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 $63. Elberta has$ 2 more than Anjou and Anjou has one-third as much as Granny Smith. 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.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: Granny Smith has $\$63$. Anjou has one-third as much as Granny Smith, and Elberta has $\$2$ more than Anjou. How many dollars does Elberta have?

Givens: Granny Smith has $\$63$; Anjou has one-third of what Granny Smith has; Elberta has $\$2$ more than Anjou; Answer choices: (A) $17$, (B) $18$, (C) $19$, (D) $21$, (E) $23$

Unknowns: The number of dollars Elberta has

Understand

Restated: Granny Smith has $\$63$. Anjou has one-third as much as Granny Smith, and Elberta has $\$2$ more than Anjou. How many dollars does Elberta have?

Givens: Granny Smith has $\$63$; Anjou has one-third of what Granny Smith has; Elberta has $\$2$ more than Anjou; Answer choices: (A) $17$, (B) $18$, (C) $19$, (D) $21$, (E) $23$

Plan

Primary tool: #4 Introduce a Variable

Three people each have an unknown amount of money, and the amounts are linked in a chain: Granny Smith $\to$ Anjou $\to$ 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.

Execute — Answer: E

#4 Introduce a Variable 6.EE.A.2 Step 1

Give each person's amount a name.
Let $G$, $A$, $E$ be Granny Smith's, Anjou's, and Elberta's dollars.
From the problem, $G = 63$, $A = \tfrac{1}{3} G$, and $E = A + 2$.

$$G = 63,\quad A = \tfrac{1}{3} G,\quad E = A + 2$$

💡 Grade 6 "write expressions with variables." The three names turn the sentence into equations you can compute with.

#4 Introduce a Variable 5.NF.B.4 Step 2

Compute Anjou's amount.
"One-third as much as Granny Smith" means $\tfrac{1}{3}$ of $63$, which is $63 \div 3$.

$$A = \tfrac{1}{3} \times 63 = 21$$

💡 Grade 5 multiplying a whole number by a fraction: $\tfrac{1}{3} \times 63$ is the same as $63 \div 3 = 21$.

#4 Introduce a Variable 3.OA.D.8 Step 3

Compute Elberta's amount.
Add $2$ to Anjou's $21$.

$$E = A + 2 = 21 + 2 = 23 \;\Rightarrow\; \textbf{(E)}$$

💡 Grade 3 two-step word problem: divide first, then add. The chain ends at Elberta.

[1] #4 6.EE.A.2 Give each person's amount a name. Let $G$, $A$, $E$ be Granny Smith's, Anjou's,

[2] #4 5.NF.B.4 Compute Anjou's amount. "One-third as much as Granny Smith" means $\tfrac{1}{3}$

[3] #4 3.OA.D.8 Compute Elberta's amount. Add $2$ to Anjou's $21$.

Review

Reasonableness: Walk the chain back up to check. If $E = 23$, then Anjou must have $E - 2 = 21$ and Granny Smith must have $3 \times 21 = 63$. 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.

Alternative: Tool #6 (Guess and Check): test each choice as Elberta's amount and see if Granny Smith ends up with $\$63$. Choice (E) $E = 23$ gives Anjou $= 23 - 2 = 21$ and Granny Smith $= 3 \times 21 = 63$ — a match. No other choice produces $63$: e.g. (D) $E = 21$ gives Granny Smith $= 3 \times 19 = 57$, and (A) $E = 17$ gives $3 \times 15 = 45$. Only (E) works.

CCSS standards used (min grade 6)

6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers (Naming each person's money with a variable ($G$, $A$, $E$) so the word sentences become the equations $A = \tfrac{1}{3} G$ and $E = A + 2$.)
5.NF.B.4 Apply and extend previous understandings of multiplication to multiply a fraction by a whole number (Computing $\tfrac{1}{3} \times 63 = 21$ to find Anjou's amount.)
3.OA.D.8 Solve two-step word problems using the four operations (Chaining the two operations (divide by $3$, then add $2$) to get from Granny Smith's $\$63$ to Elberta's $\$23$.)

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

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Review

CCSS standards used (min grade 6)

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