1. **State the problem:** We have 50 children choosing from beans, plantain, and rice. Given: - Total children = 50 - Beans = 21 - Plantain = 24 - Rice = 18 - Beans only = 3 - Plantain only = 9 - Rice only = 2 - All three items = 5 We want to find: (a) Venn diagram illustration (described verbally here). (b) Number of children who took: (i) only plantain and beans (ii) none of the three items (iii) only rice and beans 2. **Define variables for pairwise intersections only** (excluding the all three): Let: - $x$ = number of children taking only beans and plantain - $y$ = number of children taking only plantain and rice - $z$ = number of children taking only rice and beans 3. **Write equations for each set total:** For beans: $$ 3 + x + z + 5 = 21 $$ $$ x + z = 21 - 3 - 5 = 13 $$ For plantain: $$ 9 + x + y + 5 = 24 $$ $$ x + y = 24 - 9 - 5 = 10 $$ For rice: $$ 2 + y + z + 5 = 18 $$ $$ y + z = 18 - 2 - 5 = 11 $$ 4. **Solve the system:** We have: $$ x + z = 13 $$ $$ x + y = 10 $$ $$ y + z = 11 $$ Add first two equations: $$ (x + z) + (x + y) = 13 + 10 $$ $$ 2x + y + z = 23 $$ But from third equation, $y + z = 11$, so: $$ 2x + 11 = 23 $$ $$ 2x = 12 $$ $$ x = 6 $$ Then from $x + y = 10$: $$ 6 + y = 10 $$ $$ y = 4 $$ From $x + z = 13$: $$ 6 + z = 13 $$ $$ z = 7 $$ 5. **Answer the questions:** (i) Plantain and beans only = $x = 6$ (ii) Number who took none of the three items: Total taking at least one = sum of all exclusive and overlapping groups: $$ 3 + 9 + 2 + 6 + 4 + 7 + 5 = 36 $$ None took = $50 - 36 = 14$ (iii) Rice and beans only = $z = 7$ **Final answers:** (i) 6 (ii) 14 (iii) 7 A Venn diagram would have the circles labeled Beans, Plantain, Rice with the regions: - Beans only: 3 - Plantain only: 9 - Rice only: 2 - Beans and Plantain only: 6 - Plantain and Rice only: 4 - Rice and Beans only: 7 - All three: 5 - Outside all: 14