Difference between revisions of "2014 AMC 8 Problems/Problem 7"
Mathonator (talk | contribs) |
Mathonator (talk | contribs) |
||
| Line 3: | Line 3: | ||
<math>\textbf{(A) }3 : 4\qquad\textbf{(B) }4 : 3\qquad\textbf{(C) }3 : 2\qquad\textbf{(D) }7 : 4\qquad \textbf{(E) }2 : 1</math> | <math>\textbf{(A) }3 : 4\qquad\textbf{(B) }4 : 3\qquad\textbf{(C) }3 : 2\qquad\textbf{(D) }7 : 4\qquad \textbf{(E) }2 : 1</math> | ||
| − | + | ==Solution== | |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2014|num-b=6|num-a=8}} | {{AMC8 box|year=2014|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 19:48, 26 November 2014
Problem
There are four more girls than boys in Ms. Raub's class of 
students. What is the ratio of number of girls to the number of boys in her class?
Solution
See Also
| 2014 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
