AMC 8 · 2022 · #9

Easy mode Grade 4

Problem

Picture a cup of boiling water sitting in a room. The water is very hot, at $212^{\circ}\text{F}$ . The room stays at $68^{\circ}\text{F}$ the whole time.

The water cools down in a special way. Every $5$ minutes, the gap between the water's temperature and the room's temperature shrinks to half of what it was.

What is the water's temperature after $15$ minutes have passed?

$\textbf{(A) } 77 \qquad \textbf{(B) } 86 \qquad \textbf{(C) } 92 \qquad \textbf{(D) } 98 \qquad \textbf{(E) } 104$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

104

View mode:

Toolkit + CCSS Solution

Understand

Restated: A cup of water starts at $212^{\circ}\text{F}$ in a room that stays at $68^{\circ}\text{F}$. The gap between the water and the room cuts in half every $5$ minutes. What is the water's temperature after $15$ minutes?

Givens: Starting water temperature $= 212^{\circ}\text{F}$; Constant room temperature $= 68^{\circ}\text{F}$; The water-to-room temperature gap is halved every $5$ minutes; Elapsed time $= 15$ minutes; Answer choices: (A) $77$, (B) $86$, (C) $92$, (D) $98$, (E) $104$

Unknowns: The water temperature (in $^{\circ}\text{F}$) after $15$ minutes

Understand

Plan

Primary tool: #5 Look for a Pattern

Secondary: #7 Identify Subproblems, #6 Guess and Check

The rule "halve the gap every $5$ minutes" is a perfect Tool #5 (Pattern) setup: list the gap at $0, 5, 10, 15$ minutes and a clean geometric pattern $144, 72, 36, 18$ appears. Tool #7 (Identify Subproblems) splits the problem into three clean pieces — (a) find the starting gap, (b) halve it three times, (c) add the room temperature back to recover the water temperature. We avoid Tool #13 (Algebra) and any "$D_n = D_0 \cdot (\tfrac{1}{2})^n$" formula on purpose: a bright elementary student can just halve $144$ three times. Tool #6 (Guess and Check) is held in reserve as a verification pass against the multiple choices.

Execute — Answer: B

#7 Identify Subproblems 4.NBT.B.4 Step 1

Find the starting gap between the water and the room.
This is the quantity that will get halved over time, so we want it on the table before anything else.

$$212 - 68 = 144 \;^{\circ}\text{F}$$

💡 Subtracting two multi-digit whole numbers ($212 - 68$) is a Grade 4 fluency skill.

#7 Identify Subproblems 3.OA.A.3 Step 2

Count how many times the gap is halved.
The gap halves every $5$ minutes, and $15$ minutes is three $5$-minute chunks, so the gap is halved exactly $3$ times.

$$\dfrac{15 \text{ min}}{5 \text{ min}} = 3 \text{ halvings}$$

💡 "How many groups of $5$ fit in $15$?" is a Grade 3 division word-problem move.

#5 Look for a Pattern 4.OA.C.5 Step 3

Apply the halving rule three times.
Make a tiny table — this is the pattern: at each row the gap is the previous gap divided by $2$.

$$144 \;\xrightarrow{\div 2}\; 72 \;\xrightarrow{\div 2}\; 36 \;\xrightarrow{\div 2}\; 18$$

💡 Generating a sequence from the explicit rule "divide by $2$" is exactly the Grade 4 "shape or number pattern from a given rule" standard.

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

Translate the final gap back into a water temperature.
After $15$ minutes the water is $18^{\circ}\text{F}$ warmer than the room, and the room is still at $68^{\circ}\text{F}$.

$$68 + 18 = 86 \;^{\circ}\text{F} \;\Rightarrow\; \textbf{(B)}$$

💡 Adding the room temperature back to the gap is the final subproblem and a Grade 4 multi-digit addition.

[1] #7 4.NBT.B.4 Find the starting gap between the water and the room. This is the quantity that

[2] #7 3.OA.A.3 Count how many times the gap is halved. The gap halves every $5$ minutes, and $1

[3] #5 4.OA.C.5 Apply the halving rule three times. Make a tiny table — this is the pattern: at

[4] #7 4.NBT.B.4 Translate the final gap back into a water temperature. After $15$ minutes the wa

Review

Reasonableness: Common-sense check: the water cools from $212^{\circ}\text{F}$ toward $68^{\circ}\text{F}$, so the answer must be strictly between those two values — and closer to $68$ than to $212$ after $15$ minutes of fast cooling. Our answer $86^{\circ}\text{F}$ sits inside the interval $(68, 212)$ and is much nearer the room temperature, which is exactly what "the gap halves three times" should produce. Choices (D) $98$ and (E) $104$ would only happen if the gap halved fewer times; (A) $77$ would need an extra halving. Only (B) $86$ matches three halvings.

Alternative: Tool #6 (Guess and Check) on the choices works cleanly. For each candidate water temperature $T$, the implied final gap is $T - 68$. After three halvings starting from $144$, the gap must be $18$, so we need $T - 68 = 18$, i.e. $T = 86$. The other choices give gaps of $9, 24, 30, 36$ — none of which is $144$ halved three times.

CCSS standards used (min grade 4)

4.NBT.B.4 Fluently add and subtract multi-digit whole numbers (Computing the initial gap $212 - 68 = 144$ and the final water temperature $68 + 18 = 86$.)
3.OA.A.3 Solve multiplication and division word problems within 100 (Dividing $15 \div 5 = 3$ to count how many five-minute halving intervals fit in the elapsed $15$ minutes.)
4.OA.C.5 Generate a number or shape pattern following a given rule (Generating the gap sequence $144, 72, 36, 18$ by applying the rule "divide by $2$" three times.)

⭐ This AMC 8 problem only needs Grade 4 pattern-making — halving a number a few times — that you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 4)

More like this