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

m (fixed typo in answer choice)
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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>.
  
==See Also==
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== See also ==
 
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{{AMC10 box|year=2004|ab=A|num-b=5|num-a=7}}
*[[2004 AMC 10A Problems]]
 
 
 
*[[2004 AMC 10A Problems/Problem 5|Previous Problem]]
 
 
 
*[[2004 AMC 10A Problems/Problem 7|Next Problem]]
 
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Revision as of 01:39, 11 September 2007

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 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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All AMC 10 Problems and Solutions
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