2007 AMC 10A Problems/Problem 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?

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

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 the number of the 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: \[\frac{48+16x}{x+1}=20\] \[48+16x=20x+20\] \[4x=28\] \[x=7\]

$7$ people - $1$ mom = $6$ children.

Therefore, the answer is $\boxed{E}$

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).



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.



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




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
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All AMC 10 Problems and Solutions

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