2004 AMC 12B Problems/Problem 11

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 $\boxed{\mathrm{(D)}\ 13}$ .

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 $\boxed{\mathrm{(D)}\ 13}$ .

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

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2004 AMC 12B Problems/Problem 11

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