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Difference between revisions of "2007 AMC 10A Problems/Problem 10"
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<cmath>20 = \frac{48 + 16(n+1)}{n+2} \Longrightarrow 20n + 40 = 16n + 64 \Longrightarrow n = 6\ \mathrm{(E)}.</cmath>
 
<cmath>20 = \frac{48 + 16(n+1)}{n+2} \Longrightarrow 20n + 40 = 16n + 64 \Longrightarrow n = 6\ \mathrm{(E)}.</cmath>
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== Solution ==
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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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Basically, this looks like 48+16x/x+1=20
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48+16x=20x+20
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4x=28
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x=7
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7 children + 1 mom = 6 children.
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E is the answer
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"Captain Haddock"
  
 
== See also ==
 
== See also ==

Revision as of 10:45, 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

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

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
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All AMC 10 Problems and Solutions

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