AMC 10 · 2018 · #5

Grade 6 rate-ratio

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

interval-arithmeticlogical-deduction if-then-reasoningcomplementary-counting ↑ Prerequisites: interval-arithmetic

📏 Medium solution 💡 2 insights

Problem

Three hikers each make a claim about the distance d (in miles) to the nearest town: "at least 6," "at most 5," and "at most 4." Every one of the three claims is false. Determine which interval contains all possible values of d.

Pick an answer.

(A)

(0,4)

(B)

(4,5)

(C)

(4,6)

(D)

(5,6)

(E)

$(5,\infty)$

AMC 10 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Change Focus / Count the Complement

We are told what d is NOT (each claim is false), not what it is. Tool #16 (Change Focus / Count the Complement) says: flip each false claim into its true opposite. "At least 6" is false means "less than 6" is true; "at most 5" is false means "more than 5" is true, and likewise for 4. Tool #1 (Draw a Diagram) puts the three true ranges on one number line so their overlap is visible. Tool #3 (Eliminate Possibilities) then matches that overlap to the lettered intervals, dropping any choice that includes a banned value.

1STEP 1

Write each claim as an inequality

Turn each spoken claim into an inequality about d: "at least" means ≥ and "at most" means ≤.

Alice: d ≥ 6, Bob: d ≤ 5, Charlie: d ≤ 4

2STEP 2

Flip each false claim to its opposite

Each claim is false, so flip its direction: ≥ turns into ＜ and ≤ turns into ＞.

not(d ≥ 6) → d ＜ 6, not(d ≤ 5) → d ＞ 5, not(d ≤ 4) → d ＞ 4

3STEP 3

Overlap the three true ranges

All three flipped ranges must hold at once, and they overlap only in 5 ＜ d ＜ 6.

d ＜ 6 and d ＞ 5 and d ＞ 4 ⟹ 5 ＜ d ＜ 6

4STEP 4

Match the overlap to a choice

That strip 5 ＜ d ＜ 6 is the open interval (5,6), which matches choice (D).

{ d : 5 ＜ d ＜ 6 } = (5,6) → (D)

Answer

(5,6)

Test a value inside (5,6), say d = 5.5: Alice ( ≥ 6) false, Bob ( ≤ 5) false, Charlie ( ≤ 4) false — all three false, as required. Test the rejected endpoints: d = 5 makes Bob true and d = 6 makes Alice true, so both correctly fall outside. A value like d = 4.5 would make Bob's and Charlie's claims false but is below 5, so it fails d ＞ 5 — which is why choices (B), (C), and (A) that reach below 5 are wrong.

💡Key takeaway

When a claim is false, just flip it: "at least 6" becomes "less than 6," and stacking the three flipped claims on a number line traps d in the gap (5,6).

Write each claim as an inequality
Flip each false claim to its opposite
Overlap the three true ranges
Match the overlap to a choice

Write each claim as an inequality

Flip each false claim to its opposite

Overlap the three true ranges

Match the overlap to a choice

More like this