Difference between revisions of "1995 AJHSME Problems/Problem 17"
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==Solution== | ==Solution== | ||
| − | By the tables, Annville has 11 6th graders and Cleona has 34. Together they have 45 6th graders and 300 total students, so the percent is <math>\frac{45}{300}=\frac{15}{100}= | + | By the tables, Annville has <math>11</math> 6th graders and Cleona has <math>34</math>. Together they have <math>45</math> 6th graders and <math>300</math> total students, so the percent is <math>\frac{45}{300}=\frac{15}{100}= \boxed{\text{(D)}\ 15\%}</math> |
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| + | ==See Also== | ||
| + | {{AJHSME box|year=1995|num-b=16|num-a=18}} | ||
Latest revision as of 02:15, 23 December 2012
Problem
The table below gives the percent of students in each grade at Annville and Cleona elementary schools:
![\[\begin{tabular}{rccccccc}&\textbf{\underline{K}}&\textbf{\underline{1}}&\textbf{\underline{2}}&\textbf{\underline{3}}&\textbf{\underline{4}}&\textbf{\underline{5}}&\textbf{\underline{6}}\\ \textbf{Annville:}& 16\% & 15\% & 15\% & 14\% & 13\% & 16\% & 11\%\\ \textbf{Cleona:}& 12\% & 15\% & 14\% & 13\% & 15\% & 14\% & 17\%\end{tabular}\]](http://latex.artofproblemsolving.com/0/4/d/04d9fa2706aa8263094d3283d69b3b746abadb9c.png)
Annville has 100 students and Cleona has 200 students. In the two schools combined, what percent of the students are in grade 6?
Solution
By the tables, Annville has 
6th graders and Cleona has 
. Together they have 
6th graders and 
total students, so the percent is 
See Also
| 1995 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 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 | ||
