AMC 8 · 2001 · #3

Grade 6 arithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Fraction / Expression Sum Evaluation

fraction-multiplicationmulti-digit-arithmeticratio-proportion identify-subproblems ↑ Prerequisites: fraction-multiplicationmulti-digit-arithmetic

📏 Short solution 💡 2 insights

Problem

Granny Smith has

63</span>. Anjou has <span class="mk-c">one-third as much</span> as Granny Smith, and Elberta has <span class="mk-c">

2 more than Anjou. How many dollars does Elberta have?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2001 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Introduce a Variable

Three people each have an unknown amount of money, and the amounts are linked in a chain: Granny Smith → Anjou → Elberta. Tool #4 (Introduce a Variable) lets us name each amount (G, A, E) so the word "one-third" and the words "$2 more" become a short equation. With the variables in place, the chain unwinds in two arithmetic steps and Elberta's amount drops out.

1STEP 1

Name each amount — let G, A, E be the three people's dollars — so G = 63, A = $\frac{1}{3}$ G, E = A + 2.

G = 63, A =

\frac{1}{3}

G, E = A + 2

2STEP 2

One-third of Granny Smith's 63 is 63 ÷ 3, so A = 21.

A =

\frac{1}{3}

× 63 = 21

3STEP 3

Add 2 to Anjou's 21, giving E = 23 → (E).

E = A + 2 = 21 + 2 = 23 → (E)

Answer

Walk the chain back up to check. If E = 23, then Anjou must have E - 2 = 21 and Granny Smith must have 3 × 21 = 63. That matches the given

63, so the chain is consistent. The answer 23 also appears in the choices as (E), and 23 is just slightly larger than Anjou's

21, which fits the "$2 more" wording.

💡Key takeaway

Name each amount, then follow the chain: 63 ÷ 3 = 21, then 21 + 2 = 23. Naming the unknowns turns a word problem into two short calculations.

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