AMC 8 · 2007 · #7

Grade 6 arithmetic

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

mean-median-mode-rangemulti-digit-arithmetic identify-subproblems ↑ Prerequisites: multi-digit-arithmeticmean-median-mode-range

📏 Short solution 💡 2 insights

Problem

The average age of $5$ people in a room is $30$ years. An $18$ -year-old person leaves
the room. What is the average age of the four remaining people?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2007 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.

View mode:

Toolkit + CCSS Solution

Understand

Restated: Five people in a room have an average age of $30$. One of them, who is $18$, walks out. What is the average age of the four people who stay?

Givens: There are $5$ people in the room; Their average age is $30$ years; An $18$-year-old leaves; Answer choices: (A) $25$, (B) $26$, (C) $29$, (D) $33$, (E) $36$

Unknowns: The average age of the $4$ remaining people

Understand

Restated: Five people in a room have an average age of $30$. One of them, who is $18$, walks out. What is the average age of the four people who stay?

Givens: There are $5$ people in the room; Their average age is $30$ years; An $18$-year-old leaves; Answer choices: (A) $25$, (B) $26$, (C) $29$, (D) $33$, (E) $36$

Plan

Primary tool: #7 Identify Subproblems

Tool #7 (Identify Subproblems) fits because averages do not subtract directly — you cannot just take $30 - 18$. Instead the question splits into three small subproblems, each one a single Grade 6 step: (a) recover the original sum of ages from the original average, (b) subtract the leaver's age to get the new sum, (c) divide the new sum by the new count. No variable or equation is needed — the mean formula run forward and backward is enough.

Execute — Answer: D

#7 Identify Subproblems 6.SP.B.5 Step 1

Recover the original total.
The mean formula says sum $=$ average $\times$ count.
With $5$ people averaging $30$, the total of their ages is $5 \times 30$.

$$\text{original sum} = 5 \times 30 = 150 \text{ years}$$

💡 Grade 6 mean formula in reverse: knowing the average and the count pins down the total.

#7 Identify Subproblems 4.OA.A.3 Step 2

Subtract the leaver.
The $18$-year-old walks out, so the total drops by exactly $18$.
Nobody else's age changes.

$$\text{new sum} = 150 - 18 = 132 \text{ years}$$

💡 Only the total changes here — the count change is handled in the next step.

#7 Identify Subproblems 6.SP.B.5 Step 3

Re-average.
Divide the new total by the new count of $4$ people.

$$\text{new average} = \dfrac{132}{4} = 33 \;\Rightarrow\; \textbf{(D)}$$

💡 Same mean formula, applied forward this time: sum $\div$ count.

[1] #7 6.SP.B.5 Recover the original total. The mean formula says sum $=$ average $\times$ count

[2] #7 4.OA.A.3 Subtract the leaver. The $18$-year-old walks out, so the total drops by exactly

[3] #7 6.SP.B.5 Re-average. Divide the new total by the new count of $4$ people.

Review

Reasonableness: Check the direction first. The person who left was $18$, well below the group average of $30$, so pulling them out should raise the average — and $33 > 30$, which matches. Verify by running it the other way: if the $4$ remaining people average $33$, their total is $4 \times 33 = 132$. Add back the $18$-year-old to get $132 + 18 = 150$, and $150 / 5 = 30$, the original average. The numbers close the loop.

Alternative: Tool #3 (Write an Equation) gives the same answer in one line. Let $N$ be the new average. The total of all $5$ ages can be written two ways: as $5 \times 30 = 150$ from the original average, and as (sum of the $4$ who stayed) $+ 18 = 4N + 18$ from the new average plus the leaver. Setting them equal, $4N + 18 = 150$, so $4N = 132$ and $N = 33$.

CCSS standards used (min grade 6)

6.SP.B.5 Summarize numerical data sets, including reporting the number of observations and measures of center (Using the mean formula both ways: average $\times$ count $=$ sum to recover the original total of $150$, and sum $\div$ count to compute the new average $132/4 = 33$.)
4.OA.A.3 Solve multistep word problems with whole numbers using the four operations (Subtracting the leaver's age from the original total, $150 - 18 = 132$, as the middle step of the multi-step plan.)

⭐ Averages do not subtract. To handle a "someone leaves" problem, go back to the total, take the leaver's age out, then divide by the new count.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 6)

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