Difference between revisions of "2004 AMC 12A Problems/Problem 4"

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Therefore, of Bertha's daughters and granddaughters, <math>24+2=26</math> do not have daughters <math>\Rightarrow\mathrm{(E)}</math>.
 
Therefore, of Bertha's daughters and granddaughters, <math>24+2=26</math> do not have daughters <math>\Rightarrow\mathrm{(E)}</math>.
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OR
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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.
  
 
== See also ==
 
== See also ==

Revision as of 20:24, 17 January 2008

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 $\Rightarrow\mathrm{(E)}$.

OR

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.

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
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