\frac{45}{115} = 0.3913$$ (approx 39.13%) Conclusion: Less than half prefer rice, so rice is less popular compared to other foodstuffs.