Difference between revisions of "2002 AMC 8 Problems/Problem 7"
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From the bar graph, we can see that <math>5</math> students chose candy E. There are <math>6+8+4+2+5=25</math> total students in Mrs. Sawyers class. The percent that chose E is <math>\frac{5}{25} \cdot 100 = \boxed{\text{(E)}\ 20}</math>. | From the bar graph, we can see that <math>5</math> students chose candy E. There are <math>6+8+4+2+5=25</math> total students in Mrs. Sawyers class. The percent that chose E is <math>\frac{5}{25} \cdot 100 = \boxed{\text{(E)}\ 20}</math>. | ||
==See Also== | ==See Also== | ||
− | {{AMC8 box|year=2002|num-b= | + | {{AMC8 box|year=2002|num-b=6|num-a=8}} |
Revision as of 19:12, 23 December 2012
Problem
The students in Mrs. Sawyer's class were asked to do a taste test of five kinds of candy. Each student chose one kind of candy. A bar graph of their preferences is shown. What percent of her class chose candy E?
Solution
From the bar graph, we can see that students chose candy E. There are
total students in Mrs. Sawyers class. The percent that chose E is
.
See Also
2002 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 |