Difference between revisions of "2018 AMC 8 Problems/Problem 8"
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==Solution== | ==Solution== | ||
| − | The mean number of days is the total number of days divided by the number of students. The total number of days is <math>1\cdot 1+2\cdot 3+3\cdot 2+4\cdot 6+5\cdot 8+6\cdot 3+7\cdot 2=109</math>. The total number of students is <math>1+3+2+6+8+3+2=25</math>. Hence, <math>\frac{109}{25}=\boxed{\textbf{(C) } 4.36}</math> | + | 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 <math>1\cdot 1+2\cdot 3+3\cdot 2+4\cdot 6+5\cdot 8+6\cdot 3+7\cdot 2=109</math>. The total number of students is <math>1+3+2+6+8+3+2=25</math>. Hence, <math>\frac{109}{25}=\boxed{\textbf{(C) } 4.36}</math> |
==See Also== | ==See Also== | ||
Revision as of 16:53, 30 September 2019
Problem 8
Mr. Garcia asked the members of his health class how many days last week they exercised for at least 30 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]](http://latex.artofproblemsolving.com/f/7/a/f7add6258fd09de75421fb0d08e39071b65b128c.png)
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?
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 
. The total number of students is 
. Hence, 
See Also
| 2018 AMC 8 (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
