2004 AMC 12B Problems/Problem 11
Contents
Problem
All the students in an algebra class took a 
-point test. Five students scored , each student scored at least 
, and the mean score was 
. What is the smallest possible number of students in the class?
Solution
Let the number of students be 
. Then the sum of their scores is at least 
. At the same time, we need to achieve the mean , which is equivalent to achieving the sum 
.
Hence we get a necessary condition on 
: we must have 
.
This can be simplified to 
. The smallest integer for which this is true is 
.
To finish our solution, we now need to find one way how 
students could have scored on the test. We have 
points to divide among them. The five s make 
, hence we must divide the remaining 
points among the other 
students. This can be done e.g. by giving 
points to each of them.
Hence the smallest possible number of students is 
.
Solution
Let's assume that each student scores 60 points on the test. This gives (500 + 60(n))/(n+5) = 100
Solving the equation for n we see that n = 7.5.
Because of the extra five people we know that the total amount of people is 12.5, however, to adjust for the complexity of not every score being a 60 we round up to 13. Giving .
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 | |
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