1. **Stating the problem:** We have a group of students joining three competitions: Colouring (Mewarna), Singing (Menyanyi), and Cooking (Memasak). The table gives numbers for various intersections and totals. We need to complete the Venn diagram and find: (i) Number of students who joined exactly one competition. (ii) Number of students who did not join both Colouring and Cooking. 2. **Given data:** - Colouring only: $12$ - Singing only: $6$ - Cooking only: $?$ (unknown) - Colouring and Singing (but not Cooking): $?$ - Singing and Cooking (but not Colouring): $?$ - Colouring and Cooking (but not Singing): $?$ - All three (Colouring, Singing, Cooking): $5$ - Colouring total: $30$ - Singing total: $15$ - Cooking total: $35$ 3. **Define variables for unknown intersections:** Let: - $x$ = number of students in Colouring and Singing only - $y$ = number of students in Singing and Cooking only - $z$ = number of students in Colouring and Cooking only - $c$ = number of students in Cooking only 4. **Use totals to form equations:** - Colouring total: $12 + x + z + 5 = 30$ (Colouring only + Colouring & Singing + Colouring & Cooking + all three) - Singing total: $6 + x + y + 5 = 15$ - Cooking total: $c + y + z + 5 = 35$ 5. **Simplify each:** - $12 + x + z + 5 = 30 \implies x + z = 30 - 17 = 13$ - $6 + x + y + 5 = 15 \implies x + y = 15 - 11 = 4$ - $c + y + z + 5 = 35 \implies c + y + z = 30$ 6. **From the first two equations:** - $x + z = 13$ - $x + y = 4$ Subtract second from first: $$ (x + z) - (x + y) = 13 - 4 \implies z - y = 9 $$ 7. **From the third equation:** $$ c + y + z = 30 $$ Using $z = y + 9$ from above: $$ c + y + (y + 9) = 30 \implies c + 2y = 21 $$ 8. **We want to find $c$, $y$, $z$, and $x$.** From $x + y = 4$, express $x = 4 - y$. From $x + z = 13$, substitute $x$: $$ (4 - y) + z = 13 \implies z = 9 + y $$ This matches $z = y + 9$ found earlier. 9. **Since $c + 2y = 21$, $c$ and $y$ must be non-negative integers.** Try values for $y$: - If $y=6$, then $c = 21 - 12 = 9$ - Check if $x = 4 - y = 4 - 6 = -2$ (not possible) Try $y=3$: - $c = 21 - 6 = 15$ - $x = 4 - 3 = 1$ - $z = y + 9 = 3 + 9 = 12$ All are non-negative integers. 10. **Final values:** - $x = 1$ (Colouring and Singing only) - $y = 3$ (Singing and Cooking only) - $z = 12$ (Colouring and Cooking only) - $c = 15$ (Cooking only) 11. **Complete Venn diagram values:** - Colouring only: 12 - Singing only: 6 - Cooking only: 15 - Colouring & Singing only: 1 - Singing & Cooking only: 3 - Colouring & Cooking only: 12 - All three: 5 12. **(i) Number of students who joined exactly one competition:** $$ 12 + 6 + 15 = 33 $$ 13. **(ii) Number of students who did not join both Colouring and Cooking:** "Did not join both" means students who are not in the intersection of Colouring and Cooking (including those in all three). Total students in Colouring and Cooking (including all three): $$ 12 + 5 + 1 + 3 + 6 + 15 + 9 + 30 + 35 $$ But we need total students first. Sum all unique students: $$ 12 + 6 + 15 + 1 + 3 + 12 + 5 = 54 $$ Students who joined both Colouring and Cooking are those in: - Colouring & Cooking only: 12 - All three: 5 Total in both Colouring and Cooking: $$ 12 + 5 = 17 $$ Therefore, students who did not join both Colouring and Cooking: $$ 54 - 17 = 37 $$ **Final answers:** - (i) $33$ - (ii) $37$