AMC 10 · 2022 · #4
Grade 4 rate-ratioProblem
A donkey suffers an attack of hiccups and the first hiccup happens at one afternoon. Suppose that
the donkey hiccups regularly every seconds. At what time does the donkey’s th hiccup occur?
Pick an answer.
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 $4{:}58{:}15$.