Difference between revisions of "2007 AMC 10A Problems/Problem 10"
(→Solution 3) |
(→Solution 3) |
||
Line 33: | Line 33: | ||
<math>\frac{m+48+xy}{2+x}=20 ==> 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 | + | 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>. |
Revision as of 11:52, 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 , 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 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
7 people - 1 mom = 6 children.
E is the answer
Solution 3
Let be the Mom's age.
Let the number of children be and their average be
. Their age totaled up is simply
.
We have the following two equations:
, where
is the family's total age and
(Mom + Dad + Children).
The next equation is , where
is the total ages of the Mom and the children, and
is the number of people.
==>
.
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.