The Tuck Shop Change Problem
Neo runs the tuck shop at a primary school in Palapye. On Monday morning the cash box holds P250 in coins: only P5, P2 and P1 coins. During break, 60 learners each buy one item. 25 learners buy a fat cake for P3, 20 buy a juice carton for P4, and 15 buy a packet of peanuts for P2. Every learner pays with a single P10 note. Neo must give correct change to each learner using only the coins already in the box, and she cannot ask anyone to swap money. By the end of break she has served all 60 learners successfully.
1. Calculate the total value of coins Neo must have paid out as change, and the total value of the P10 notes she received. Then explain why the starting coins in the box matter, not just the total value.
Read the worked answer
Notes received: 60 × P10 = P600.
Change per item:
- Fat cake P3: change P7 each, 25 × P7 = P175
- Juice P4: change P6 each, 20 × P6 = P120
- Peanuts P2: change P8 each, 15 × P8 = P120
Total change = P175 + P120 + P120 = P415.
Total value is not enough on its own. Neo needs the right denominations at the right moment. For example, a P7 change can be P5+P2, but if she has already used all her P5 coins on earlier learners, later learners cannot be served even though the box still holds plenty of value in P1 and P2 coins. The order of serving and the mix of coins decide whether the plan works. So the correct judgement is to check coin supply against the exact change amounts, not just the total.
2. Neo claims she can serve the learners in any order and still succeed. Is she right? Justify your answer with a specific ordering that would fail if she is wrong, or a reason why no ordering fails if she is right.
Read the worked answer
She is not automatically right. The outcome depends on the coin mix, which the setup does not fully fix.
Consider a coin mix of P250 made of 40 × P5, 20 × P2, 10 × P1. Total = 200 + 40 + 10 = P250.
If Neo serves all 25 fat-cake learners first, she pays P7 each. The natural P7 is P5+P2. After 20 such payments she has used all 20 P2 coins. The 21st fat-cake learner needs P7 but only P5 and P1 coins remain; she can pay P5+P1+P1 = P7, using two P1 coins. She has 10 P1 coins, so she can cover 5 more learners this way, reaching exactly 25. Then juice learners need P6 = P5+P1. She has no P1 coins left and no P2 coins. She cannot make P6 with only P5 coins. Failure.
So order matters. A safer strategy is to give change using the largest coins possible and to watch which denominations are running low, or to serve learners needing smaller change first. The judgement being tested is that feasibility is a sequencing problem, not just an arithmetic one.