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

Revision as of 10:51, 24 December 2019

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 $\frac{m+xy}{1+x)=16$ (Error compiling LaTeX. Unknown error_msg), 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.

@@ Line 31: / Line 31: @@
 <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> (Mom + Dad + Children).
-<math>\frac{m+48+xy}{2+x}=20 ==> </math>m+48+xy=40+40x<math>
+<math>\frac{m+48+xy}{2+x}=20 ==> m+48+xy=40+40x</math>
-The next equation is </math>\frac{m+xy}{1+x)=16<math>, where </math>m+xy<math> is the total ages of the Mom and the children, and </math>1+x<math> is the number of people.
+The next equation is <math>\frac{m+xy}{1+x)=16</math>, where <math>m+xy</math> is the total ages of the Mom and the children, and <math>1+x</math> is the number of people.
-</math>\frac{m+xy}{1+x}=16<math> ==> </math>m+xy=16+16x<math>.
+<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.
+We know the value for <math>m+xy</math>, so we substitute the value back in the first equation.
-</math>m+48+xy=40+40x<math> ==> </math>(16+16x)+48=40+40x<math>.
+<math>m+48+xy=40+40x</math> ==> <math>(16+16x)+48=40+40x</math>.
-</math>x=6<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}$ children.
+Earlier, we set <math>x</math> to be the number of children. Therefore, there are <math>\boxed{\text{(E)}  6}</math> 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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