AMC 8 · 2022 · #7
Grade 6 rate-ratioProblem
When the World Wide Web first became popular in the 1990s, download speeds reached a maximum of about 56 kilobits per second. Approximately how many minutes would the download of a 4.2-megabyte song have taken at that speed? (Note that there are 8000 kilobits in a megabyte.)
Pick an answer.
AMC 8 2022 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: In the $1990$s the fastest dial-up internet sent data at about $56$ kilobits per second. A song is $4.2$ megabytes large, and $1$ megabyte is $8000$ kilobits. Roughly how many minutes would the song have taken to download at $56$ kilobits per second?
Givens: Download speed = $56$ kilobits per second (kbps); Song size = $4.2$ megabytes (MB); Unit conversion: $1$ MB $= 8000$ kilobits (kb); Answer choices (minutes): (A) $0.6$, (B) $10$, (C) $1800$, (D) $7200$, (E) $36000$
Unknowns: The approximate download time of the song, expressed in minutes
Understand
Restated: In the $1990$s the fastest dial-up internet sent data at about $56$ kilobits per second. A song is $4.2$ megabytes large, and $1$ megabyte is $8000$ kilobits. Roughly how many minutes would the song have taken to download at $56$ kilobits per second?
Givens: Download speed = $56$ kilobits per second (kbps); Song size = $4.2$ megabytes (MB); Unit conversion: $1$ MB $= 8000$ kilobits (kb); Answer choices (minutes): (A) $0.6$, (B) $10$, (C) $1800$, (D) $7200$, (E) $36000$
Plan
Primary tool: #8 Analyze the Units
Secondary: #7 Identify Subproblems
The core relationship is $\text{time} = \text{size} / \text{speed}$, but the size is in megabytes, the speed is in kilobits per second, and the answer is in minutes. Tool #8 (Analyze the Units) is the natural primary tool: track "MB", "kb", "s", and "min" through every operation so the wrong unit can never sneak into the final answer. Tool #7 (Identify Subproblems) splits the calculation into three clean pieces — (1) convert the file size to kilobits, (2) divide by the rate to get seconds, (3) convert seconds to minutes — so each step does exactly one job and the arithmetic stays small.
Execute — Answer: B
5.NBT.B.7 Step 1 - Convert the song's size from megabytes to kilobits.
- The given conversion factor is $8000$ kb per $1$ MB, so multiply the size in MB by $8000$.
💡 Multiplying a decimal ($4.2$) by a whole number ($8000$) is exactly the Grade 5 "decimals to hundredths" skill, and the "MB" units cancel to leave "kb".
5.NBT.B.6 Step 2 - Divide the file size in kilobits by the download rate to get the time in seconds.
- Spotting that $56 \times 6 = 336$ makes the arithmetic clean: $33{,}600 = 336 \times 100 = 56 \times 6 \times 100$.
💡 Dividing a four-digit number by a two-digit number is the Grade 5 long-division standard; the "kb" cancels and only "s" is left, which Tool #8 confirms is correct.
5.MD.A.1 Step 3 - Convert the download time from seconds to minutes.
- Since $1$ minute $= 60$ seconds, divide the total seconds by $60$.
💡 Converting between standard units of time within one system (seconds $\to$ minutes) is the Grade 5 measurement-conversion standard.
6.RP.A.3 Step 4 - Cross-check the result against the overall rate $\text{time} = \text{size} / \text{speed}$.
- The full chain of units is $\text{MB} \to \text{kb} \to \text{s} \to \text{min}$, every "unwanted" unit cancels, and only "min" is left, which matches what the problem asks for.
💡 Treating $56$ kb/s as a unit rate and using it together with another rate ($8000$ kb/MB) is Grade 6 rate reasoning across multiple unit conversions.
5.NBT.B.7 Convert the song's size from megabytes to kilobits. The given conversion factor 5.NBT.B.6 Divide the file size in kilobits by the download rate to get the time in seconds 5.MD.A.1 Convert the download time from seconds to minutes. Since $1$ minute $= 60$ secon 6.RP.A.3 Cross-check the result against the overall rate $\text{time} = \text{size} / \te Review
Reasonableness: $10$ minutes feels right for a song on $1990$s dial-up: today the same $4.2$ MB song downloads in well under a second over broadband, but $56$ kbps is about $1000$ times slower than a basic modern connection, so multiplying a fraction of a second by roughly $1000$ lands in the ten-minute range. The other choices fail an order-of-magnitude check: (A) $0.6$ min $= 36$ s is way too fast, while (C) $1800$ min $= 30$ hr and (D) $7200$ min $= 5$ days are way too slow for a single song.
Alternative: Tool #3 (Eliminate Possibilities) on the answer choices: convert each candidate back into seconds and check whether $33{,}600$ kb fits into that time at $56$ kb/s. Only (B) $10$ min $= 600$ s satisfies $56 \times 600 = 33{,}600$ kb, so all other choices are eliminated without a fresh calculation.
CCSS standards used (min grade 6)
5.NBT.B.7Add, subtract, multiply, and divide decimals to hundredths (Multiplying the decimal file size by the whole-number conversion factor: $4.2 \times 8000 = 33{,}600$ kilobits.)5.NBT.B.6Find whole-number quotients with up to four-digit dividends and two-digit divisors (Dividing the file size in kilobits by the download speed: $33{,}600 \div 56 = 600$ seconds.)5.MD.A.1Convert among different-sized standard measurement units within a given system (Converting the download time from seconds to minutes: $600 \text{ s} \div 60 = 10$ minutes.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Combining two rates ($8000$ kb per MB and $56$ kb per second) through unit-cancellation reasoning to chain MB $\to$ kb $\to$ s $\to$ min and obtain $10$ minutes.)
⭐ This AMC 8 problem only needs Grade 6 rate reasoning — chaining unit conversions until just "minutes" is left — that you already know!
⭐ This AMC 8 problem only needs Grade 6 rate reasoning — chaining unit conversions until just "minutes" is left — that you already know!
More like this
Same archetype — closest grade level first.