AMC 10 Daily Practice Round 4

Complete problem set with solutions and individual problem pages

Problem 17 Easy

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 grand-daughters have no children? (2004 AMC 10A Problems, Problem #6)

  • A.

    22

  • B.

    23

  • C.

    24

  • D.

    25

  • E.

    26

Answer:E

Solution 1: Since Bertha has 6 daughters, she has 30-6=24 granddaughters, of which none have daughters. Of Bertha's daughters, \dfrac{24}{6}=4 have daughters, so 6-4=2 do not have daughters. Therefore, of Bertha's daughters and granddaughters, 24+2=26 do not have daughters. \boxed{ (\text{E}) 26}.

Solution 2: Bertha has 30-6=24 granddaughters, none of whom have any daughters. The granddaughters are the children of \dfrac{24}{6}=4 of Bertha's daughters, so the number of women having no daughters is 30-4=\boxed {26}.

Draw a tree diagram and see that the answer can be found in the sum of 6+6 granddaughters, 5+5 daughters, and 4 more daughters. Adding them together gives the answer of \boxed{ (\text{E}) 26}.