Difference between revisions of "2007 AMC 10A Problems/Problem 10"
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== Solution 2 == | == Solution 2 == | ||
− | Let x be number of children | + | 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> |
− | + | This statement, written as an equation, is: <cmath>\frac{48+16x}{x+1}=20</cmath> | |
− | < | + | <cmath>48+16x=20x+20</cmath> |
− | 4x=28 | + | <cmath>4x=28</cmath> |
− | x=7</ | + | <cmath>x=7</cmath> |
− | 7 people - 1 mom = 6 children. | + | <math>7</math> people - <math>1</math> mom = <math>6</math> children. |
− | + | Therefore, the answer is <math>\boxed{E}</math> | |
==Solution 3== | ==Solution 3== | ||
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Let <math>m</math> be the Mom's age. | Let <math>m</math> be the Mom's age. | ||
− | Let the number of children be <math>x</math> and their average be <math>y</math>. Their age totaled up is simply <math>xy</math>. | + | 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: | 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> | + | <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>\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. |
− | Earlier, we set < | + | <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 20:54, 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 , the father is
years old, and the average age of the mother and children is
. How many children are in the family?
Solution 1
Let be the number of children. Then the total ages of the family is
, and the total number of people in the family is
. So
Solution 2
Let be the number of the children and the mom. The father, who is
, plus the sum of the ages of the kids and mom divided by the number of kids and mom plus
(for the dad) =
. This is because the average age of the entire family is
This statement, written as an equation, is:
people -
mom =
children.
Therefore, the answer is
Solution 3
Let be the Mom's age.
Let the number of children be and their average age be
. Their age totaled up is simply
.
We have the following two equations:
, where
is the family's total age and
is the total number of people in the family.
The next equation is , where
is the total age of the Mom and the children, and
is the number of children along with the Mom.
.
We know the value for , so we substitute the value back in the first equation.
.
.
Earlier, we set to be the number of children. Therefore, there are
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.