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2005 AMC 10A Problem 6

2005 AMC 10A Problem 6AMC 10 Step by Step

Name: Date: September 13, 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

Convert each average to a total: both groups sum to 600, so all 50 numbers sum to 1200 and average 24.

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 $\boxed{\textbf{(B)}\ 24}$ .

Why this works

The mean of a merged collection is the merged sum over the merged count, which weights each group's average by its size. Here the larger group has the smaller average, so the combined mean sits below the midpoint $25$ . Always pass through totals when combining averages.

The trap

Averaging the two averages to get 25, ignoring that the groups have different sizes.

Common mistakes

Averaging the two averages to get 25, ignoring that the groups have different sizes.
Dividing the total $1200$ by the wrong count (for example by $20$ or $30$ ) instead of by all $50$ numbers.

Techniques

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