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

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== Solution 2 ==
 
== Solution 2 ==
  
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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.
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Let <math>x</math> be the number of the children and the mom. The father, who is <math>48</math>, plus the sum of the ages of the kids and mom divided by the number of kids and mom plus <math>1</math> (for the dad) = <math>20</math>. This is because the average age of the entire family is <math>20.</math>
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Basically, this looks like 48+16x/x+1=20
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This statement, written as an equation, is: <cmath>\frac{48+16x}{x+1}=20</cmath>
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48+16x=20x+20
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<cmath>48+16x=20x+20</cmath>
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4x=28
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<cmath>4x=28</cmath>
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x=7
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<cmath>x=7</cmath>
  
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7 children + 1 mom = 6 children.
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<math>7</math> people - <math>1</math> mom = <math>6</math> children.
  
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E is the answer
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Therefore, the answer is <math>\boxed{E}</math>
  
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"Captain Haddock"
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==Solution 3==
 +
 
 +
Let <math>m</math> be the Mom's age.
 +
 
 +
Let the number of children be <math>x</math> and their average age be <math>y</math>. Their age totaled up is simply <math>xy</math>.
 +
 
 +
We have the following two equations:
 +
 
 +
<math>\frac{m+48+xy}{2+x}=20</math>, where <math>m+48+xy</math> is the family's total age and <math>2+x</math> is the total number of people in the family.
 +
 
 +
<math>\frac{m+48+xy}{2+x}=20</math> 
 +
 
 +
<math>m+48+xy=40+20x</math>
 +
 
 +
The next equation is <math>\frac{m+xy}{1+x}=16</math>, where <math>m+xy</math> is the total age of the Mom and the children, and <math>1+x</math> is the number of children along with the Mom.
 +
 
 +
<math>\frac{m+xy}{1+x}=16</math>
 +
 
 +
<math>m+xy=16+16x</math>.
 +
 
 +
We know the value for <math>m+xy</math>, so we substitute the value back in the first equation.
 +
 
 +
<math>m+48+xy=40+20x</math>
 +
 
 +
<math>(16+16x)+48=40+20x</math>.
 +
 
 +
<math>x=6</math>.
 +
 
 +
Earlier, we set <math>x</math> to be the number of children. Therefore, there are <math>\boxed{\text{(E)}  6}</math> children.
  
 
== See also ==
 
== See also ==

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.

See also

2007 AMC 10A (Problems • Answer Key • Resources)
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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