AMC 8 · 2005 · #2
Grade 6 arithmeticProblem
Karl bought five folders from Pay-A-Lot at a cost of each.
Pay-A-Lot had a 20%-off sale the following day. How much could
Karl have saved on the purchase by waiting a day?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: Karl bought $5$ folders at $\$2.50$ each. The next day, the store had a $20\%$-off sale. How much money could Karl have saved by waiting one day?
Givens: Number of folders bought $= 5$; Price per folder $= \$2.50$; Next-day sale $= 20\%$ off; Answer choices: (A) $\$1.00$, (B) $\$2.00$, (C) $\$2.50$, (D) $\$2.75$, (E) $\$5.00$
Unknowns: The savings (in dollars) Karl would have made by buying during the sale
Understand
Restated: Karl bought $5$ folders at $\$2.50$ each. The next day, the store had a $20\%$-off sale. How much money could Karl have saved by waiting one day?
Givens: Number of folders bought $= 5$; Price per folder $= \$2.50$; Next-day sale $= 20\%$ off; Answer choices: (A) $\$1.00$, (B) $\$2.00$, (C) $\$2.50$, (D) $\$2.75$, (E) $\$5.00$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #16 Change Focus
The question splits cleanly into two small jobs (Tool #7, Identify Subproblems): first find the original total cost of all $5$ folders, then take $20\%$ of that total to get the savings. Tool #16 (Change Focus) gives a faster shortcut: since $20\% = \tfrac{1}{5}$, taking $20\%$ of "$5$ folders" is the same as the cost of $1$ folder. Both routes land on the same answer; the subproblem route confirms it with a direct calculation.
Execute — Answer: C
5.NBT.B.7 Step 1 - First subproblem: find the original total cost.
- Five folders at $\$2.50$ each is a multiplication.
💡 Multiplying a count by a per-item price is the Grade 5 decimal multiplication move.
6.RP.A.3 Step 2 - Second subproblem: find $20\%$ of the total cost.
- Convert $20\%$ to the fraction $\tfrac{1}{5}$ or the decimal $0.20$ and multiply by the total.
💡 "Percent of a quantity" is the Grade 6 percent-as-rate-per-100 calculation: $20\%$ of $\$12.50$ means $\tfrac{20}{100} \times \$12.50$.
5.NBT.B.7 First subproblem: find the original total cost. Five folders at $\$2.50$ each is 6.RP.A.3 Second subproblem: find $20\%$ of the total cost. Convert $20\%$ to the fraction Review
Reasonableness: Quick sanity check: $20\%$ of $\$10$ would be $\$2$, so $20\%$ of a number a bit bigger than $\$10$ should be a bit bigger than $\$2$. The answer $\$2.50$ fits, and it sits between (B) $\$2.00$ and (D) $\$2.75$ in the right place. The two extreme choices fail: (A) $\$1.00$ would be only $8\%$ of $\$12.50$, and (E) $\$5.00$ would be $40\%$ — neither matches the $20\%$ discount.
Alternative: Tool #16 (Change Focus): instead of computing $20\%$ of the dollar total, notice that $20\% = \tfrac{1}{5}$, and Karl bought exactly $5$ folders. So $20\%$ off the purchase equals the cost of $\tfrac{1}{5} \times 5 = 1$ folder, which is $\$2.50$. Same answer (C), no multiplication of a decimal needed.
CCSS standards used (min grade 6)
5.NBT.B.7Multiply decimals to hundredths using strategies based on place value (Computing the original total cost $5 \times \$2.50 = \$12.50$ with decimal multiplication.)6.RP.A.3Find a percent of a quantity as a rate per $100$ (Taking $20\%$ of $\$12.50$ to get the $\$2.50$ discount that Karl would have saved.)
⭐ $20\%$ off five folders is the same as one folder free — that shortcut turns this AMC 8 problem into a Grade 6 percent-of-a-quantity exercise.
⭐ $20\%$ off five folders is the same as one folder free — that shortcut turns this AMC 8 problem into a Grade 6 percent-of-a-quantity exercise.