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2007 AMC 10A Problem 10

2007 AMC 10A Problem 10AMC 10 Step by Step

Name: Date: September 13, 2026

Ratios, Percents & Averages · difficulty 2 of 5 · about 2 min · position P1-10

The Dunbar family consists of a mother, a father, and some children. The average age of the members of the family is $20$ , the father is $48$ years old, and the average age of the mother and children is $16$ . How many children are in the family?

A)
$2$
B)
$3$
C)
$4$
D)
$5$
E)
$6$

Answer

Key insight

Convert each average into a total: the whole family sums to 20(n+2) and the mother plus children sum to 16(n+1); the difference is the father's 48.

Solution

Let there be $n$ children, so the family has $n + 2$ members and the mother-plus-children group has $n + 1$ members.

Turn each average into a total of ages:
- whole family: $20(n + 2)$ ;
- mother and children: $16(n + 1)$ .

The two totals differ by exactly the father's age:
$20(n+2) - 16(n+1) = 48 \quad\Longrightarrow\quad 4n + 24 = 48 \quad\Longrightarrow\quad n = 6.$

The answer is $\boxed{\textbf{(E)}\ 6}$ .

Why this works

Averages are hard to combine directly, but totals add. Whenever a problem gives an average of a group and an average of a subgroup, multiply each by its size to get sums, and the missing member is the difference. This is the standard move for "average of the rest" problems.

Alternative approach

Excess method: relative to the family average $20$ , the father is $+28$ and each of the other $n + 1$ people averages $-4$ . Balancing, $4(n+1) = 28$ , so $n + 1 = 7$ and $n = 6$ .

The trap

Using n for the number of people in the mother-and-children group and then forgetting to subtract the mother at the end.

Common mistakes

Using $n$ for the number of people in the mother-and-children group and then forgetting to subtract the mother at the end.
Averaging the averages, e.g. assuming the father's age plus $16$ somehow averages to $20$ , instead of working with totals.

Techniques

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