Difference between revisions of "2014 AMC 8 Problems/Problem 7"
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==Solution 2== | ==Solution 2== | ||
| − | To make the amount of boys and girls equal, 28 - 4 = 24. 24/2 = 12. The girls would need to be 12 + the 4 that we subtracted = 16. The boys would be 12. The ratio of girls to boys would be 16 : 12, but simplified would be 4 : 3. Thus, the answer is | + | To make the amount of boys and girls equal, 28 - 4 = 24. 24/2 = 12. The girls would need to be 12 + the 4 that we subtracted = 16. The boys would be 12. The ratio of girls to boys would be 16 : 12, but simplified would be 4 : 3. Thus, the answer is 4 : 3 |
—-MiracleMaths | —-MiracleMaths | ||
Revision as of 19:23, 20 December 2021
Contents
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 1
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 
.
Solution 2
To make the amount of boys and girls equal, 28 - 4 = 24. 24/2 = 12. The girls would need to be 12 + the 4 that we subtracted = 16. The boys would be 12. The ratio of girls to boys would be 16 : 12, but simplified would be 4 : 3. Thus, the answer is 4 : 3
—-MiracleMaths
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. 
