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

Revision as of 10:47, 16 February 2016

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 children + 1 mom = 6 children.

E is the answer

"Captain Haddock"

2007 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 9: / Line 9: @@
 <cmath>20 = \frac{48 + 16(n+1)}{n+2} \Longrightarrow 20n + 40 = 16n + 64 \Longrightarrow n = 6\ \mathrm{(E)}.</cmath>
-== Solution ==
+== 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.

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

Revision as of 10:47, 16 February 2016

Contents

Problem

Solution 1

Solution 2

See also