AMC8 2026

AMC8 2026 · Q10

AMC8 2026 · Q10. It mainly tests Logic puzzles.

Five runners completed the grueling Xmarathon: Luke, Melina, Nico, Olympia, and Pedro. Nico finished $11$ minutes behind Pedro. Olympia finished $2$ minutes ahead of Melina, but $3$ minutes behind Pedro. Olympia finished $6$ minutes ahead of Luke. Which runner finished fourth?

五名跑者完成了艰苦的X马拉松比赛：Luke、Melina、Nico、Olympia 和 Pedro。 Nico 比 Pedro 晚了 $11$ 分钟到达。 Olympia 比 Melina 早 $2$ 分钟到达，但比 Pedro 晚 $3$ 分钟到达。 Olympia 比 Luke 早 $6$ 分钟到达。哪位跑者排名第四？

(A) \text{Luke} \text{Luke}

(B) \text{Melina} \text{Melina}

(D) \text{Olympia} \text{Olympia}

(E) \text{Pedro} \text{Pedro}

Answer

Correct choice: (A)

正确答案：（A）

Solution

Let the respective places be $L$, $M$, $N$, $O$, and $P$ respectively. Then, $P = 11+N$, $O-2 = M$, $O+3 = P$, and $O = 6+L$. Then, substitution for $P$ gives us $O+3 = N+11$, in which $O=N+8$. Then, $N+6 = M$, $N+8=6+L \Rightarrow N+2=L$. So $N$ finished m minutes behind everyone (sad), and thus was last. Now, $O=6+L$, in which $L+9 = P$, $L+7 = M$, and thus $\boxed{\text{Luke}}$ finished fourth.

设他们分别的时间为 $L$、$M$、$N$、$O$ 和 $P$。则有 $P = 11 + N$，$O - 2 = M$，$O + 3 = P$，以及 $O = 6 + L$。将 $P$ 代入得 $O + 3 = N + 11$，即 $O = N + 8$。又有 $N + 6 = M$，$N + 8 = 6 + L \Rightarrow N + 2 = L$。因此，$N$ 比其他人晚到（悲伤），也就是说他是最后一名。由 $O = 6 + L$，得到 $L + 9 = P$，$L + 7 = M$，因此排第四的是 $\boxed{\text{Luke}}$。

Topics

LM.001 Logic puzzles