Difference between revisions of "2019 AMC 8 Problems/Problem 23"
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<math>\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14</math> | <math>\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14</math> | ||
− | ==Solution | + | ==Solution== |
− | + | Since <math>\frac{\text{total points}}{4}</math> and <math>\frac{2(\text{total points})}{7}</math> are integers, we have <math>28 | \text{total points}</math>. We see that the number of points scored by the other team members is less than or equal to <math>14</math> and greater than or equal to <math>0</math>. We let the total number of points be <math>t</math> and the total number of points scored by the other team members, which means that <math>\frac{t}{4} + \frac{2t}{7} + 15 + x = t \quad \implies \quad 0 \le \frac{13t}{28} - 15 = x \le 14</math>, which means <math>15 \le \frac{13t}{28} \le 29</math>. The only value of <math>t</math> that satisfies all conditions listed is <math>56</math>, so <math>x = \boxed{11}</math>. | |
==See Also== | ==See Also== |
Revision as of 17:50, 20 November 2019
Problem 23
After Euclid High School's last basketball game, it was determined that of the team's points were scored by Alexa and
were scored by Brittany. Chelsea scored
points. None of the other
team members scored more than
points What was the total number of points scored by the other
team members?
Solution
Since and
are integers, we have
. We see that the number of points scored by the other team members is less than or equal to
and greater than or equal to
. We let the total number of points be
and the total number of points scored by the other team members, which means that
, which means
. The only value of
that satisfies all conditions listed is
, so
.
See Also
2019 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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.