AMC 10 · 2010 · #4
Grade 5 arithmeticA book that is to be recorded onto compact discs takes 412 minutes to read aloud. Each disc can hold up to 56 minutes of reading. Assume that the smallest possible number of discs is used and that each disc contains the same length of reading. How many minutes of reading will each disc contain?
Pick an answer.
AMC 10 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A book takes $412$ minutes to read aloud, and it must be split across compact discs where each disc holds at most $56$ minutes. We use as few discs as possible, and every disc must carry the exact same amount of reading. We want the number of minutes of reading on each disc.
Givens: The whole book is $412$ minutes long.; Each disc can hold at most $56$ minutes.; The number of discs is the smallest possible.; Every disc contains the same length of reading.; Answer choices: (A) $50.2$, (B) $51.5$, (C) $52.4$, (D) $53.8$, (E) $55.2$.
Unknowns: How many minutes of reading each disc holds when the total is split evenly across the fewest discs.
Understand
Restated: A book takes $412$ minutes to read aloud, and it must be split across compact discs where each disc holds at most $56$ minutes. We use as few discs as possible, and every disc must carry the exact same amount of reading. We want the number of minutes of reading on each disc.
Givens: The whole book is $412$ minutes long.; Each disc can hold at most $56$ minutes.; The number of discs is the smallest possible.; Every disc contains the same length of reading.; Answer choices: (A) $50.2$, (B) $51.5$, (C) $52.4$, (D) $53.8$, (E) $55.2$.
Plan
Primary tool: #14 Extreme Principle
Secondary: #8 Analyze the Units, #3 Eliminate Possibilities
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 $412 \div 56$ 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.
Execute — Answer: B
4.OA.A.3 Step 1 Find the fewest discs
- Divide the total time by the disc capacity to see how many discs are needed.
- $412 \div 56 = 7.357\ldots$.
- You cannot buy a fraction of a disc, and $7$ discs hold only $7 \times 56 = 392$ minutes, which is less than $412$.
- So $7$ is too few; you must round up to $8$ discs, which can hold $8 \times 56 = 448$ minutes — enough.
💡 When a division does not come out even and you need whole containers, you always round up so nothing is left over.
5.NBT.B.7 Step 2 Share the time evenly
- With the disc count locked at $8$ and every disc holding the same amount, split the $412$ minutes into $8$ equal parts.
- $412 \div 8 = 51.5$ minutes per disc.
💡 Equal sharing is just dividing the whole by the number of parts.
5.NBT.B.7 Step 3 Check and read off the answer
- Confirm this fits: $51.5 \le 56$, so each disc stays under its limit, and $51.5 \times 8 = 412$ rebuilds the whole book.
- Among the choices, only $51.5$ matches, so the answer is $\textbf{(B)}\ 51.5$.
💡 A good answer must both fit under the capacity and add back up to the total.
4.OA.A.3 Divide the total time by the disc capacity to see how many discs are needed. $41 5.NBT.B.7 With the disc count locked at $8$ and every disc holding the same amount, split 5.NBT.B.7 Confirm this fits: $51.5 \le 56$, so each disc stays under its limit, and $51.5 Review
Reasonableness: The per-disc time must be at most $56$ and, since $8$ discs is the fewest possible, the load per disc should be close to the $56$ limit rather than far below it. $51.5$ minutes is just under $56$, and using one fewer disc would demand $412 \div 7 \approx 58.9$ minutes per disc, which overflows the $56$ cap — confirming $8$ really is the minimum. Multiplying back, $8 \times 51.5 = 412$, so no reading is lost. Everything checks out with (B).
Alternative: Use Tool #3 (Eliminate Possibilities) directly on the choices. If the reading is split evenly into a whole number of equal discs, then $412$ divided by the minutes-per-disc must be a whole number. Test each option: $412 \div 50.2$, $412 \div 52.4$, $412 \div 53.8$, and $412 \div 55.2$ are all non-integers, but $412 \div 51.5 = 8$ exactly. Only (B) yields a whole number of discs, and $8$ is also the smallest count that keeps each disc at or below $56$ minutes.
CCSS standards used (min grade 5)
4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Dividing $412$ by $56$ and rounding up to the next whole disc because $7$ discs cannot hold all $412$ minutes.)5.NBT.B.7Add, subtract, multiply, and divide decimals to hundredths (Dividing $412$ by $8$ to get $51.5$ minutes per disc, and checking $51.5 \times 8 = 412$.)
⭐ You cannot buy part of a disc, so round the number of discs up to $8$, then share the $412$ minutes equally: $412 \div 8 = 51.5$ minutes on each disc.
⭐ You cannot buy part of a disc, so round the number of discs up to $8$, then share the $412$ minutes equally: $412 \div 8 = 51.5$ minutes on each disc.
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