Difference between revisions of "2008 AMC 8 Problems/Problem 15"
| Line 7: | Line 7: | ||
\textbf{(D)}\ 56\qquad | \textbf{(D)}\ 56\qquad | ||
\textbf{(E)}\ 72</math> | \textbf{(E)}\ 72</math> | ||
| + | |||
| + | ==Solution== | ||
| + | The total number of points from the first <math>8</math> games is <math>7+4+3+6+8+3+1+5=37</math>. To make this a multiple of <math>9</math> by scoring less than <math>10</math> points, Theresa must score <math>8</math> points to have a total of <math>45</math> points. To make a multiple of <math>10</math>, she must score <math>5</math> points. The product of these two numbers of points is <math>8 \cdot 5 = \boxed{\textbf{(B)}\ 40}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2008|num-b=14|num-a=16}} | {{AMC8 box|year=2008|num-b=14|num-a=16}} | ||
Revision as of 02:40, 25 December 2012
Problem
In Theresa's first 
basketball games, she scored 
and 
points. In her ninth game, she scored fewer than 
points and her points-per-game average for the nine games was an integer. Similarly in her tenth game, she scored fewer than points and her points-per-game average for the games was also an integer. What is the product of the number of points she scored in the ninth and tenth games?
Solution
The total number of points from the first games is 
. To make this a multiple of 
by scoring less than points, Theresa must score points to have a total of 
points. To make a multiple of , she must score points. The product of these two numbers of points is 
.
See Also
| 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 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
