AMC 8 · 2012 · #3

Grade 4 arithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Fraction / Expression Sum Evaluation

unit-conversionmulti-digit-arithmeticplace-value identify-subproblems ↑ Prerequisites: multi-digit-arithmetic

📏 Medium solution 💡 2 insights

Problem

On February 13 The Oshkosh Northwester listed the length of daylight as 10 hours and 24 minutes, the sunrise was $6:57\textsc{am}$ , and the sunset as $8:15\textsc{pm}$ . The length of daylight and sunrise were correct, but the sunset was wrong. When did the sun really set?

Pick an answer.

(A)

5:10 PM

(B)

5:21 PM

(C)

5:41 PM

(D)

5:57 PM

(E)

6:03 PM

AMC 8 2012 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.

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Toolkit + CCSS Solution

Understand

Restated: On February 13, the newspaper reports daylight as $10$ hours $24$ minutes, sunrise at $6{:}57$ AM, and sunset at $8{:}15$ PM. The daylight length and sunrise are correct, but the printed sunset is wrong. Find the actual sunset time.

Givens: Sunrise time: $6{:}57$ AM (correct); Length of daylight: $10$ hours $24$ minutes (correct); Printed sunset: $8{:}15$ PM (wrong — ignore); Answer choices: (A) $5{:}10$ PM, (B) $5{:}21$ PM, (C) $5{:}41$ PM, (D) $5{:}57$ PM, (E) $6{:}03$ PM

Unknowns: The correct sunset time, written in $12$-hour PM format

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #3 Eliminate Possibilities

Sunset = sunrise + daylight is a single sum, but it is cleaner to split it into two subproblems (Tool #7): add the hours, then add the minutes, then carry. After computing, Tool #3 (Eliminate Possibilities) lets us check the answer against the five choices — a free verification on any AMC multiple-choice problem. Tool #8 (Units) is implicitly used, but the bookkeeping here is hours vs. minutes, not a unit conversion across systems, so #7 captures the structure best.

Execute — Answer: B

#7 Identify Subproblems 4.MD.A.2 Step 1

Set up the relationship.
Daylight is the time from sunrise to sunset, so the correct sunset equals the correct sunrise plus the correct daylight length.
Ignore the printed (wrong) sunset.

$$\text{sunset} = 6{:}57 \text{ AM} + 10 \text{ h } 24 \text{ min}$$

💡 Picking the right relationship before computing is the first move of Tool #7 — name the subproblems before doing them.

#7 Identify Subproblems 4.NBT.B.4 Step 2

Subproblem 1: add the hours.
$6$ hours (from $6{:}57$) plus $10$ hours of daylight gives $16$ hours.

$$6 + 10 = 16 \text{ h}$$

💡 Treating hours as their own column is exactly the "add place by place" Grade 4 algorithm.

#7 Identify Subproblems 4.NBT.B.4 Step 3

Subproblem 2: add the minutes.
$57$ minutes plus $24$ minutes is $81$ minutes.

$$57 + 24 = 81 \text{ min}$$

💡 Same place-value addition, this time in the minutes column.

#7 Identify Subproblems 4.MD.A.1 Step 4

Carry.
$81$ minutes is more than $60$, so peel off $60$ minutes as $1$ hour.
That leaves $21$ minutes and bumps the hour count from $16$ to $17$.

$81 \text{ min} = 60 \text{ min} + 21 \text{ min} = 1 \text{ h } 21 \text{ min}$, so total $= 17 \text{ h } 21 \text{ min}$

💡 Trading $60$ minutes for $1$ hour is the same idea as carrying $10$ ones into a ten — Grade 4 unit conversion within the time system.

#3 Eliminate Possibilities 4.MD.A.2 Step 5

Convert $17{:}21$ in $24$-hour format to PM time by subtracting $12$ from the hours.
Then check the answer choices: $5{:}21$ PM is choice (B).

$$17 - 12 = 5 \;\Rightarrow\; 17{:}21 = 5{:}21 \text{ PM} \;\Rightarrow\; \textbf{(B)}$$

💡 Matching the computed result against the five choices is Tool #3 (Eliminate Possibilities) — only (B) survives, the rest are off by $11$, $20$, $36$, or $42$ minutes.

[1] #7 4.MD.A.2 Set up the relationship. Daylight is the time from sunrise to sunset, so the cor

[2] #7 4.NBT.B.4 Subproblem 1: add the hours. $6$ hours (from $6{:}57$) plus $10$ hours of daylig

[3] #7 4.NBT.B.4 Subproblem 2: add the minutes. $57$ minutes plus $24$ minutes is $81$ minutes.

[4] #7 4.MD.A.1 Carry. $81$ minutes is more than $60$, so peel off $60$ minutes as $1$ hour. Tha

[5] #3 4.MD.A.2 Convert $17{:}21$ in $24$-hour format to PM time by subtracting $12$ from the ho

Review

Reasonableness: Quick sanity check: $10$ hours $24$ minutes after $6{:}57$ AM should land in the late afternoon. Round to $11$ hours after $7$ AM, that is roughly $6$ PM, so a sunset near $5$ to $6$ PM is plausible. The exact value $5{:}21$ PM sits in that window. Reverse-check: from $5{:}21$ PM back to $6{:}57$ AM, count from $6{:}57$ AM to $5{:}21$ PM as $5$ hours $3$ minutes to noon, plus $5$ hours $21$ minutes after noon, totaling $10$ hours $24$ minutes. Matches.

Alternative: Tool #11 (Work Backwards): start from each PM choice, subtract $10$ h $24$ min, and see which one gives $6{:}57$ AM. $5{:}21$ PM $-$ $10$ h $24$ min $=$ $6{:}57$ AM exactly, picking out (B). The other choices land at $6{:}46$, $7{:}17$, $7{:}33$, or $7{:}39$ AM — none match the given sunrise.

CCSS standards used (min grade 4)

4.MD.A.2 Solve word problems involving distances, time, liquid volumes, and money (Recognizing that sunset = sunrise + daylight and computing the elapsed-time word problem in hours and minutes.)
4.NBT.B.4 Fluently add and subtract multi-digit whole numbers using the standard algorithm (Adding the hours ($6 + 10 = 16$) and the minutes ($57 + 24 = 81$) column by column before regrouping.)
4.MD.A.1 Know relative sizes of measurement units within one system; convert larger units to smaller (Trading $60$ minutes for $1$ hour to rewrite $16$ h $81$ min as $17$ h $21$ min, and converting $17{:}21$ from $24$-hour to $12$-hour PM format.)

⭐ If you can add hours, add minutes, and carry $60$ minutes as $1$ hour, you have all the Grade 4 skills this AMC 8 problem needs.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 4)

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