Difference between revisions of "2007 AMC 10A Problems/Problem 10"
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<math>\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6</math> | <math>\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6</math> | ||
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Let <math>n</math> be the number of children. Then the total ages of the family is <math>48 + 16(n+1)</math>, and the total number of people in the family is <math>n+2</math>. So | Let <math>n</math> be the number of children. Then the total ages of the family is <math>48 + 16(n+1)</math>, and the total number of people in the family is <math>n+2</math>. So | ||
Revision as of 11:46, 16 February 2016
Contents
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
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 children + 1 mom = 6 children.
E is the answer
"Captain Haddock"
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.