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

m (Solution)
(Solution)
 
Line 9: Line 9:
  
 
==Solution==
 
==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*5</math> is <math>40</math>. The answer is <math>\boxed{B, 40}</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>, multiplying these together you get <math>8\cdot5</math> is <math>40</math>. The answer is <math>\boxed{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 12:47, 21 February 2019

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 closest multiple of $9$ is $45$. $45-37=8$ now we have to add a number to get a multiple of 10. The next multiple is $50$ we added $5$, multiplying these together you get $8\cdot5$ is $40$. The answer is $\boxed{B, 40}$

See Also

2008 AMC 8 (ProblemsAnswer KeyResources)
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

Invalid username
Login to AoPS