1. **Problem 2: Venn Diagram for condiment orders** We have 60 customers and three condiments: Eggs (E), Salad (S), and Gari (G). Given: - $n(E) = 26$, $n(S) = 22$, $n(G) = 25$ - $n(E \cap S \text{ only}) = 6$, $n(E \cap G \text{ only}) = 4$, $n(S \cap G \text{ only}) = 1$ - $n(\text{none}) = 4$ 2. **Draw the Venn diagram** Let $x = n(E \cap S \cap G)$ be the number who ordered all three condiments. 3. **Use set theory to find $x$** Total customers ordered at least one condiment: $$60 - 4 = 56$$ 4. **The principle of inclusion-exclusion for three sets:** $$n(E \cup S \cup G) = n(E) + n(S) + n(G) - n(E \cap S) - n(E \cap G) - n(S \cap G) + n(E \cap S \cap G)$$ We know: $$n(E \cap S) = n(E \cap S \text{ only}) + x = 6 + x$$ $$n(E \cap G) = 4 + x$$ $$n(S \cap G) = 1 + x$$ 5. Substitute values: $$56 = 26 + 22 + 25 - (6 + x) - (4 + x) - (1 + x) + x$$ Simplify: $$56 = 73 - 6 - x - 4 - x - 1 - x + x$$ $$56 = 73 - 11 - 2x$$ $$56 = 62 - 2x$$ 6. Solve for $x$: $$2x = 62 - 56 = 6$$ $$x = 3$$ So, 3 customers ordered all three condiments. 7. **Find number of Salad only** $$n(S \text{ only}) = n(S) - n(E \cap S) - n(S \cap G) + n(E \cap S \cap G)$$ Remember intersections counted twice are added back, but "only" excludes intersections. So, $$n(S \text{ only}) = 22 - (6 + 3) - (1 + 3) + 3 = 22 - 9 - 4 + 3 = 12$$ 8. **Find customers with only one condiment** $$n(E \text{ only}) = 26 - (6 + 4 + 3) = 26 - 13 = 13$$ $$n(S \text{ only}) = 12$$ (from above) $$n(G \text{ only}) = 25 - (4 + 1 + 3) = 25 - 8 = 17$$ Total only one condiment: $$13 + 12 + 17 = 42$$ --- 9. **Problem 2c: Find $k$ and equation of line QR** - Points: $P(2,-3)$, $Q(3,1)$, $R(-1,k)$ - $PQ$ is perpendicular to $QR$. 10. **Find vectors:** $$\overrightarrow{PQ} = (3 - 2, 1 - (-3)) = (1, 4)$$ $$\overrightarrow{QR} = (-1 - 3, k - 1) = (-4, k - 1)$$ 11. **Perpendicular vectors satisfy:** $$\overrightarrow{PQ} \cdot \overrightarrow{QR} = 0$$ So, $$1 \times (-4) + 4 \times (k - 1) = 0$$ $$-4 + 4k -4 = 0$$ $$4k - 8 = 0$$ $$4k = 8$$ $$k = 2$$ 12. **Find equation of line QR with $k=2$** Points: $Q(3,1)$ and $R(-1,2)$ 13. **Slope of QR:** $$m = \frac{2 - 1}{-1 - 3} = \frac{1}{-4} = -\frac{1}{4}$$ 14. **Equation using point-slope form:** $$y - 1 = -\frac{1}{4}(x - 3)$$ Simplify: $$y - 1 = -\frac{1}{4}x + \frac{3}{4}$$ $$y = -\frac{1}{4}x + \frac{3}{4} + 1 = -\frac{1}{4}x + \frac{7}{4}$$ --- 15. **Problem 3a: Right triangle with hypotenuse $(2x + 3)$ cm, other sides $x$ and $(x + 7)$ cm** 16. **Use Pythagoras theorem:** $$x^2 + (x + 7)^2 = (2x + 3)^2$$ 17. Expand each term: $$x^2 + (x^2 + 14x + 49) = 4x^2 + 12x + 9$$ $$2x^2 + 14x + 49 = 4x^2 + 12x + 9$$ 18. Rearrange all to one side: $$0 = 4x^2 + 12x + 9 - 2x^2 - 14x - 49$$ $$0 = 2x^2 - 2x - 40$$ 19. Simplify: $$x^2 - x - 20 = 0$$ 20. **Solve quadratic:** Factors of -20 that sum to -1 are -5 and 4. So, $$(x - 5)(x + 4) = 0$$ $$x = 5$$ or $$x = -4$$ (discard negative side length) 21. **Calculate area:** Area $= \frac{1}{2} \times x \times (x + 7) = \frac{1}{2} \times 5 \times 12 = 30$ cm$^2$ --- **Final answers:** - Number customers with all three condiments: $3$ - Number customers with Salad only: $12$ - Number customers with only one condiment: $42$ - Value of $k = 2$ - Equation of line $QR: y = -\frac{1}{4}x + \frac{7}{4}$ - Length $x = 5$ cm - Area of triangle = $30$ cm$^2$