Competition · AMC preparation · step 4 of 4
AMC 8 · 2001 · #3
Grade 6 arithmeticPick an answer.
AMC 8 2001 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Name each amount
Name each amount — let G, A, E be the three people's dollars — so G = 63, A = G, E = A + 2.
Grade 6 "write expressions with variables." The three names turn the sentence into equations you can compute with.
6.EE.A.2Introduce A VariableFind Anjou's amount
One-third of Granny Smith's 63 is 63 ÷ 3, so A = 21.
Grade 5 multiplying a whole number by a fraction: 1/3 × 63 is the same as 63 ÷ 3 = 21.
One-third of Granny Smith's $63 works out to $21 for Anjou.
▸ Why?
"One-third as much" means Granny Smith's $63 is cut into 3 equal parts and Anjou keeps one of those parts.
▸ Why?
Cutting the money into 3 equal parts leaves no gaps or overlaps, so the 3 parts add back to the whole $63 and each part is that whole shared evenly by 3.
▸ Why?
One of those 3 equal parts is $63 ÷ 3, and that quotient is $21.
▸ Why?
63 ÷ 3 = 21 because 3 groups of 21 rebuild 63, and dividing by 3 simply undoes multiplying by 3.
Find Elberta's amount
Add 2 to Anjou's 21, giving E = 23 → (E).
Grade 3 two-step word problem: divide first, then add. The chain ends at Elberta.
3.OA.D.8Introduce A VariableName each amount, then follow the chain: 63 ÷ 3 = 21, then 21 + 2 = 23. Naming the unknowns turns a word problem into two short calculations.
- Name each amount
- Find Anjou's amount
- Find Elberta's amount
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