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 $7+4+3+6+8+3+1+5=37$ . 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 $8 \cdot 5 = \boxed{\textbf{(B)}\ 40}$ .

2008 AMC 8 (Problems • Answer Key • Resources)
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
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2008 AMC 8 Problems/Problem 15

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