Difference between revisions of "2008 AMC 8 Problems/Problem 15"

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==Problem 15==
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==Problem==
 
In Theresa's first <math>8</math> basketball games, she scored <math>7, 4, 3, 6, 8, 3, 1</math> and <math>5</math> points. In her ninth game, she scored fewer than <math>10</math> points and her points-per-game average for the nine games was an integer. Similarly in her tenth game, she scored fewer than <math>10</math> points and her points-per-game average for the <math>10</math> games was also an integer. What is the product of the number of points she scored in the ninth and tenth games?
 
In Theresa's first <math>8</math> basketball games, she scored <math>7, 4, 3, 6, 8, 3, 1</math> and <math>5</math> points. In her ninth game, she scored fewer than <math>10</math> points and her points-per-game average for the nine games was an integer. Similarly in her tenth game, she scored fewer than <math>10</math> points and her points-per-game average for the <math>10</math> games was also an integer. What is the product of the number of points she scored in the ninth and tenth games?
  

Revision as of 13:05, 9 December 2012

Problem

In Theresa's first $8$ basketball games, she scored $7, 4, 3, 6, 8, 3, 1$ and $5$ points. In her ninth game, she scored fewer than $10$ points and her points-per-game average for the nine games was an integer. Similarly in her tenth game, she scored fewer than $10$ points and her points-per-game average for the $10$ games was also an integer. What is the product of the number of points she scored in the ninth and tenth games?

$\textbf{(A)}\ 35\qquad \textbf{(B)}\ 40\qquad \textbf{(C)}\ 48\qquad \textbf{(D)}\ 56\qquad \textbf{(E)}\ 72$

See Also

2008 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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All AJHSME/AMC 8 Problems and Solutions