Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 3)
m (→‎Solution 2)
Line 11: Line 11:
 
== Solution 2 ==
 
== 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.
+
Let <math>x</math> be the number of the children + the mom. The father, who is <math>48</math>, plus the number of 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>
−
Basically, this looks like 48+16x/x+1=20
+
Basically, this looks like: <cmath>\frac{48+16x}{x+1}=20</cmath>
−
<math>48+16x=20x+20
+
<cmath>48+16x=20x+20</cmath>
−
4x=28
+
<cmath>4x=28</cmath>
−
x=7</math>
+
<cmath>x=7</cmath>
  
−
7 people - 1 mom = 6 children.
+
<math>7</math> people - <math>1</math> mom = <math>6</math> children.
  
−
E is the answer
+
Therefore, the answer is <math>\boxed{E}</math>
  
 
==Solution 3==
 
==Solution 3==

Revision as of 18:13, 24 October 2020

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 + 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).

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

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

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 (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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_10A_Problems/Problem_10&oldid=135772"