AMC 8 · 2011 · #5
Grade 4 arithmeticProblem
What time was it 2011 minutes after midnight on January 1, 2011?
Pick an answer.
AMC 8 2011 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: Starting from midnight at the beginning of January 1, 2011, advance the clock by $2011$ minutes. What date and time does the clock show?
Givens: Start time: $\text{January 1, 2011, at 12:00 AM (midnight)}$; Elapsed amount: $2011$ minutes; $1$ hour $= 60$ minutes; $1$ day $= 24$ hours; Answer choices: (A) Jan 1, 9:31 PM, (B) Jan 1, 11:51 PM, (C) Jan 2, 3:11 AM, (D) Jan 2, 9:31 AM, (E) Jan 2, 6:01 PM
Unknowns: The clock date and time after $2011$ minutes have elapsed from the starting midnight
Understand
Restated: Starting from midnight at the beginning of January 1, 2011, advance the clock by $2011$ minutes. What date and time does the clock show?
Givens: Start time: $\text{January 1, 2011, at 12:00 AM (midnight)}$; Elapsed amount: $2011$ minutes; $1$ hour $= 60$ minutes; $1$ day $= 24$ hours; Answer choices: (A) Jan 1, 9:31 PM, (B) Jan 1, 11:51 PM, (C) Jan 2, 3:11 AM, (D) Jan 2, 9:31 AM, (E) Jan 2, 6:01 PM
Plan
Primary tool: #8 Analyze the Units
Secondary: #7 Identify Subproblems
The given quantity is in minutes but the answer mixes days, hours, and minutes, so the work is a unit-conversion cascade — exactly Tool #8 (Analyze the Units). Tool #7 (Identify Subproblems) splits the single "add $2011$ minutes" task into three clean subproblems: (a) convert $2011$ minutes into hours plus leftover minutes, (b) convert the resulting hours into days plus leftover hours, (c) place the leftover hours and minutes on the clock starting at midnight. Each subproblem is just division with remainder.
Execute — Answer: D
4.MD.A.1 Step 1 - Convert $2011$ minutes into hours and leftover minutes.
- Because $1$ hour $= 60$ minutes, divide $2011$ by $60$: the quotient is the number of whole hours and the remainder is the leftover minutes.
💡 Trading $60$ minutes for $1$ hour is the standard minutes-to-hours conversion taught in Grade 4.
4.NBT.B.6 Step 2 - Convert the $33$ hours into days and leftover hours.
- Because $1$ day $= 24$ hours, divide $33$ by $24$: the quotient is the number of whole days and the remainder is the leftover hours.
💡 Finding a whole-number quotient and remainder when dividing $33$ by $24$ is the Grade 4 division-with-remainder skill.
4.OA.A.3 Step 3 - Combine the pieces.
- The total elapsed time is $1$ day $+ 9$ hours $+ 31$ minutes.
- Stating it as one expression makes the next step (a single calendar/clock add) easy.
💡 Splitting the trip into "days", "hours", and "minutes" subproblems is the Tool #7 subproblems move, and assembling the parts is the Grade 4 multi-step word-problem skill.
3.MD.A.1 Step 4 - Add the elapsed time to the start time.
- Start at January 1, 2011, $12{:}00$ AM.
- Adding $1$ day moves the date to January 2, 2011, $12{:}00$ AM.
- Adding $9$ hours gives January 2 at $9{:}00$ AM.
- Adding $31$ minutes gives January 2 at $9{:}31$ AM.
💡 Adding elapsed time to a clock and rolling the date over at midnight is the Grade 3 elapsed-time skill.
4.MD.A.1 Convert $2011$ minutes into hours and leftover minutes. Because $1$ hour $= 60$ 4.NBT.B.6 Convert the $33$ hours into days and leftover hours. Because $1$ day $= 24$ hour 4.OA.A.3 Combine the pieces. The total elapsed time is $1$ day $+ 9$ hours $+ 31$ minutes 3.MD.A.1 Add the elapsed time to the start time. Start at January 1, 2011, $12{:}00$ AM. Review
Reasonableness: A quick sanity bracket: $1440$ minutes $= 24$ hours $= 1$ full day, so $2011$ minutes is more than $1$ day but less than $2$ days — the answer must land on January 2. Subtract: $2011 - 1440 = 571$ minutes past midnight on Jan 2. That is $571 \div 60 = 9$ hours R $31$ min, so $9{:}31$ AM. Matches (D).
Alternative: Tool #6 (Guess and Check) on the answer choices: convert each choice back to "minutes after Jan 1 midnight" and look for $2011$. (A) Jan 1, $9{:}31$ PM $= 21 \times 60 + 31 = 1291$. (B) Jan 1, $11{:}51$ PM $= 23 \times 60 + 51 = 1431$. (C) Jan 2, $3{:}11$ AM $= 1440 + 3 \times 60 + 11 = 1631$. (D) Jan 2, $9{:}31$ AM $= 1440 + 9 \times 60 + 31 = 2011$ — the only match. (E) Jan 2, $6{:}01$ PM $= 1440 + 18 \times 60 + 1 = 2521$.
CCSS standards used (min grade 4)
3.MD.A.1Tell and write time to the nearest minute and solve elapsed time problems (Adding $1$ day $9$ hours $31$ minutes onto a starting clock-and-calendar time of Jan 1, $12{:}00$ AM to land on Jan 2, $9{:}31$ AM.)4.MD.A.1Know relative sizes of measurement units and convert larger to smaller units (Using the fact that $1$ hour $= 60$ minutes and $1$ day $= 24$ hours to move between minutes, hours, and days.)4.NBT.B.6Find whole-number quotients and remainders with up to four-digit dividends (Computing $2011 \div 60 = 33$ R $31$ and $33 \div 24 = 1$ R $9$ to break the elapsed time into days, hours, and minutes.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Stringing the two divisions together and assembling the days/hours/minutes pieces into a single elapsed-time expression.)
⭐ This AMC 8 problem is just two Grade 4 divisions ($2011 \div 60$, then $33 \div 24$) and adding the leftovers to a clock — you can already do it!
⭐ This AMC 8 problem is just two Grade 4 divisions ($2011 \div 60$, then $33 \div 24$) and adding the leftovers to a clock — you can already do it!
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