Difference between revisions of "1997 AHSME Problems/Problem 11"
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Thus, the tenth game must have at least <math>113 - 84 = 29</math> points, and the answer is <math>\boxed{C}</math>. | 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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| + | == See also == | ||
| + | {{AHSME box|year=1997|num-b=10|num-a=12}} | ||
Revision as of 08:31, 9 August 2011
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 | ||
