2019 AMC 8

Complete problem set with solutions and individual problem pages

Problem 23 Hard

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?

  • A.

    10

  • B.

    11

  • C.

    12

  • D.

    13

  • E.

    14

Answer:B

Solution 1

Given the information above, we start with the equation \frac{t}{4}+\frac{2t}{7} + 15 + x = t, where t is the total number of points scored and x\leqslant 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 \geqslant 0 and divide by 28 to obtain x + 15 = 13u. Then, it is easy to see u=2 (t=56) is the only candidate remaining, giving x=\boxed{\textbf{(B)} 11}.

 

Solution 2

We first start by setting the total number of points as 28, since \text{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\cdot\frac{1}{4}+28\cdot\frac{2}{7} + 15 = 30 (> 28) ). Next, we can see that the total number of points scored is 56 as, if it is more than or equal to 84, at least one of the others will score more than 2 points. With this, we have that Alexa, Brittany, and Chelsea score: 56\cdot\frac{1}{4}+56\cdot\frac{2}{7} + 15 = 45, and thus, the other seven players would have scored a total of 56-45 = \boxed{\textbf{(B)} 11}.

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