1997 AHSME Problems/Problem 11

Problem

In the sixth, seventh, eighth, and ninth basketball games of the season, a player scored $23$ , $14$ , $11$ , and $20$ points, respectively. Her points-per-game average was higher after nine games than it was after the first five games. If her average after ten games was greater than $18$ , what is the least number of points she could have scored in the tenth game?

$\textbf{(A)}\ 26\qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 28\qquad\textbf{(D)}\ 29\qquad\textbf{(E)}\ 30$

Solution

The sum of the scores for games $6$ through $9$ is $68$ . The average in these four games is $\frac{68}{4} = 17$ .

The total points in all ten games is greater than $10\cdot 18 = 180$ . Thus, it must be at least $181$ .

There are at least $181 - 68 = 113$ points in the other six games: games $1-5$ and game $10$ .

Games $1-5$ must have an average of less than $17$ . Thus we cannot put more than $16 + 17 + 17 + 17 + 17 = 84$ points in those five games.

Thus, the tenth game must have at least $113 - 84 = 29$ points, and the answer is $\boxed{D}$ .

1997 AHSME (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 • 26 • 27 • 28 • 29 • 30
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1997 AHSME Problems/Problem 11

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