Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2004 AMC 12B Problems/Problem 11"

(→‎See Also)
m
Line 24: Line 24:
 
To finish our solution, we now need to find one way how <math>13</math> students could have scored on the test. We have <math>13\cdot 76 = 988</math> points to divide among them. The five <math>100</math>s make <math>500</math>, hence we must divide the remaining <math>488</math> points among the other <math>8</math> students. This can be done e.g. by giving <math>61</math> points to each of them.  
 
To finish our solution, we now need to find one way how <math>13</math> students could have scored on the test. We have <math>13\cdot 76 = 988</math> points to divide among them. The five <math>100</math>s make <math>500</math>, hence we must divide the remaining <math>488</math> points among the other <math>8</math> students. This can be done e.g. by giving <math>61</math> points to each of them.  
  
−
Hence the smallest possible number of students is <math>\boxed{13}</math>.
+
Hence the smallest possible number of students is <math>\boxed{\mathrm{(D)}\ 13}</math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 21:11, 29 December 2014

Problem

All the students in an algebra class took a $100$-point test. Five students scored $100$, each student scored at least $60$, and the mean score was $76$. What is the smallest possible number of students in the class?

$\mathrm{(A)}\ 10 \qquad \mathrm{(B)}\ 11 \qquad \mathrm{(C)}\ 12 \qquad \mathrm{(D)}\ 13 \qquad \mathrm{(E)}\ 14$

Solution

Let the number of students be $n\geq 5$. Then the sum of their scores is at least $5\cdot 100 + (n-5)\cdot 60$. At the same time, we need to achieve the mean $76$, which is equivalent to achieving the sum $76n$.

Hence we get a necessary condition on $n$: we must have $5\cdot 100 + (n-5)\cdot 60 \leq 76n$. This can be simplified to $200 \leq 16n$. The smallest integer $n$ for which this is true is $n=13$.

To finish our solution, we now need to find one way how $13$ students could have scored on the test. We have $13\cdot 76 = 988$ points to divide among them. The five $100$s make $500$, hence we must divide the remaining $488$ points among the other $8$ students. This can be done e.g. by giving $61$ points to each of them.

Hence the smallest possible number of students is $\boxed{\mathrm{(D)}\ 13}$.

See Also

2004 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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 AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_12B_Problems/Problem_11&oldid=66709"