These constructed practice examples show how an explanation can uncover an error and lead to a more complete answer.
Fractions: answer “how many remain?”
Question: A school has 240 tickets. It sells 3/8 of them. How many tickets remain?
A child might calculate 90 and stop. That calculation finds the tickets sold, so there is still a step to complete.
- Divide the whole into eight equal parts: 240 ÷ 8 = 30 tickets in each eighth.
- Three of those parts are sold: 3 × 30 = 90 tickets.
- Subtract the sold tickets: 240 − 90 = 150 tickets remain.
A strip divided into eight equal parts gives another explanation: if three parts are sold, five remain. Finding 5/8 of 240 also gives 150. Label each result “sold” or “remaining” to keep the calculation attached to the question.
Try a changed example: If 2/5 of 150 tickets are sold, how many remain? Each fifth is 30 tickets, so 60 are sold and 90 remain. Explaining why three fifths remain shows more understanding than repeating the first calculation.
Measurement: count complete pieces
Question: A ribbon is 2 metres long. How many complete 35 cm pieces can be cut from it, and how much ribbon is left?
First put the lengths in the same unit: 2 metres = 200 cm. Five pieces use 5 × 35 = 175 cm; six would need 6 × 35 = 210 cm. The answer is five complete pieces, with 25 cm left.
Rounding up to six pieces would require more ribbon than we have. The word “complete” tells us how to interpret the amount left over. Comparing multiples of 35 solves the problem without needing a long-division method.
Try a changed example: With the same 2-metre ribbon, make each piece 40 cm long. Five pieces use all 200 cm, so there are five complete pieces and no ribbon left. The number of pieces is unchanged, but the remainder is different.