2008 AMC 8 Problems/Problem 15

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 $7+4+3+6+8+3+1+5=37$ . We have to make this a multiple of $9$ by scoring less than $10$ points. The closest multiple of $9$ is $45$ . $45-37=8$ Now we have to add a number to get a multiple of 10. The next multiple is $50$ We added $5$ . $8*5$ is $40$ The answer is B, 40

2008 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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All AJHSME/AMC 8 Problems and Solutions

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2008 AMC 8 Problems/Problem 15

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