2019 AMC 8 Problems/Problem 23

Problem 23

After Euclid High School's last basketball game, it was determined that $\frac{1}{4}$ of the team's points were scored by Alexa and $\frac{2}{7}$ were scored by Brittany. Chelsea scored $15$ points. None of the other $7$ team members scored more than $2$ points What was the total number of points scored by the other $7$ team members?

$\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14$

Solution 1

Since $\frac{\text{total points}}{4}$ and $\frac{2(\text{total points})}{7}$ are integers, we have $28 | \text{total points}$ . We see that the number of points scored by the other team members is less than or equal to $14$ and greater than or equal to $0$ . We let the total number of points be $t$ and the total number of points scored by the other team members, which means that $\frac{t}{4} + \frac{2t}{7} + 15 + x = t \quad \implies \quad 0 \le \frac{13t}{28} - 15 = x \le 14$ , which means $15 \le \frac{13t}{28} \le 29$ . The only value of $t$ that satisfies all conditions listed is $56$ , so $x=\boxed{\textbf{(B)} 11}$ .

Solution 2

Starting from the above equation $\frac{t}{4}+\frac{2t}{7} + 15 + x = t$ where $t$ is the total number of points scored and $x\le 14$ is the number of points scored by the remaining 7 team members, we can simplify to obtain the Diophantine equation $x+15 = \frac{13}{28}t$ , or $28x+28\cdot 15=13t$ . Since $t$ is necessarily divisible by 28, let $t=28u$ where $u \ge 0$ and divide by 28 to obtain $x + 15 = 13u$ . Then it is easy to see $u=2$ ( $t=56$ ) is the only candidate, giving $x=\boxed{\textbf{(B)} 11}$ . -scrabbler94

Solution 3

Fakesolve: We first start by setting the total number of points as $28$ , since $lcm(4,7) = 28$ . However, we see that this does not work since we surpass the number of points just with the information given ( $28*\frac{1}{4}+28*\frac{2}{7} + 15 > 28$ ). Next, we assume that the total number of points scored is $56$ . With this, we have that Alexa, Brittany, and Chelsea score: $56*\frac{1}{4}+56*\frac{2}{7} + 15 = 45$ , and thus, the other seven players would have scored a total of $56-45 = \boxed{\textbf{(B)} 11}$ (We see that this works since we could have $4$ of them score $2$ points, and the other $3$ of them score $1$ point) -aops5234

2019 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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2019 AMC 8 Problems/Problem 23

Contents

Problem 23

Solution 1

Solution 2

Solution 3

See Also