AMC10 2004 A

AMC10 2004 A · Q6

AMC10 2004 A · Q6. It mainly tests Arithmetic misc, Basic counting (rules of product/sum).

Bertha has 6 daughters and no sons. Some of her daughters have 6 daughters, and the rest have none. Bertha has a total of 30 daughters and granddaughters, and no great-granddaughters. How many of Bertha’s daughters and granddaughters have no daughters?

Bertha 有 6 个女儿，没有儿子。她的一些女儿有 6 个女儿，其余的没有。Bertha 总共有 30 个女儿和孙女，没有曾孙女。Bertha 的女儿和孙女中，有多少人没有女儿？

(A) 22 22

(B) 23 23

(D) 25 25

(E) 26 26

Answer

Correct choice: (E)

正确答案：（E）

Solution

Bertha has $30-6=24$ granddaughters, none of whom have any daughters. The granddaughters are the children of $24/6=4$ of Bertha’s daughters, so the number of women having no daughters is $30-4=26$.

Bertha 有 $30-6=24$ 个孙女，这些孙女都没有女儿。孙女们是 Bertha 6 个女儿中 $24/6=4$ 个的子女，因此没有女儿的人数是 $30-4=26$。

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