AMC 10 · 2021 · #7
Grade 5 arithmeticProblem
Tom has a collection of 13 snakes, 4 of which are purple and 5 of which are happy. He observes that
all of his happy snakes can add,
none of his purple snakes can subtract, and
all of his snakes that can't subtract also can't add.
Which of these conclusions can be drawn about Tom's snakes?
Pick an answer.
AMC 10 2021 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: Tom owns $13$ snakes. Some are purple, some are happy, some can add, some can subtract. We are told three facts: every happy snake can add; no purple snake can subtract; and every snake that can't subtract also can't add. Which of the five answer-choice conclusions is forced by these three facts?
Givens: $13$ snakes total; $4$ purple, $5$ happy (the counts are decoys — not needed); Fact 1: happy $\Rightarrow$ can-add; Fact 2: purple $\Rightarrow$ cannot-subtract; Fact 3: cannot-subtract $\Rightarrow$ cannot-add; Choices: (A) purple $\Rightarrow$ can-add (B) purple $\Rightarrow$ happy (C) can-add $\Rightarrow$ purple (D) happy $\Rightarrow$ not purple (E) happy $\Rightarrow$ cannot-subtract
Unknowns: Which conclusion follows from the three facts
Understand
Restated: Tom owns $13$ snakes. Some are purple, some are happy, some can add, some can subtract. We are told three facts: every happy snake can add; no purple snake can subtract; and every snake that can't subtract also can't add. Which of the five answer-choice conclusions is forced by these three facts?
Givens: $13$ snakes total; $4$ purple, $5$ happy (the counts are decoys — not needed); Fact 1: happy $\Rightarrow$ can-add; Fact 2: purple $\Rightarrow$ cannot-subtract; Fact 3: cannot-subtract $\Rightarrow$ cannot-add; Choices: (A) purple $\Rightarrow$ can-add (B) purple $\Rightarrow$ happy (C) can-add $\Rightarrow$ purple (D) happy $\Rightarrow$ not purple (E) happy $\Rightarrow$ cannot-subtract
Plan
Primary tool: #3 Eliminate Possibilities
Secondary: #1 Draw a Diagram, #7 Identify Subproblems
Five answer choices, all multiple-choice — Tool #3 (Eliminate Possibilities) tests each in turn against the three given facts. Tool #1 (Draw a Diagram) sketches the facts as arrows between properties so the chain is visible. Tool #7 (Identify Subproblems) splits the work: first chain facts 2 and 3, then bring in the contrapositive of fact 1, then read off which arrow lands on the answer.
Execute — Answer: D
5.G.B.3 Step 1 - Draw four boxes — happy, purple, can-add, can-subtract — and turn each "if-then" fact into an arrow.
- Fact 1: happy → can-add.
- Fact 2: purple → not-subtract.
- Fact 3: not-subtract → not-add.
💡 An if-then statement is just an arrow from one property box to another.
5.G.B.3 Step 2 - Chain facts 2 and 3: purple → not-subtract → not-add.
- So every purple snake cannot add.
💡 Arrow A→B and arrow B→C give a combined arrow A→C — like a transitive chain.
5.G.B.3 Step 3 - Take fact 1 (happy → can-add) and read it backwards.
- The contrapositive: if a snake cannot add, it is not happy.
- So not-add → not-happy.
💡 "All A are B" and "anything that is not B is not A" say the exact same thing.
5.G.B.3 Step 4 - Chain Step 2 and Step 3: purple → not-add → not-happy.
- So every purple snake is not happy, which is the same as saying every happy snake is not purple.
💡 Combine two arrows then flip with the contrapositive to land on the answer's phrasing.
5.G.B.3 Step 5 - Compare to the choices.
- (A) purple → can-add: false — Step 2 said the opposite.
- (B) purple → happy: not forced.
- (C) can-add → purple: not forced (happy snakes can add and happy snakes are not purple, so a can-add snake might not be purple).
- (D) happy → not purple: matches Step 4 exactly.
- (E) happy → cannot-subtract: not forced (fact 2 only says purple snakes cannot subtract).
- Only (D) is forced.
💡 Walk down the five options and keep only the one that the chain forces.
5.G.B.3 Draw four boxes — happy, purple, can-add, can-subtract — and turn each "if-then" 5.G.B.3 Chain facts 2 and 3: purple → not-subtract → not-add. So every purple snake cann 5.G.B.3 Take fact 1 (happy → can-add) and read it backwards. The contrapositive: if a sn 5.G.B.3 Chain Step 2 and Step 3: purple → not-add → not-happy. So every purple snake is 5.G.B.3 Compare to the choices. (A) purple → can-add: false — Step 2 said the opposite. Review
Reasonableness: Test by constructing a concrete example. Suppose a snake is happy. By fact 1 it can add. If it were also purple, then by fact 2 it cannot subtract, and by fact 3 it cannot add — but we just said it can add. Contradiction. So a happy snake cannot be purple — confirming (D). The counts $13, 4, 5$ never entered the argument, exactly as the problem implied by their roundness.
Alternative: Tool #4 (Matrix Logic) — build a $2 \times 2 \times 2 \times 2$ table of all $16$ possible (happy, purple, add, subtract) combinations, then strike out every row that violates fact 1, 2, or 3. Survivors with happy = yes all have purple = no, confirming (D). More systematic but slower than the contrapositive chain.
CCSS standards used (min grade 5)
5.G.B.3Understand that attributes belonging to a category apply to all subcategories (Treating "all happy snakes can add" as the inclusion of the happy category inside the can-add category, then chaining inclusions and using their reverse (contrapositive) to deduce happy $\subseteq$ not-purple.)
⭐ This AMC 10 problem only needs Grade 5 category reasoning you already know — turn each "all $A$ are $B$" into an arrow, chain purple → not-subtract → not-add and flip happy → can-add to its contrapositive, and the chain gives happy → not-purple, which is choice (D).
⭐ This AMC 10 problem only needs Grade 5 category reasoning you already know — turn each "all $A$ are $B$" into an arrow, chain purple → not-subtract → not-add and flip happy → can-add to its contrapositive, and the chain gives happy → not-purple, which is choice (D).
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