AMC 8 · 2003 · #17

Easy mode Grade 1

Problem

There are six children. They come from two families. Each family has exactly three siblings.

Each child has one eye color (blue or brown) and one hair color (black or blond). The table below shows each child's eye color and hair color.

$\begin{array}{c|c|c}\text{Child}&\text{Eye Color}&\text{Hair Color}\\ \hline\text{Benjamin}&\text{Blue}&\text{Black}\\ \hline\text{Jim}&\text{Brown}&\text{Blonde}\\ \hline\text{Nadeen}&\text{Brown}&\text{Black}\\ \hline\text{Austin}&\text{Blue}&\text{Blonde}\\ \hline\text{Tevyn}&\text{Blue}&\text{Black}\\ \hline\text{Sue}&\text{Blue}&\text{Blonde}\\ \hline\end{array}$

Here is the rule. Any two children from the same family must share at least one feature: either the same eye color, or the same hair color (or both).

Jim has two siblings in this list. Which two are they?

Pick an answer.

(A)

Nadeen and Austin

(B)

Benjamin and Sue

(C)

Benjamin and Austin

(D)

Nadeen and Tevyn

(E)

Austin and Sue

AMC 8 2003 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.

View mode:

Toolkit + CCSS Solution

Understand

Restated: Six children — Benjamin, Jim, Nadeen, Austin, Tevyn, Sue — come from two families of three siblings each. Each child is recorded with eye color (blue or brown) and hair color (black or blond). Children in the same family share at least one of those two traits. Given Jim has brown eyes and blond hair, which two of the others are Jim's siblings?

Givens: Two families, each with exactly $3$ siblings ($6$ children total); Each child has eye color $\in$ {Blue, Brown} and hair color $\in$ {Black, Blond}; Same-family siblings share at least one trait (eye color or hair color); Traits: Benjamin (Blue, Black), Jim (Brown, Blond), Nadeen (Brown, Black), Austin (Blue, Blond), Tevyn (Blue, Black), Sue (Blue, Blond); Answer choices: (A) Nadeen and Austin, (B) Benjamin and Sue, (C) Benjamin and Austin, (D) Nadeen and Tevyn, (E) Austin and Sue

Unknowns: The two children who are in Jim's family

Understand

Plan

Primary tool: #4 Use Matrix Logic

Secondary: #3 Eliminate Possibilities

This is a "who belongs with whom?" puzzle with two attributes per child — exactly the trigger for Tool #4 (Matrix Logic). Build a small table (child by trait), highlight Jim's traits (Brown, Blond), and the candidates for Jim's family must share at least one highlighted trait with Jim. That immediately narrows the pool. Tool #3 (Eliminate Possibilities) then handles the answer choices: any choice containing a child who shares no trait with Jim is out, and the remaining choice must also have its two named children share a trait with each other so all three siblings are pairwise compatible.

Execute — Answer: E

#4 Use Matrix Logic K.MD.B.3 Step 1

Lay the data out as a matrix and mark Jim's row.
Jim is Brown-eyed and Blond-haired, so a sibling of Jim must also have Brown eyes or Blond hair (or both).
Scan the table and tag each other child with whether they share a trait with Jim.

$$\begin{array}{l|c|c|c}\text{Child}&\text{Eyes}&\text{Hair}&\text{Shares with Jim?}\\\hline \text{Benjamin}&\text{Blue}&\text{Black}&\text{No}\\ \text{Nadeen}&\textbf{Brown}&\text{Black}&\text{Yes (eyes)}\\ \text{Austin}&\text{Blue}&\textbf{Blond}&\text{Yes (hair)}\\ \text{Tevyn}&\text{Blue}&\text{Black}&\text{No}\\ \text{Sue}&\text{Blue}&\textbf{Blond}&\text{Yes (hair)}\end{array}$$

💡 Sorting the children into "matches Jim" vs "does not match Jim" is the Kindergarten move: classify objects by a shared attribute.

#3 Eliminate Possibilities K.MD.B.3 Step 2

Benjamin and Tevyn share neither trait with Jim, so they cannot be in Jim's family.
Cross off any answer choice that names them.
That eliminates (B) Benjamin and Sue, (C) Benjamin and Austin, and (D) Nadeen and Tevyn.

$$\text{Eliminate: (B), (C), (D)} \Rightarrow \text{remaining: (A), (E)}$$

💡 If a child shares no trait with Jim, putting them in Jim's family would break the rule immediately.

#4 Use Matrix Logic 1.MD.C.4 Step 3

Decide between (A) Nadeen and Austin and (E) Austin and Sue.
The rule says all three siblings must pairwise share a trait, so Jim's two siblings must also share a trait with each other.
Check each finalist pair.

$\text{(A) Nadeen + Austin}:\;\text{Brown/Black vs Blue/Blond}\Rightarrow \text{no shared trait}\;\text{✗}$ \quad $\text{(E) Austin + Sue}:\;\text{Blue/Blond vs Blue/Blond}\Rightarrow \text{both share (eyes and hair)}\;\text{✓}$

💡 The matrix forces the pair-with-each-other check too: a sibling group is only valid if every pair inside it shares a trait.

[1] #4 K.MD.B.3 Lay the data out as a matrix and mark Jim's row. Jim is Brown-eyed and Blond-hai

[2] #3 K.MD.B.3 Benjamin and Tevyn share neither trait with Jim, so they cannot be in Jim's fami

[3] #4 1.MD.C.4 Decide between (A) Nadeen and Austin and (E) Austin and Sue. The rule says all t

Review

Reasonableness: Verify the family split. Jim's family is {Jim, Austin, Sue}: Jim–Austin share Blond hair, Jim–Sue share Blond hair, Austin–Sue share both Blue eyes and Blond hair. The other family is {Benjamin, Nadeen, Tevyn}: Benjamin–Tevyn share Blue eyes and Black hair, Benjamin–Nadeen share Black hair, Nadeen–Tevyn share Black hair. Every pair in both families shares at least one trait, so the split is consistent. A handy shortcut also confirms it: the hair-color column has exactly three Blonds (Jim, Austin, Sue) and three Blacks (Benjamin, Nadeen, Tevyn) — splitting by hair color alone already produces two valid families of three, and that matches (E).

Alternative: Tool #1 (Draw a Diagram) — a Venn-style two-circle picture. Draw a circle for "Brown eyes" and a circle for "Blond hair"; place each child in the right region. Jim sits in the intersection. Sue and Austin sit in the Blond-only region (Blue eyes, Blond hair). Nadeen sits in the Brown-only region (Brown eyes, Black hair). Benjamin and Tevyn sit outside both circles. The two children closest to Jim — sharing the Blond region — are Austin and Sue, matching (E).

CCSS standards used (min grade 1)

K.MD.B.3 Classify objects into given categories and count the numbers in each category (Sorting the six children by shared traits with Jim (Brown eyes, Blond hair) so non-matching children are immediately ruled out.)
1.MD.C.4 Organize, represent, and interpret data with up to three categories (Reading the trait table to check pairwise compatibility — both that each candidate shares a trait with Jim, and that Jim's two siblings share a trait with each other.)

⭐ Two traits per child, one shared trait per sibling pair — this AMC 8 problem is a Kindergarten "sort by category" task in disguise, and answer (E) Austin and Sue falls out the moment you split by hair color.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Review

CCSS standards used (min grade 1)

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