2004 AMC 12A Problems/Problem 4

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, she 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. $\boxed{\mathrm{(E)}\ 26}$

Solution 2

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 $\boxed{\mathrm{(E)}\ 26}$ .

2004 AMC 12A (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2004 AMC 12A Problems/Problem 4

Contents

Problem

Solution

Solution 2

See also