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{ | + | Thus, the tenth game must have at least <math>113 - 84 = 29</math> points, and the answer is <math>\boxed{D}</math>. |
== See also == | == See also == | ||
{{AHSME box|year=1997|num-b=10|num-a=12}} | {{AHSME box|year=1997|num-b=10|num-a=12}} | ||
+ | {{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 ,, , and 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 , what is the least number of points she could have scored in the tenth game?
Solution
The sum of the scores for games through is . The average in these four games is .
The total points in all ten games is greater than . Thus, it must be at least .
There are at least points in the other six games: games and game .
Games must have an average of less than . Thus we cannot put more than points in those five games.
Thus, the tenth game must have at least points, and the answer is .
See also
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 | ||
All AHSME Problems and Solutions |
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