← Sensim Lab Sensim Math

AMC 10 · 2022 · #4

Easy mode Grade 4
📗 View original problem →

Problem

A donkey starts hiccuping at 4:004:00 in the afternoon. That first hiccup happens right at 4:004:00.

After that, the donkey hiccups again every 55 seconds, like clockwork. So the second hiccup is 55 seconds after 4:004:00, the third hiccup is 1010 seconds after 4:004:00, and so on.

At what time does the 700700th hiccup happen?

(A) 15 seconds after 4:58\textbf{(A) }15 \text{ seconds after } 4:58

(B) 20 seconds after 4:58\textbf{(B) }20 \text{ seconds after } 4:58

(C) 25 seconds after 4:58\textbf{(C) }25 \text{ seconds after } 4:58

(D) 30 seconds after 4:58\textbf{(D) }30 \text{ seconds after } 4:58

(E) 35 seconds after 4:58\textbf{(E) }35 \text{ seconds after } 4:58

Pick an answer.

(A)
$15 \text{ seconds after } 4:58$
(B)
$20 \text{ seconds after } 4:58$
(C)
$25 \text{ seconds after } 4:58$
(D)
$30 \text{ seconds after } 4:58$
(E)
$35 \text{ seconds after } 4:58$
View mode:

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

#9 Solve an Easier Related Problem 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).
$$n=1 \to 0\text{ s}, \; n=2 \to 5\text{ s}, \; n=3 \to 10\text{ s}, \; n=4 \to 15\text{ s}$$

💡 Three small cases reveal the rule "elapsed seconds $= 5 \cdot (\text{hiccup number} - 1)$" — Grade 4 "generate a pattern from a rule".

#5 Look for a Pattern 4.OA.C.5 Step 2
  • Apply the pattern to $n = 700$.
  • Number of gaps between the first and the $700$th hiccup is $700 - 1 = 699$.
$$\text{gaps} = 700 - 1 = 699$$

💡 The fencepost rule: with $n$ posts there are $n - 1$ fence sections — same idea applies to hiccups and the gaps between them.

#9 Solve an Easier Related Problem 4.NBT.B.5 Step 3

Multiply gaps by seconds per gap to get total elapsed time.

$$699 \times 5 = 3495 \text{ seconds}$$

💡 $699 \times 5 = (700 - 1) \times 5 = 3500 - 5 = 3495$ — Grade 4 "multiply multi-digit by one-digit".

#8 Analyze the Units 4.MD.A.1 Step 4

Convert $3495$ seconds to minutes and remainder seconds using $60$ s = $1$ min.

$$3495 \div 60 = 58 \text{ R } 15 \;\Rightarrow\; 58\text{ min } 15\text{ s}$$

💡 Convert smaller units to larger units — Grade 4 "relative sizes of measurement units". $60 \times 58 = 3480$, then $3495 - 3480 = 15$.

#8 Analyze the Units 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).
$$4{:}00{:}00 + 58\text{ min } 15\text{ s} = 4{:}58{:}15 \;\Rightarrow\; \textbf{(A)}$$

💡 Add minutes and seconds to a clock time — Grade 4 "distances, time, money word problems".

[1] #9 4.OA.C.5 Try small cases to anchor the rule. Hiccup $1$ is at $0$ s after $4{:}00$ — no g
[2] #5 4.OA.C.5 Apply the pattern to $n = 700$. Number of gaps between the first and the $700$th
[3] #9 4.NBT.B.5 Multiply gaps by seconds per gap to get total elapsed time.
[4] #8 4.MD.A.1 Convert $3495$ seconds to minutes and remainder seconds using $60$ s = $1$ min.
[5] #8 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.5 Generate 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.5 Multiply 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.1 Know 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.2 Solve 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 $4{:}58{:}15$.

More like this

Same archetype — closest grade level first.