Printable

2005 AMC 10A Problem 6

2005 AMC 10A Problem 6AMC 10 Step by Step

Name: Date: October 28, 2026

Statistics & Data · difficulty 1 of 5 · about 1 min · position P1-10

The average (mean) of $20$ numbers is $30$ , and the average of $30$ other numbers is $20$ . What is the average of all $50$ numbers?

A) $23$
B) $24$
C) $25$
D) $26$
E) $27$

Answer

Key insight

Turn each average into a total: 20 times 30 is 600 and 30 times 20 is 600. All 50 numbers add to 1200, so the average is 24.

Solution

You cannot just average the two averages, because the groups have different sizes. Instead, go back to totals.

1. First group. $20$ numbers with average $30$ add up to $20 \times 30 = 600$ .
2. Second group. $30$ numbers with average $20$ add up to $30 \times 20 = 600$ .
3. All together. The $50$ numbers add up to $600 + 600 = 1200$ .
4. New average. Divide the total by how many numbers there are:
$\frac{1200}{50} = 24.$

The answer is $\boxed{\textbf{(B)}\ 24}$ .

Why this works

An average is a total divided by a count, so to combine groups, add the totals and add the counts. The bigger group pulls the combined average toward its own average. Here the bigger group has average $20$ , so the answer lands below the halfway value $25$ .

The trap

Averaging 30 and 20 to get 25. That ignores that the group with average 20 has more numbers in it.

Concise solution

Averages combine through sums, not directly. The first group of $20$ numbers has total $20 \cdot 30 = 600$ ; the second group of $30$ numbers has total $30 \cdot 20 = 600$ .

All $50$ numbers therefore add to $1200$ , and their mean is
$\frac{1200}{50} = 24.$

The answer is $\boxed{\textbf{(B)}\ 24}$ .

Review in the AoPS books

Prealgebra, Chapter 13: Data and Statistics
Introduction to Algebra, Chapter 6: Ratios and Percents
Introduction to Algebra, Chapter 7: Proportion (§7.1, §7.2, §7.3)

Techniques

Set up the equation/formula and compute; no special trick needed