A zoo trip costs R90. Adult tickets cost R20 and child tickets cost R10. Find all possible combinations of adults and children that could have visited the zoo.

Here are the solutions for the problems.

11 May 2026: Challenge - A visit to the zoo

Let $A$ be the number of adults and $C$ be the number of children.
The cost for an adult ticket is R20 and for a child ticket is R10.
The total cost is R90.
The equation is: $20A + 10C = 90$.
We can simplify this by dividing by 10: $2A + C = 9$.

We need to find whole number solutions for $A$ and $C$.

• Step 1: Find possible combinations.

  • If $A = 1$: $2(1) + C = 9 \Rightarrow 2 + C = 9 \Rightarrow C = 7$. (1 adult, 7 children)
  • If $A = 2$: $2(2) + C = 9 \Rightarrow 4 + C = 9 \Rightarrow C = 5$. (2 adults, 5 children)
  • If $A = 3$: $2(3) + C = 9 \Rightarrow 6 + C = 9 \Rightarrow C = 3$. (3 adults, 3 children)
  • If $A = 4$: $2(4) + C = 9 \Rightarrow 8 + C = 9 \Rightarrow C = 1$. (4 adults, 1 child)
  • If $A = 5$: $2(5) + C = 9 \Rightarrow 10 + C = 9 \Rightarrow C = -1$. (Not possible)

• Step 2: Provide one answer. Let's choose the first valid combination. How many are children? $\boxed{7}$ How many are adults? $\boxed{1}$

• Step 3: List other possible answers. Is there another answer? Yes. Adults $\boxed{2}$ Children $\boxed{5}$ Adults $\boxed{3}$ Children $\boxed{3}$ Adults $\boxed{4}$ Children $\boxed{1}$

12 May 2026: Homework - Parcel weight

One parcel weighs 3 kg.

• a. 2 Parcel: