Difference between revisions of "2002 AMC 8 Problems/Problem 7"

(Problem 7)
 
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== Problem ==
  
== Problem 7 ==
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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?
  
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?
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<asy>
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real[] r={6, 8, 4, 2, 5};
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int i;
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for(i=0; i<5; i=i+1) {
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filldraw((4i,0)--(4i+3,0)--(4i+3,2r[i])--(4i,2r[i])--cycle, black, black);
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}
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draw(origin--(19,0)--(19,16)--(0,16)--cycle, linewidth(0.9));
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for(i=1; i<8; i=i+1) {
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draw((0,2i)--(19,2i));
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}
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label("$0$", (0,2*0), W);
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label("$1$", (0,2*1), W);
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label("$2$", (0,2*2), W);
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label("$3$", (0,2*3), W);
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label("$4$", (0,2*4), W);
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label("$5$", (0,2*5), W);
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label("$6$", (0,2*6), W);
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label("$7$", (0,2*7), W);
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label("$8$", (0,2*8), W);
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label("$A$", (0*4+1.5, 0), S);
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label("$B$", (1*4+1.5, 0), S);
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label("$C$", (2*4+1.5, 0), S);
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label("$D$", (3*4+1.5, 0), S);
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label("$E$", (4*4+1.5, 0), S);
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label("SWEET TOOTH", (9.5,18), N);
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label("Kinds of candy", (9.5,-2), S);
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label(rotate(90)*"Number of students", (-2,8), W);</asy>
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<math>\text{(A)}\ 5 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 20</math>
  
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==Solution==
  
[[Solution]]
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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>.
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==See Also==
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{{AMC8 box|year=2002|num-b=6|num-a=8}}
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{{MAA Notice}}

Latest revision as of 00:41, 5 July 2013

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]

$\text{(A)}\ 5 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 20$

Solution

From the bar graph, we can see that $5$ students chose candy E. There are $6+8+4+2+5=25$ total students in Mrs. Sawyers class. The percent that chose E is $\frac{5}{25} \cdot 100 = \boxed{\text{(E)}\ 20}$.

See Also

2002 AMC 8 (ProblemsAnswer KeyResources)
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

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