2007 AMC 10A Problems/Problem 10

Problem

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?

$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$

Solution 1

Let $n$ be the number of children. Then the total ages of the family is $48 + 16(n+1)$, and the total number of people in the family is $n+2$. So

\[20 = \frac{48 + 16(n+1)}{n+2} \Longrightarrow 20n + 40 = 16n + 64 \Longrightarrow n = 6\ \mathrm{(E)}.\]

Solution 2

Let x be number of children+the mom. The father, who is 48, plus the number of kids and mom divided by the number of kids and mom plus 1 (for the dad)=20. This is because the average age of the entire family is 20. Basically, this looks like 48+16x/x+1=20 $48+16x=20x+20 4x=28 x=7$

7 people - 1 mom = 6 children.

E is the answer

Solution 3

Let $m$ be the Mom's age.

Let the number of children be $x$ and their average be $y$. Their age totaled up is simply $xy$.

We have the following two equations:

$\frac{m+48+xy}{2+x}=20$, where $m+48+xy$ is the family's total age and $2+x$ (Mom + Dad + Children).

$\frac{m+48+xy}{2+x}=20$

$m+48+xy=40+40x$

The next equation is $\frac{m+xy}{1+x}=16$, where $m+xy$ is the total ages of the Mom and the children, and $1+x$ is the number of people.

$\frac{m+xy}{1+x}=16$

$m+xy=16+16x$.

We know the value for $m+xy$, so we substitute the value back in the first equation.

$m+48+xy=40+40x$

$(16+16x)+48=40+40x$.

$x=6$.

Earlier, we set $x$ to be the number of children. Therefore, there are $\boxed{\text{(E)}  6}$ children.

See also

2007 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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