Difference between revisions of "2008 AMC 8 Problems/Problem 15"
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− | The total number of points from the first <math>8</math> games is <math>7+4+3+6+8+3+1+5=37</math>. We have to make this a multiple of <math>9</math> by scoring less than <math>10</math> points. The closest multiple of <math>9</math> is <math>45</math>. <math>45-37=8</math> | + | The total number of points from the first <math>8</math> games is <math>7+4+3+6+8+3+1+5=37</math>. We have to make this a multiple of <math>9</math> by scoring less than <math>10</math> points. The closest multiple of <math>9</math> is <math>45</math>. <math>45-37=8</math> now we have to add a number to get a multiple of 10. The next multiple is <math>50</math> we added <math>5</math> <math>8*5</math> is <math>40</math> The answer is B, 40 |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2008|num-b=14|num-a=16}} | {{AMC8 box|year=2008|num-b=14|num-a=16}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 19:51, 21 October 2018
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 . We have to make this a multiple of by scoring less than points. The closest multiple of is . now we have to add a number to get a multiple of 10. The next multiple is we added is The answer is B, 40
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 |
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