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.

Fresh day Elisa, let's solve. 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 2 + C = 9 C = 7. (1 adult, 7 children) If A = 2: 2(2) + C = 9 4 + C = 9 C = 5. (2 adults, 5 children) If A = 3: 2(3) + C = 9 6 + C = 9 C = 3. (3 adults, 3 children) If A = 4: 2(4) + C = 9 8 + C = 9 C = 1. (4 adults, 1 child) If A = 5: 2(5) + C = 9 10 + C = 9 C = -1. (Not possible) Step 2: Provide one answer. Let's choose the first valid combination. How many are children? 7 How many are adults? 1 Step 3: List other possible answers. Is there another answer? Yes. Adults 2 Children 5 Adults 3 Children 3 Adults 4 Children 1 12 May 2026: Homework - Parcel weight One parcel weighs 3 kg. a. 2 Parcel: Step 1: Multiply the weight of one parcel by the number of parcels. 2 × 3\, kg = 6\, kg The weight of 2 parcels is 6\, kg. b. 3 Parcel: Step 1: Multiply the weight of one parcel by the number of parcels. 3 × 3\, kg = 9\, kg The weight of 3 parcels is 9\, kg. c. 4 Parcel: Step 1: Multiply the weight of one parcel by the number of parcels. 4 × 3\, kg = 12\, kg The weight of 4 parcels is 12\, kg. 13 May 2026: Homework - Subtract without a calculator a. 68 - 24 Step 1: Subtract the ones digits: 8 - 4 = 4. Step 2: Subtract the tens digits: 6 - 2 = 4. 68 - 24 = 44 b. 60 - 38 Step 1: Subtract the ones digits. Since 0 < 8, borrow from the tens place. The 6 becomes 5, and the 0 becomes 10. So, 10 - 8 = 2. Step 2: Subtract the tens digits: 5 - 3 = 2. 60 - 38 = 22 Drop the next question!