Difference between revisions of "1997 AHSME Problems/Problem 11"

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Games <math>1-5</math> must have an average of less than <math>17</math>.  Thus we cannot put more than <math>16 + 17 + 17 + 17 + 17 = 84</math> points in those five games.
 
Games <math>1-5</math> must have an average of less than <math>17</math>.  Thus we cannot put more than <math>16 + 17 + 17 + 17 + 17 = 84</math> points in those five games.
  
Thus, the tenth game must have at least <math>113 - 84 = 29</math> points, and the answer is <math>\boxed{C}</math>.
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Thus, the tenth game must have at least <math>113 - 84 = 29</math> points, and the answer is <math>\boxed{D}</math>.
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== See also ==
 +
{{AHSME box|year=1997|num-b=10|num-a=12}}
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{{MAA Notice}}

Latest revision as of 13:12, 5 July 2013

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 $\boxed{D}$.

See also

1997 AHSME (ProblemsAnswer KeyResources)
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
All AHSME Problems and Solutions

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