2018 AMC 8 Problems/Problem 8

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Problem 8

Mr. Garcia asked the members of his health class how many days last week they exercised for at least 30 minutes. Ayush only exercised for 2 minutes. The results are summarized in the following bar graph, where the heights of the bars represent the number of students.

[asy] size(8cm); void drawbar(real x, real h) {   fill((x-0.15,0.5)--(x+0.15,0.5)--(x+0.15,h)--(x-0.15,h)--cycle,gray); } draw((0.5,0.5)--(7.5,0.5)--(7.5,5)--(0.5,5)--cycle); for (real i=1; i<5; i=i+0.5) {   draw((0.5,i)--(7.5,i),gray); } drawbar(1.0,1.0); drawbar(2.0,2.0); drawbar(3.0,1.5); drawbar(4.0,3.5); drawbar(5.0,4.5); drawbar(6.0,2.0); drawbar(7.0,1.5); for (int i=1; i<8; ++i) {   label("$"+string(i)+"$",(i,0.25)); } for (int i=1; i<9; ++i) {   label("$"+string(i)+"$",(0.5,0.5*(i+1)),W); } label("Number of Days of Exercise",(4,-0.1)); label(rotate(90)*"Number of Students",(-0.1,2.75)); [/asy] What was the mean number of days of exercise last week, rounded to the nearest hundredth, reported by the students in Mr. Garcia's class?

$\textbf{(A) } 3.50 \qquad \textbf{(B) } 3.57 \qquad \textbf{(C) } 4.36 \qquad \textbf{(D) } 4.50 \qquad \textbf{(E) } 5.00$

Solution

The mean, or average number of days is the total number of days divided by the number of students. The total number of days is $1\cdot 1+2\cdot 3+3\cdot 2+4\cdot 6+5\cdot 8+6\cdot 3+7\cdot 2=109$. The total number of students is $1+3+2+6+8+3+2=25$. Hence, $\frac{109}{25}=\boxed{\textbf{(C) } 4.36}$ Ayush Agrawal is teh GOAT.

See Also

2018 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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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