AMC 8 · 1999 · #6
Easy mode Grade 1Problem
Five friends — Bo, Coe, Flo, Jo, and Moe — each have a different amount of money.
Here is what we know:
- Jo has less money than Flo. Bo also has less money than Flo.
- Bo has more money than Moe. Coe also has more money than Moe.
- Jo has more money than Moe, but less money than Bo.
One of these five friends has the smallest amount. Which one?
Pick an answer.
AMC 8 1999 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
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."
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