AMC 10 · 2010 · #4
Grade 5 arithmeticPick an answer.
AMC 10 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The phrase "smallest possible number of discs" is a boundary question, so Tool #14 (Extreme Principle) is the lead: find the fewest whole discs whose capacity can still cover 412 minutes. Because 412 ÷ 56 is not a whole number, we round up to the next whole disc. Tool #8 (Analyze the Units) keeps the bookkeeping honest — we are tracking minutes per disc, so once the disc count is fixed we simply share the 412 minutes equally. Tool #3 (Eliminate Possibilities) is a fast cross-check: only one answer choice can equal 412 split into a whole number of equal parts.
Find the fewest discs
412 ÷ 56 = 7.357…, and 7 discs hold only 392 minutes, so round the count up to 8 discs.
When a division does not come out even and you need whole containers, you always round up so nothing is left over.
When the division does not come out even and whole containers are needed, you always round up.
▸ Why?
The leftover after filling whole containers still has to go somewhere, so one more is required.
▸ Why?
Any smaller count leaves material behind, so the rounded-up count is the smallest that works.
Share the time evenly
Every disc carries the same amount, so split 412 minutes into 8 equal parts: 412 ÷ 8 = 51.5 minutes per disc.
Equal sharing is just dividing the whole by the number of parts.
5.NBT.B.7Analyze The UnitsCheck and read off the answer
Check it fits: 51.5 ≤ 56 stays under the limit and 51.5 × 8 = 412 rebuilds the book, so the answer is (B).
A good answer must both fit under the capacity and add back up to the total.
5.NBT.B.7Eliminate PossibilitiesYou cannot buy part of a disc, so round the number of discs up to 8, then share the 412 minutes equally: 412 ÷ 8 = 51.5 minutes on each disc.
- Find the fewest discs
- Share the time evenly
- Check and read off the answer