AMC 10 · 2022 · #4
Easy mode Grade 4Problem
A donkey starts hiccuping at 4:00 in the afternoon. That first hiccup happens right at 4:00.
After that, the donkey hiccups again every 5 seconds, like clockwork. So the second hiccup is 5 seconds after 4:00, the third hiccup is 10 seconds after 4:00, and so on.
At what time does the 700th hiccup happen?
Pick an answer.
AMC 10 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: A donkey's first hiccup happens at exactly $4{:}00$ in the afternoon and then one hiccup every $5$ seconds. At what wall-clock time does the $700$th hiccup occur?
Givens: The $1$st hiccup is at $4{:}00{:}00$ PM; Hiccups are evenly spaced $5$ seconds apart; We want the time of the $700$th hiccup; Answer choices: (A) $15$ s after $4{:}58$, (B) $20$ s after $4{:}58$, (C) $25$ s after $4{:}58$, (D) $30$ s after $4{:}58$, (E) $35$ s after $4{:}58$
Unknowns: The clock time (minutes and seconds past $4{:}00$) of the $700$th hiccup
Understand
Restated: A donkey's first hiccup happens at exactly $4{:}00$ in the afternoon and then one hiccup every $5$ seconds. At what wall-clock time does the $700$th hiccup occur?
Givens: The $1$st hiccup is at $4{:}00{:}00$ PM; Hiccups are evenly spaced $5$ seconds apart; We want the time of the $700$th hiccup; Answer choices: (A) $15$ s after $4{:}58$, (B) $20$ s after $4{:}58$, (C) $25$ s after $4{:}58$, (D) $30$ s after $4{:}58$, (E) $35$ s after $4{:}58$
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #5 Look for a Pattern, #8 Analyze the Units
Jumping straight to the $700$th hiccup is hard to picture. Tool #9 (Easier Related Problem) says: try $n = 2$, $n = 3$, $n = 4$ first and see the rule that connects the hiccup number to elapsed seconds. Tool #5 (Look for a Pattern) makes the rule explicit — the $n$th hiccup happens $5(n - 1)$ seconds after $4{:}00$. Tool #8 (Analyze the Units) finishes the job by converting total seconds to minutes-and-seconds for the clock answer.
Execute — Answer: A
4.OA.C.5 Step 1 - Try small cases to anchor the rule.
- Hiccup $1$ is at $0$ s after $4{:}00$ — no gap yet.
- Hiccup $2$ is after one $5$-second gap ($5$ s).
- Hiccup $3$ is after two gaps ($10$ s).
- Hiccup $4$ is after three gaps ($15$ s).
💡 Three small cases reveal the rule "elapsed seconds $= 5 \cdot (\text{hiccup number} - 1)$" — Grade 4 "generate a pattern from a rule".
4.OA.C.5 Step 2 - Apply the pattern to $n = 700$.
- Number of gaps between the first and the $700$th hiccup is $700 - 1 = 699$.
💡 The fencepost rule: with $n$ posts there are $n - 1$ fence sections — same idea applies to hiccups and the gaps between them.
4.NBT.B.5 Step 3 Multiply gaps by seconds per gap to get total elapsed time.
💡 $699 \times 5 = (700 - 1) \times 5 = 3500 - 5 = 3495$ — Grade 4 "multiply multi-digit by one-digit".
4.MD.A.1 Step 4 Convert $3495$ seconds to minutes and remainder seconds using $60$ s = $1$ min.
💡 Convert smaller units to larger units — Grade 4 "relative sizes of measurement units". $60 \times 58 = 3480$, then $3495 - 3480 = 15$.
4.MD.A.2 Step 5 - Add the elapsed time to the start time $4{:}00{:}00$.
- The $700$th hiccup is at $4{:}58{:}15$, which is $15$ seconds after $4{:}58$ — choice (A).
💡 Add minutes and seconds to a clock time — Grade 4 "distances, time, money word problems".
4.OA.C.5 Try small cases to anchor the rule. Hiccup $1$ is at $0$ s after $4{:}00$ — no g 4.OA.C.5 Apply the pattern to $n = 700$. Number of gaps between the first and the $700$th 4.NBT.B.5 Multiply gaps by seconds per gap to get total elapsed time. 4.MD.A.1 Convert $3495$ seconds to minutes and remainder seconds using $60$ s = $1$ min. 4.MD.A.2 Add the elapsed time to the start time $4{:}00{:}00$. The $700$th hiccup is at $ Review
Reasonableness: Order-of-magnitude check. $700$ hiccups at one every $5$ seconds is roughly $700 \times 5 = 3500$ seconds, which is just under $3600 = 60 \times 60$ — so just under one hour. Starting at $4{:}00$, we land just before $5{:}00$, near $4{:}58$ or so. Our computed $4{:}58{:}15$ matches. Also the fencepost adjustment ($699$ instead of $700$) saves us $5$ seconds — that's why the answer is $15$ s and not $20$ s past $4{:}58$, ruling out choice (B).
Alternative: Tool #13 (Convert to Algebra). Let $t_n$ denote the time of the $n$th hiccup as seconds past $4{:}00$. Linear model $t_n = 5(n - 1)$. Plug $n = 700$: $t_{700} = 5 \cdot 699 = 3495$ s. Convert: $3495 = 58 \cdot 60 + 15$. Same answer, just stated as an explicit formula.
CCSS standards used (min grade 4)
4.OA.C.5Generate a number or shape pattern following a given rule (Building the rule "elapsed seconds $= 5(n - 1)$" from small cases ($n = 1, 2, 3, 4$).)4.NBT.B.5Multiply a whole number of up to four digits by a one-digit whole number (Computing $699 \times 5 = 3495$ to get total elapsed seconds.)4.MD.A.1Know relative sizes of measurement units and convert larger to smaller units (Knowing $1\text{ min} = 60\text{ s}$ to convert $3495$ seconds into $58$ min $15$ s.)4.MD.A.2Solve word problems involving distances, time, liquid volumes, and money (Adding $58$ min $15$ s to the start time $4{:}00{:}00$ to land on $4{:}58{:}15$.)
⭐ This AMC 10 problem only needs Grade 4 "with $n$ events there are $n-1$ gaps" — the $700$th hiccup happens after $699 \times 5 = 3495$ seconds, which is $58$ minutes $15$ seconds, landing at $4{:}58{:}15$.
⭐ This AMC 10 problem only needs Grade 4 "with $n$ events there are $n-1$ gaps" — the $700$th hiccup happens after $699 \times 5 = 3495$ seconds, which is $58$ minutes $15$ seconds, landing at $4{:}58{:}15$.
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