Difference between revisions of "2004 AMC 10A Problems/Problem 13"

 
(Solution)
(7 intermediate revisions by 6 users not shown)
Line 2: Line 2:
 
At a party, each man danced with exactly three women and each woman danced with exactly two men.  Twelve men attended the party.  How many women attended the party?
 
At a party, each man danced with exactly three women and each woman danced with exactly two men.  Twelve men attended the party.  How many women attended the party?
  
<math> \mathrm{(A) \ } 8 \qquad \mathrm{(B) \ } 12 \qquad \mathrm{(C) \ } 16 \qquad \mathrm{(D) \ } 18 {2}\qquad \mathrm{(E) \ } 24  </math>
+
<math> \mathrm{(A) \ } 8 \qquad \mathrm{(B) \ } 12 \qquad \mathrm{(C) \ } 16 \qquad \mathrm{(D) \ } 18 \qquad \mathrm{(E) \ } 24  </math>
  
 
==Solution==
 
==Solution==
If each man danced with 3 women, then there were a total of <math>3\times12=36</math> pairs of a man and a women.  However, each women only danced with 2 men, so there must have been <math>\frac{36}2=18</math> women <math>\Rightarrow\mathrm{(D)}</math>.
+
If each man danced with <math>3</math> women, then there will be a total of <math>3\times12=36</math> pairs of men and women.  However, each woman only danced with <math>2</math> men, so there must have been <math>\frac{36}2 \Longrightarrow \boxed{\mathrm{(D)}\ 18}</math> women.
  
==See Also==
+
== See also ==
 +
{{AMC10 box|year=2004|ab=A|num-b=12|num-a=14}}
  
*[[2004 AMC 10A Problems]]
+
[[Category:Introductory Algebra Problems]]
 
+
{{MAA Notice}}
*[[2004 AMC 10A Problems/Problem 12|Previous Problem]]
 
 
 
*[[2004 AMC 10A Problems/Problem 14|Next Problem]]
 

Revision as of 20:53, 5 January 2016

Problem

At a party, each man danced with exactly three women and each woman danced with exactly two men. Twelve men attended the party. How many women attended the party?

$\mathrm{(A) \ } 8 \qquad \mathrm{(B) \ } 12 \qquad \mathrm{(C) \ } 16 \qquad \mathrm{(D) \ } 18 \qquad \mathrm{(E) \ } 24$

Solution

If each man danced with $3$ women, then there will be a total of $3\times12=36$ pairs of men and women. However, each woman only danced with $2$ men, so there must have been $\frac{36}2 \Longrightarrow \boxed{\mathrm{(D)}\ 18}$ women.

See also

2004 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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