Difference between revisions of "2014 CEMC Gauss (Grade 8) Problems/Problem 10"
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<math> \textbf{(A)}\ 2 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 6 \qquad\textbf{(E)}\ 8 </math> | <math> \textbf{(A)}\ 2 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 6 \qquad\textbf{(E)}\ 8 </math> | ||
==Solution 1== | ==Solution 1== | ||
| − | Let there be <math>3x</math> girls and <math>5x</math> boys in the class, as the ratio of girls to boys is <math>3:5</math>. The total is then <math>3x + 5x = 8x</math>. | + | Let there be <math>3x</math> girls and <math>5x</math> boys in the class, as the ratio of girls to boys is <math>3:5</math>. The total number of students in the class is then <math>3x + 5x = 8x</math>. |
We know that the class has <math>24</math> students in total. Thus, we have: | We know that the class has <math>24</math> students in total. Thus, we have: | ||
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~anabel.disher | ~anabel.disher | ||
| + | {{CEMC box|year=2014|competition=Gauss (Grade 8)|num-b=9|num-a=11}} | ||
Latest revision as of 11:32, 18 October 2025
Problem
The ratio of the number of girls to the number of boys in a class of 
students is 
. How many fewer girls than boys are in the class?
Solution 1
Let there be 
girls and 
boys in the class, as the ratio of girls to boys is . The total number of students in the class is then 
.
We know that the class has students in total. Thus, we have:
There is then 
fewer girls than boys, or 
fewer girls.
~anabel.disher
| 2014 CEMC Gauss (Grade 8) (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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 | ||
| CEMC Gauss (Grade 8) | ||
