AMC 8 · 1999 · #6
Grade 1 logicProblem
Bo, Coe, Flo, Jo, and Moe have different amounts of money. Neither Jo nor Bo has as much money as Flo. Both Bo and Coe have more than Moe. Jo has more than Moe, but less than Bo. Who has the least amount of money?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: Five people — Bo, Coe, Flo, Jo, Moe — have different amounts of money. We are given a handful of "more than / less than" clues that compare some of them in pairs. The question asks who holds the smallest amount.
Givens: Five people, all with different amounts: Bo, Coe, Flo, Jo, Moe; Clue 1: Neither Jo nor Bo has as much as Flo, so $J < F$ and $B < F$; Clue 2: Both Bo and Coe have more than Moe, so $B > M$ and $C > M$; Clue 3: Jo has more than Moe but less than Bo, so $M < J < B$; Answer choices: (A) Bo, (B) Coe, (C) Flo, (D) Jo, (E) Moe
Unknowns: Which of the five people has the least amount of money
Understand
Restated: Five people — Bo, Coe, Flo, Jo, Moe — have different amounts of money. We are given a handful of "more than / less than" clues that compare some of them in pairs. The question asks who holds the smallest amount.
Givens: Five people, all with different amounts: Bo, Coe, Flo, Jo, Moe; Clue 1: Neither Jo nor Bo has as much as Flo, so $J < F$ and $B < F$; Clue 2: Both Bo and Coe have more than Moe, so $B > M$ and $C > M$; Clue 3: Jo has more than Moe but less than Bo, so $M < J < B$; Answer choices: (A) Bo, (B) Coe, (C) Flo, (D) Jo, (E) Moe
Plan
Primary tool: #3 Eliminate Possibilities
Secondary: #1 Draw a Diagram
There are only five candidates for "least", so the fastest path is Tool #3 (Eliminate Possibilities): find one clue that beats each candidate and cross them off. The moment four names are crossed off, the fifth must be the answer. Tool #1 (Draw a Diagram) supports the count by stacking the clues into a single chain $M < J < B < F$ (plus a side note $M < C$), so the ordering is visible at a glance and nothing gets missed.
Execute — Answer: E
1.NBT.B.3 Step 1 - Translate each clue into a $<$ or $>$ statement.
- Use the first letter of each name: $B, C, F, J, M$.
- Clue 1 gives $J < F$ and $B < F$.
- Clue 2 gives $B > M$ and $C > M$.
- Clue 3 gives $M < J$ and $J < B$.
💡 Grade 1 introduces the symbols $<$ and $>$ for comparing amounts. Every clue in the problem is exactly one of those comparisons rewritten with a symbol.
K.MD.A.2 Step 2 - Eliminate Bo.
- From $B > M$, Bo has more money than Moe, so Bo is not the smallest.
💡 Kindergarten compare-two: if Bo has more than Moe, then Bo cannot be the one with the smallest pile.
K.MD.A.2 Step 3 - Eliminate Coe.
- From $C > M$, Coe also has more than Moe, so Coe is not the smallest either.
💡 Same compare-two reasoning: a single clue showing Coe beats Moe is enough to rule Coe out.
K.MD.A.2 Step 4 - Eliminate Jo.
- From $M < J$, Jo has more than Moe, so Jo is not the smallest.
💡 One direct comparison again: Jo $>$ Moe means Jo cannot hold the smallest amount.
1.MD.A.1 Step 5 - Eliminate Flo using transitivity.
- We already know $J > M$, and Clue 1 says $F > J$.
- Chain them: $F > J > M$, so $F > M$.
- Flo has more than Moe, so Flo is not the smallest.
💡 Grade 1 "order three objects" is exactly this move: if Flo $>$ Jo and Jo $>$ Moe, then Flo $>$ Moe without ever comparing them directly.
1.MD.A.1 Step 6 - Bo, Coe, Jo, Flo have all been knocked out.
- By elimination, Moe must be the one with the least money.
- We can also see this visually by stacking the chain: $M < J < B < F$, with $M < C$ noted on the side — Moe sits at the bottom of every chain.
💡 Lining the five names up on a number line makes Moe's bottom spot obvious — the same transitive ordering Grade 1 uses for "shortest to tallest".
1.NBT.B.3 Translate each clue into a $<$ or $>$ statement. Use the first letter of each na K.MD.A.2 Eliminate Bo. From $B > M$, Bo has more money than Moe, so Bo is not the smalles K.MD.A.2 Eliminate Coe. From $C > M$, Coe also has more than Moe, so Coe is not the small K.MD.A.2 Eliminate Jo. From $M < J$, Jo has more than Moe, so Jo is not the smallest. 1.MD.A.1 Eliminate Flo using transitivity. We already know $J > M$, and Clue 1 says $F > 1.MD.A.1 Bo, Coe, Jo, Flo have all been knocked out. By elimination, Moe must be the one Review
Reasonableness: Plug in any concrete amounts that fit the chain $M < J < B < F$ and $M < C$ — for example $M = 1, J = 2, B = 3, F = 4, C = 5$ (all different, all clues satisfied). The smallest value is $1$, held by Moe. Try another assignment such as $M = 10, J = 20, B = 30, F = 40, C = 100$: still smallest is Moe. Every clue is satisfied and Moe always ends up at the bottom, so (E) is robust.
Alternative: Tool #1 (Draw a Diagram) on its own: draw a vertical "money ladder" with bigger amounts higher up. Place Moe at the bottom, then walk through the clues — $J > M$ puts Jo above Moe; $J < B$ puts Bo above Jo; $B < F$ puts Flo above Bo; $C > M$ puts Coe somewhere above Moe (exact spot doesn't matter). Every name sits above Moe on the ladder, so Moe is at the bottom: answer (E).
CCSS standards used (min grade 1)
K.MD.A.2Directly compare two objects with a measurable attribute in common (Reading each single clue (e.g., $B > M$, $C > M$, $J > M$) as a direct two-person comparison to cross that person off the "least" list.)1.NBT.B.3Compare two two-digit numbers using symbols (Rewriting every English clue with the symbols $<$ and $>$ so the comparisons can be lined up and chained.)1.MD.A.1Order three objects by length and compare lengths indirectly (Chaining $F > J$ with $J > M$ to deduce $F > M$ without a direct Flo-vs-Moe clue, and then ordering all five names into $M < J < B < F$.)
⭐ If a clue says someone has more money than Moe, that person cannot be the least — and Bo, Coe, Jo, and Flo all get knocked out by exactly that rule, leaving Moe. The only Grade 1 idea you need is "order objects from least to greatest, using the chain rule."
⭐ If a clue says someone has more money than Moe, that person cannot be the least — and Bo, Coe, Jo, and Flo all get knocked out by exactly that rule, leaving Moe. The only Grade 1 idea you need is "order objects from least to greatest, using the chain rule."