Difference between revisions of "2007 AMC 10A Problems/Problem 10"
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− | 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 | + | 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>\frac{m+xy}{1+x}=16</math> |
Revision as of 12:49, 3 June 2021
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.