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

m (moved video solution to bottom)
m (title fix)
 
Line 10: Line 10:
 
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>
 
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>
  
**Alcumus**
+
=== 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>.
 
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 2 ===
+
=== 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>.
 
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>.
  

Latest revision as of 12:56, 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.

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?

$\mathrm{(A) \ } 22 \qquad \mathrm{(B) \ } 23 \qquad \mathrm{(C) \ } 24 \qquad \mathrm{(D) \ } 25 \qquad \mathrm{(E) \ } 26$

Solutions

Solution 1

Since Bertha has $6$ daughters, she 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. $\boxed{\mathrm{(E)}\ 26}$

Solution 2 (From Alcumus)

Bertha has $30 - 6 = 24$ granddaughters, none of whom have any daughters. The granddaughters are the children of $24/6 = 4$ of Bertha's daughters, so the number of women having no daughters is $30 - 4 = \boxed{26}$.

Solution 3

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. Adding them together gives the answer of $\boxed{\mathrm{(E)}\ 26}$.

Video Solution

https://youtu.be/HfOIeW-TGhc

Education, the Study of Everything

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS