2004 AMC 12A Problems/Problem 4

Revision as of 19:05, 3 December 2007 by Azjps (talk | contribs) (duplicate)
The following problem is from both the 2004 AMC 12A #4 and 2004 AMC 10A #6, so both problems redirect to this page.

Problem

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 daughters?

$\mathrm{(A) \ } 22 \qquad \mathrm{(B) \ } 23 \qquad \mathrm{(C) \ } 24 \qquad \mathrm{(D) \ } 25 \qquad \mathrm{(E) \ } 26$

Solution

Since Bertha has 6 daughters, Bertha has $30-6=24$ granddaughters, of which none have daughters. Of Bertha's daughters, $\frac{24}6=4$ have daughters, so $6-4=2$ do not have daughters.

Therefore, of Bertha's daughters and granddaughters, $24+2=26$ do not have daughters $\Rightarrow\mathrm{(E)}$.

See also

2004 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions
2004 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions