AMC10 2007 A

AMC10 2007 A · Q10

AMC10 2007 A · Q10. It mainly tests Linear equations, Averages (mean).

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?

Dunbar一家有母亲、父亲和一些孩子。全家成员的平均年龄是$20$岁，父亲$48$岁，母亲和孩子们的平均年龄是$16$岁。家里有多少孩子？

(A) 2 2

(B) 3 3

(D) 5 5

(E) 6 6

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): Let $N$ represent the number of children in the family and $T$ represent the sum of the ages of all the family members. The average age of the members of the family is 20, and the average age of the members when the 48-year-old father is not included is 16, so $20=\dfrac{T}{N+2}$ and $16=\dfrac{T-48}{N+1}$. This implies that $20N+40=T$ and $16N+16=T-48$, so $20N+40=16N+64$. Hence $4N=24$ and $N=6$.

答案（E）：设 $N$ 表示家庭中孩子的数量，$T$ 表示全体家庭成员年龄之和。家庭成员的平均年龄为 20；当不包含 48 岁的父亲时，成员的平均年龄为 16，所以 $20=\dfrac{T}{N+2}$ 且 $16=\dfrac{T-48}{N+1}$。这意味着 $20N+40=T$ 且 $16N+16=T-48$，因此 $20N+40=16N+64$。所以 $4N=24$，从而 $N=6$。

Topics