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

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Hence the smallest possible number of students is <math>\boxed{\mathrm{(D)}\ 13}</math>.
 
Hence the smallest possible number of students is <math>\boxed{\mathrm{(D)}\ 13}</math>.
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== Solution ==
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 +
Let's assume that each student scores 60 points on the test. This gives
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(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 <math>\boxed{\mathrm{(D)}\ 13}</math>.
  
 
== See Also ==
 
== See Also ==

Latest revision as of 22:54, 24 June 2024

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

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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