Difference between revisions of "2007 AMC 10A Problems/Problem 10"

Latest revision as of 19:54, 19 August 2023

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 the number of the children and the mom. The father, who is $48$ , plus the sum of the ages of the 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 $20.$ This statement, written as an equation, is: $\frac{48+16x}{x+1}=20$ $48+16x=20x+20$ $4x=28$ $x=7$

$7$ 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 age 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 $2+x$ is the total number of people in the family.

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

$m+48+xy=40+20x$

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

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

$m+xy=16+16x$ .

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

$m+48+xy=40+20x$

$(16+16x)+48=40+20x$ .

$x=6$ .

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

@@ Line 7: / Line 7: @@
 Let <math>n</math> be the number of children. Then the total ages of the family is <math>48 + 16(n+1)</math>, and the total number of people in the family is <math>n+2</math>. So
-<math></math>20 = \frac{48 + 16(n+1)}{n+2} \Longrightarrow 20n + 40 = 16n + 64 \Longrightarrow n = 6 <math>\boxed{E}</math>
+<cmath>20 = \frac{48 + 16(n+1)}{n+2} \Longrightarrow 20n + 40 = 16n + 64 \Longrightarrow n = 6\ \mathrm{(E)}.</cmath>
 == Solution 2 ==

2007 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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