Difference between revisions of "2004 AMC 12A Problems/Problem 4"
(duplicate) |
(→Solution) |
||
| Line 10: | Line 10: | ||
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>. | ||
| + | |||
| + | 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 == | == 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?
Solution
Since Bertha has 6 daughters, Bertha 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 
.
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 (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 | ||
