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>. | ||
| + | |||
| + | ==Video Solution by WhyMath== | ||
| + | https://youtu.be/A7uh_ZTDegE | ||
| + | |||
==See Also== | ==See Also== | ||
| − | {{AMC8 box|year=2002|num-b= | + | {{AMC8 box|year=2002|num-b=6|num-a=8}} |
| + | {{MAA Notice}} | ||
Latest revision as of 13:28, 29 October 2024
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?
![[asy] real[] r={6, 8, 4, 2, 5}; int i; for(i=0; i<5; i=i+1) { filldraw((4i,0)--(4i+3,0)--(4i+3,2r[i])--(4i,2r[i])--cycle, black, black); } draw(origin--(19,0)--(19,16)--(0,16)--cycle, linewidth(0.9)); for(i=1; i<8; i=i+1) { draw((0,2i)--(19,2i)); } label("$0$", (0,2*0), W); label("$1$", (0,2*1), W); label("$2$", (0,2*2), W); label("$3$", (0,2*3), W); label("$4$", (0,2*4), W); label("$5$", (0,2*5), W); label("$6$", (0,2*6), W); label("$7$", (0,2*7), W); label("$8$", (0,2*8), W); label("$A$", (0*4+1.5, 0), S); label("$B$", (1*4+1.5, 0), S); label("$C$", (2*4+1.5, 0), S); label("$D$", (3*4+1.5, 0), S); label("$E$", (4*4+1.5, 0), S); label("SWEET TOOTH", (9.5,18), N); label("Kinds of candy", (9.5,-2), S); label(rotate(90)*"Number of students", (-2,8), W);[/asy]](http://latex.artofproblemsolving.com/b/8/c/b8c9bfef9339a2c3cb17bb5eacdfa7d91c09cf4e.png)
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 
.
Video Solution by WhyMath
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
