Difference between revisions of "2004 AMC 12A Problems/Problem 4"
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{{duplicate|[[2004 AMC 12A Problems|2004 AMC 12A #4]] and [[2004 AMC 10A Problems/Problem 6|2004 AMC 10A #6]]}} | {{duplicate|[[2004 AMC 12A Problems|2004 AMC 12A #4]] and [[2004 AMC 10A Problems/Problem 6|2004 AMC 10A #6]]}} | ||
− | ==Problem== | + | == 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? | 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? | ||
− | <math> \mathrm{(A) \ } 22 \qquad \mathrm{(B) \ } 23 \qquad \mathrm{(C) \ } 24 \qquad \mathrm{(D) \ } 25 \qquad \mathrm{(E) \ } 26 | + | <math>\mathrm{(A) \ } 22 \qquad \mathrm{(B) \ } 23 \qquad \mathrm{(C) \ } 24 \qquad \mathrm{(D) \ } 25 \qquad \mathrm{(E) \ } 26</math> |
− | ==Solution== | + | == Solutions == |
− | Since Bertha has 6 daughters, | + | === Solution 1 === |
+ | 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 (From Alcumus) === | |
+ | Bertha has <math>30 - 6 = 24</math> granddaughters, none of whom have any daughters. The granddaughters are the children of <math>24/6 = 4</math> of Bertha's daughters, so the number of women having no daughters is <math>30 - 4 = \boxed{26}</math>. | ||
− | + | === Solution 3 === | |
+ | 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>. | ||
− | + | == Video Solution == | |
+ | https://youtu.be/HfOIeW-TGhc | ||
+ | |||
+ | Education, the Study of Everything | ||
== See also == | == See also == | ||
Line 20: | Line 26: | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 13:56, 19 January 2021
- 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?
Solutions
Solution 1
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 (From Alcumus)
Bertha has granddaughters, none of whom have any daughters. The granddaughters are the children of of Bertha's daughters, so the number of women having no daughters is .
Solution 3
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 .
Video Solution
Education, the Study of Everything
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.