AMC 8 · 1999 · #13

Grade 6 arithmetic

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

mean-median-mode-rangemulti-digit-arithmeticlinear-equations-one-var identify-subproblems ↑ Prerequisites: multi-digit-arithmeticmean-median-mode-range

📏 Short solution 💡 2 insights

Problem

The average age of the 40 members of a computer science camp is 17 years. There are 20 girls, 15 boys, and 5 adults. If the average age of the girls is 15 and the average age of the boys is 16, what is the average age of the adults?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 1999 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.

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Toolkit + CCSS Solution

Understand

Restated: A computer science camp has $40$ members whose average age is $17$. The members split into $20$ girls (average age $15$), $15$ boys (average age $16$), and $5$ adults. Find the average age of the adults.

Givens: Total members: $40$, with overall average age $17$; $20$ girls with average age $15$; $15$ boys with average age $16$; $5$ adults with unknown average age; Answer choices: (A) $26$, (B) $27$, (C) $28$, (D) $29$, (E) $30$

Unknowns: The average age of the $5$ adults

Understand

Plan

Primary tool: #11 Find an Invariant

Secondary: #9 Solve an Easier Problem

The total sum of all $40$ ages is the invariant here (Tool #11): you can compute it two ways — directly from the overall average ($17 \times 40$), or by adding the three subgroup sums (girls $+$ boys $+$ adults). Both must give the same number, which pins down the adults' total. Tool #9 (Solve an Easier Problem) splits the work into three small "average $\times$ count" calculations instead of one big algebra setup.

Execute — Answer: C

#11 Find an Invariant 6.SP.B.5 Step 1

Find the total sum of all $40$ ages from the overall average.
Average $\times$ count gives the total.

$$\text{Total sum} = 17 \times 40 = 680$$

💡 Mean $=$ sum $\div$ count, so sum $=$ mean $\times$ count. This is the Grade 6 definition of average rearranged.

#9 Solve an Easier Problem 6.SP.B.5 Step 2

Find the sum of the girls' ages and the sum of the boys' ages separately.
Each subgroup also obeys sum $=$ average $\times$ count.

$$\text{Girls' sum} = 15 \times 20 = 300, \quad \text{Boys' sum} = 16 \times 15 = 240$$

💡 Breaking the big group into two easier pieces keeps the arithmetic small and avoids any algebra.

#11 Find an Invariant 6.EE.B.7 Step 3

Use the invariant.
The three subgroup sums (girls $+$ boys $+$ adults) must equal the total sum $680$.
Subtract to find the adults' total age.

$$\text{Adults' sum} = 680 - 300 - 240 = 140$$

💡 Same grand total, two different ways of counting — what's missing from the second way must be the adults' share.

#9 Solve an Easier Problem 6.SP.B.5 Step 4

Divide the adults' total age by the number of adults to get their average age.

$$\text{Adults' average} = \dfrac{140}{5} = 28 \;\Rightarrow\; \textbf{(C)}$$

💡 Back to the Grade 6 mean formula: average $=$ sum $\div$ count, with sum $= 140$ and count $= 5$.

[1] #11 6.SP.B.5 Find the total sum of all $40$ ages from the overall average. Average $\times$ c

[2] #9 6.SP.B.5 Find the sum of the girls' ages and the sum of the boys' ages separately. Each s

[3] #11 6.EE.B.7 Use the invariant. The three subgroup sums (girls $+$ boys $+$ adults) must equa

[4] #9 6.SP.B.5 Divide the adults' total age by the number of adults to get their average age.

Review

Reasonableness: Plug $28$ back in and rebuild the overall average. Total ages $= 20(15) + 15(16) + 5(28) = 300 + 240 + 140 = 680$. Divide by $40$ members: $680 \div 40 = 17$, matching the given overall average. Also, the adults' average ($28$) is well above the kids' averages ($15$ and $16$), which makes sense — only $5$ adults have to pull the camp average up from the mid-teens to $17$, so each adult must be much older than the kids.

Alternative: Tool #13 (Use Algebra): let $x$ be the adults' average. Write the weighted-average equation $\dfrac{20(15) + 15(16) + 5x}{40} = 17$. Solve: $300 + 240 + 5x = 680$, so $5x = 140$ and $x = 28$. Same answer, but heavier setup than the invariant-and-subtract path.

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 mean $=$ sum $\div$ count (and its rearrangement sum $=$ mean $\times$ count) for the whole camp and for each subgroup.)
6.EE.B.7 Solve real-world and mathematical problems by writing and solving equations of the form $x + p = q$ (Reading $300 + 240 + (\text{adults' sum}) = 680$ as a one-step equation and subtracting to isolate the adults' total.)

⭐ Total age stays the same no matter how you split the camp — turn each "average" into a sum, add the pieces, and the missing piece pops right out.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 6)

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