Six Ways to Spend $33.75
The school store sells notebooks for $2.25 each and pens for $1.50 each; there is no tax and nothing else is sold. A class's receipt shows a total of $33.75, and the clerk remembers that the class bought between 15 and 20 items in all. The treasurer wants to know exactly how many of each were bought so that the cost can be split fairly. A parent adds one more memory: the class bought at least twice as many pens as notebooks. The treasurer is uneasy - a receipt plus two half-remembered clues feels like a thin basis for splitting a bill twelve ways.
1. Ignoring the parent's clue, how many different notebook/pen combinations fit the receipt and the 15-to-20 item range? List them all and show why the list is complete.
Read the worked answer
Set up. Let n = notebooks, p = pens. Then 2.25n + 1.50p = 33.75. Multiply by 4: 9n + 6p = 135, so 3n + 2p = 45.
Integer restrictions. 2p is even and 45 is odd, so 3n is odd, so n must be odd. Also n + p must be between 15 and 20.
Enumerate odd n: | n | p = (45 − 3n)/2 | n + p | in range? | | 5 | 15 | 20 | yes | | 7 | 12 | 19 | yes | | 9 | 9 | 18 | yes | | 11 | 6 | 17 | yes | | 13 | 3 | 16 | yes | | 15 | 0 | 15 | yes | n = 3 gives p = 18 and a total of 21 items - out of range; larger odd n gives a negative p. So the list is complete.
Six combinations satisfy the receipt and the clerk's range: the receipt alone does not answer the question.
2. Apply the parent's clue (pens at least twice the notebooks). Which combination survives, and why is an inequality a more powerful clue here than another exact number would be?
Read the worked answer
Test each candidate against p ≥ 2n:
- n = 5, p = 15: 15 ≥ 10 ✔
- n = 7, p = 12: 12 < 14 ✘
- n = 9, p = 9: 9 < 18 ✘
- n = 11, p = 6: ✘
- n = 13, p = 3: ✘
- n = 15, p = 0: ✘
Answer: 5 notebooks and 15 pens. Check: 5 × $2.25 = $11.25 and 15 × $1.50 = $22.50, total $33.75 ✔; 20 items, inside 15–20 ✔.
The clue is powerful because it is a direction: as n rises by 2, p falls by 3, so the ratio pens-to-notebooks drops steadily and the inequality cuts off the entire upper half of the list at once, instead of merely adjusting one entry.
3. The parent says she may have meant 'at most twice as many pens.' Does the answer stay unique? Then identify a single extra fact that would decide the question outright.
Read the worked answer
'At most twice as many pens' does not settle it. p ≤ 2n keeps n = 7, 9, 11, 13, 15 (12 ≤ 14, 9 ≤ 18, 6 ≤ 22, 3 ≤ 26, 0 ≤ 30) and drops only n = 5. So one word - 'at least' versus 'at most' - is the difference between a unique answer and five candidates. The treasurer is right to be uneasy: a misheard comparison flips the conclusion.
A single fact that decides it: the total number of items. The six candidates have totals 20, 19, 18, 17, 16, 15 - all different. If the clerk recalled only 'there were exactly 18 items', then n = 9 and p = 9 follows immediately. Knowing the total, the number of notebooks, or the number of pens would each pin the combination; the money total alone never does.