Difference between revisions of "2014 AMC 8 Problems/Problem 7"
m (→Solution) |
|||
| Line 4: | Line 4: | ||
<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== | ==Solution== | ||
| − | We can set up an equation with <math>x</math> being the number of girls in the class. The number of boys in the class is equal to <math>x-4</math>. Since the total number of students is equal to <math>28</math>, we get <math>x+x-4=28</math>. Solving this equation, we get <math>x=16</math>. There are <math>16-4=12</math> boys in our class, and our answer is <math>16:12=\boxed{4:3}</math>. | + | We can set up an equation with <math>x</math> being the number of girls in the class. The number of boys in the class is equal to <math>x-4</math>. Since the total number of students is equal to <math>28</math>, we get <math>x+x-4=28</math>. Solving this equation, we get <math>x=16</math>. There are <math>16-4=12</math> boys in our class, and our answer is <math>16:12=\boxed{4:3}</math>. So the answer is <math>\boxed{B}</math> |
==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 15:41, 3 November 2015
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
We can set up an equation with 
being the number of girls in the class. The number of boys in the class is equal to 
. Since the total number of students is equal to , we get 
. Solving this equation, we get 
. There are 
boys in our class, and our answer is 
. So the answer is 
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. 
