AMC 10 · 2005 · #6
Grade 6 arithmeticThe average (mean) of 20 numbers is 30, and the average of 30 other numbers is 20. What is the average of all 50 numbers?
Pick an answer.
AMC 10 2005 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.
Toolkit + CCSS Solution
Understand
Restated: One group of twenty numbers has an average of thirty. A separate group of thirty numbers has an average of twenty. Put all fifty numbers together and find the average of the whole collection.
Givens: A first group of $20$ numbers has average (mean) $30$; A second group of $30$ numbers has average (mean) $20$; The two groups are combined into one collection of $20 + 30 = 50$ numbers; Answer choices: (A) $23$, (B) $24$, (C) $25$, (D) $26$, (E) $27$
Unknowns: The average (mean) of all $50$ numbers taken together
Understand
Restated: One group of twenty numbers has an average of thirty. A separate group of thirty numbers has an average of twenty. Put all fifty numbers together and find the average of the whole collection.
Givens: A first group of $20$ numbers has average (mean) $30$; A second group of $30$ numbers has average (mean) $20$; The two groups are combined into one collection of $20 + 30 = 50$ numbers; Answer choices: (A) $23$, (B) $24$, (C) $25$, (D) $26$, (E) $27$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #16 Change Focus / Count the Complement, #3 Eliminate Possibilities
Averages cannot be added together, but totals can, so Tool #16 (Change Focus) shifts attention from the two averages to the two hidden totals. Tool #7 (Identify Subproblems) then splits the work into three clean pieces: the total of the first group, the total of the second group, and the average of everything combined. Tool #3 (Eliminate Possibilities) guards the main trap, because simply averaging the two averages $30$ and $20$ gives $25$, which is choice (C) but is wrong since the two groups are not the same size.
Execute — Answer: B
6.SP.B.5 Step 1 Turn the first average into a total
- An average is the total sum shared equally among the numbers, so the total sum is the average multiplied by how many numbers there are.
- The first group has $20$ numbers averaging $30$, so its total sum is $20 \times 30 = 600$.
- The average by itself hides this total, so recovering it is the first move.
💡 An average is just the pile shared out evenly, so multiplying it back by the count rebuilds the whole pile.
6.SP.B.5 Step 2 Turn the second average into a total
- Do the same for the second group.
- It has $30$ numbers averaging $20$, so its total sum is $30 \times 20 = 600$.
- Notice both groups happen to total $600$, even though one has more numbers each worth less and the other has fewer numbers each worth more.
💡 The same rebuild-the-pile move works for the second group regardless of its different size.
6.SP.A.3 Step 3 Add the totals and count all the numbers
- Now the two totals can be added, since real sums combine directly.
- The grand total of all the numbers is $600 + 600 = 1200$, and the number of values is $20 + 30 = 50$.
- This grand total over the full count is exactly what a combined average needs.
💡 Totals and counts stack up cleanly when groups merge, even though averages do not.
6.SP.A.3 Step 4 Divide to get the combined average
- The average of all $50$ numbers is the grand total divided by the count: $1200 \div 50 = 24$.
- Watch the trap: averaging the two given averages gives $(30 + 20)\div 2 = 25$, which is choice (C).
- That shortcut fails because the group averaging $20$ has more numbers ($30$ of them), so it pulls the combined average below the midpoint, down to $24$.
- The answer is $24$, choice (B).
💡 The bigger group tugs the shared average toward its own value, so the mean leans toward the twenty-average side.
6.SP.B.5 An average is the total sum shared equally among the numbers, so the total sum i 6.SP.B.5 Do the same for the second group. It has $30$ numbers averaging $20$, so its tot 6.SP.A.3 Now the two totals can be added, since real sums combine directly. The grand tot 6.SP.A.3 The average of all $50$ numbers is the grand total divided by the count: $1200 \ Review
Reasonableness: The combined average must land somewhere between the two group averages $20$ and $30$, and $24$ does. Because the larger group (the $30$ numbers) averages the smaller value $20$, the result should sit below the plain midpoint $25$, and $24$ is indeed just under $25$. If the group sizes were swapped, the pull would go the other way and the average would rise above $25$ instead. Choice (C) $25$ is exactly the midpoint you would get by wrongly averaging the two averages, which is why it is the tempting wrong answer.
Alternative: Use a weighted average directly: weight each group average by its share of the $50$ numbers. That gives $\frac{20}{50}\times 30 + \frac{30}{50}\times 20 = \frac{2}{5}\times 30 + \frac{3}{5}\times 20 = 12 + 12 = 24$, confirming choice (B) without computing the grand total separately.
CCSS standards used (min grade 6)
6.SP.B.5Summarize numerical data sets, relating the number of observations to their measure of center (Reversing the mean-equals-sum-over-count relationship to rebuild each group's total sum ($20\times 30$ and $30\times 20$) from its average and size.)6.SP.A.3Recognize that a measure of center summarizes all values of a data set with a single number (Combining the two totals over the full count of $50$ to compute the single mean of the whole collection and seeing why averaging the averages fails.)
⭐ You cannot average two averages when the groups are different sizes; turn each average back into a total, add the totals, and divide by how many numbers there are in all.
⭐ You cannot average two averages when the groups are different sizes; turn each average back into a total, add the totals, and divide by how many numbers there are in all.
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