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

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==See Also==
 
==See Also==
 
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Revision as of 01:38, 5 July 2013

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?

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

Solution

The total number of points from the first $8$ games is $7+4+3+6+8+3+1+5=37$. To make this a multiple of $9$ by scoring less than $10$ points, Theresa must score $8$ points to have a total of $45$ points. To make a multiple of $10$, she must score $5$ points. The product of these two numbers of points is $8 \cdot 5 = \boxed{\textbf{(B)}\ 40}$.

See Also

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

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