AMC 8 · 2020 · #6

Grade 1 logic

Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In

logical-deductionsystematic-enumeration caseworkif-then-reasoningsystematic-enumeration ↑ Prerequisites: logical-deduction

📏 Medium solution 💡 3 insights

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Problem

Five friends — Aaron, Darren, Karen, Maren, Sharon — sit in a five-car train with one seat per car. Cars are numbered 1 (front) through 5 (back). Using the clues, figure out who sits in car 3 (the middle car).

Pick an answer.

(A)

$text{Aaron}$

(B)

$text{Darren}$

(C)

$text{Karen}$

(D)

$text{Maren}$

(E)

$text{Sharon}$

AMC 8 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

The most restrictive clue is the SA block (Sharon immediately followed by Aaron), and Maren is already pinned to car 5. So the only freedom for S and A is which pair of adjacent cars in 1–4 they occupy — that gives exactly three cases. Tool #2 (Systematic List) lays out those three placements in order so none is missed. Tool #3 (Eliminate Possibilities) then tests each case against the Darren and Karen clues and crosses out the ones that break a rule. A small Tool #1 picture (five boxes labeled 1–5) keeps the bookkeeping concrete.

1STEP 1

Draw five boxes for cars 1–5 (front to back) and place the one fixed fact: Maren in car 5.

[ _ , _ , _ , _ , M ]

2STEP 2

Sharon-then-Aaron is an adjacent pair SA that must fit in cars 1–4 (car 5 is Maren); list Sharon's start car in order: 1, 2, 3.

Case 1: [S,A,,,M] Case 2: [,S,A,,M] Case 3: [,,S,A,M]

3STEP 3

Case 1 [S,A,,,M]: Aaron in car 2 needs Darren in front, but the only earlier car is car 1 — Sharon's — so Case 1 is eliminated.

car(D) ＜ 2 → car(D) = 1, but car 1 = S. Contradiction.

4STEP 4

Case 2 [,S,A,,M]: Darren must take car 1, Karen car 4; |4-1|=3 ＞ 1, so [D, S, A, K, M] satisfies every clue.

[D, S, A, K, M] , |4-1| = 3 ＞ 1 ✓

5STEP 5

Case 3 [,,S,A,M]: Karen and Darren must fill cars 1 and 2, which are adjacent (|1-2|=1), so both sub-cases fail.

D=1, K=2: |2-1|=1 ✗ D=2, K=1: |1-2|=1 ✗

6STEP 6

Only Case 2 survives, so [D, S, A, K, M] is the unique seating and the middle car (car 3) holds Aaron.

Middle car = 3 → Aaron → (A)

Answer

text{Aaron}

Plug the arrangement [D, S, A, K, M] back into every clue. Maren is in car 5 ✓. Aaron (car 3) is directly behind Sharon (car 2) ✓. Darren (car 1) is in front of Aaron (car 3) ✓. Between Karen (car 4) and Darren (car 1) sit Sharon and Aaron — two people, more than one ✓. All four clues hold, and Cases 1 and 3 were each ruled out by a specific clue, so the answer is unique. Aaron in the middle is consistent.

💡Key takeaway

This AMC 8 problem only needs Grade 1 number comparison (1 ＜ 2 ＜ 3) and Kindergarten position words (in front of, behind, middle) — you already know all of these!

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