Difference between revisions of "2004 AMC 12A Problems/Problem 4"
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==Solution== | ==Solution== | ||
| − | Since Bertha has 6 daughters, | + | Since Bertha has <math>6</math> daughters, she has <math>30-6=24</math> granddaughters, of which none have daughters. Of Bertha's daughters, <math>\frac{24}6=4</math> have daughters, so <math>6-4=2</math> do not have daughters. Therefore, of Bertha's daughters and granddaughters, <math>24+2=26</math> do not have daughters. <math>\boxed{\mathrm{(E)}\ 26}</math> |
| − | + | ==Solution 2== | |
| − | + | Draw a tree diagram and see that the answer can be found in the sum of <math>6+6</math> granddaughters, <math>5+5</math> daughters, and <math>4</math> more daughters. Adding them together gives the answer of <math>\boxed{\mathrm{(E)}\ 26}</math>. | |
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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 23:17, 20 July 2014
- The following problem is from both the 2004 AMC 12A #4 and 2004 AMC 10A #6, so both problems redirect to this page.
Contents
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?
Solution
Since Bertha has 
daughters, she has 
granddaughters, of which none have daughters. Of Bertha's daughters, 
have daughters, so 
do not have daughters. Therefore, of Bertha's daughters and granddaughters, 
do not have daughters. 
Solution 2
Draw a tree diagram and see that the answer can be found in the sum of 
granddaughters, 
daughters, and 
more daughters. Adding them together gives the answer of .
See also
| 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. 
