Difference between revisions of "2008 AMC 8 Problems/Problem 15"
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In Theresa's first <math>8</math> basketball games, she scored <math>7, 4, 3, 6, 8, 3, 1</math> and <math>5</math> points. In her ninth game, she scored fewer than <math>10</math> points and her points-per-game average for the nine games was an integer. Similarly in her tenth game, she scored fewer than <math>10</math> points and her points-per-game average for the <math>10</math> games was also an integer. What is the product of the number of points she scored in the ninth and tenth games? | In Theresa's first <math>8</math> basketball games, she scored <math>7, 4, 3, 6, 8, 3, 1</math> and <math>5</math> points. In her ninth game, she scored fewer than <math>10</math> points and her points-per-game average for the nine games was an integer. Similarly in her tenth game, she scored fewer than <math>10</math> points and her points-per-game average for the <math>10</math> games was also an integer. What is the product of the number of points she scored in the ninth and tenth games? | ||
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\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>. 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>, multiplying these together you get <math>8\cdot5</math> is <math>40</math>. The answer is <math>\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}} | ||
+ | {{MAA Notice}} |
Latest revision as of 19:24, 8 August 2021
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
, multiplying these together you get
is
. The answer 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.