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Difference between revisions of "2008 AMC 8 Problems/Problem 15"

m (Undo revision 159798 by Raina0708 (talk) I messaged Raina0708 for the use of LaTeX. Deletion should not have been made.)
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==Solution==
 
==Solution==
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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>. 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{(B)\ 40}</math>
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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 smallest multiple of <math>9</math> that is greater than <math>37</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 of <math>10</math> is <math>50</math>, which means that we add <math>5</math>. Multiplying these together, you get <math>8\cdot5 = </math> <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}}
 
{{MAA Notice}}

Latest revision as of 17:20, 27 October 2024

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 smallest multiple of $9$ that is greater than $37$ is $45$. $45-37=8$. Now we have to add a number to get a multiple of 10. The next multiple of $10$ is $50$, which means that we add $5$. Multiplying these together, you get $8\cdot5 =$ $\boxed{\textbf{(B)}\ 40}$.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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