AMC 10 · 2004 · #15
Grade 4 arithmeticPatty has 20 coins consisting of nickels and dimes. If her nickels were dimes and her dimes were nickels, she would have 70 cents more. How much are her coins worth?
Pick an answer.
AMC 10 2004 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.
Toolkit + CCSS Solution
Understand
Restated: Patty has $20$ coins that are a mix of nickels ($5$ cents) and dimes ($10$ cents). If every nickel were turned into a dime and every dime were turned into a nickel, the money she has would be $70$ cents more than it is now. Find the total value of her coins right now.
Givens: Patty has $20$ coins in all.; Every coin is either a nickel worth $5$ cents or a dime worth $10$ cents.; Swapping every nickel to a dime and every dime to a nickel raises the total by $70$ cents.; Answer choices: (A) $\textdollar 1.15$, (B) $\textdollar 1.20$, (C) $\textdollar 1.25$, (D) $\textdollar 1.30$, (E) $\textdollar 1.35$.
Unknowns: The total value of Patty's $20$ coins as they are now.
Understand
Restated: Patty has $20$ coins that are a mix of nickels ($5$ cents) and dimes ($10$ cents). If every nickel were turned into a dime and every dime were turned into a nickel, the money she has would be $70$ cents more than it is now. Find the total value of her coins right now.
Givens: Patty has $20$ coins in all.; Every coin is either a nickel worth $5$ cents or a dime worth $10$ cents.; Swapping every nickel to a dime and every dime to a nickel raises the total by $70$ cents.; Answer choices: (A) $\textdollar 1.15$, (B) $\textdollar 1.20$, (C) $\textdollar 1.25$, (D) $\textdollar 1.30$, (E) $\textdollar 1.35$.
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #8 Analyze the Units, #6 Guess and Check
Chasing the two unknown counts head-on invites algebra, but the clean move is Tool #16 (Change Focus): stop staring at the totals and watch what one swap *does*. Turning a nickel into a dime adds exactly $5$ cents, and turning a dime into a nickel loses exactly $5$ cents. So the whole $70$-cent jump is built out of $5$-cent steps, and dividing tells you how many more nickels than dimes there are. Tool #8 (Analyze the Units) keeps every quantity in cents so nothing gets misread, and Tool #6 (Guess and Check) lets you confirm the split lands on $20$ coins.
Execute — Answer: A
4.MD.A.2 Step 1 See what one swap does
- A nickel is $5$ cents and a dime is $10$ cents, a gap of $5$ cents.
- When a nickel becomes a dime, the money goes up by $5$ cents.
- When a dime becomes a nickel, the money goes down by $5$ cents.
- Every single coin that changes moves the total by exactly $5$ cents, up or down.
💡 Nickels and dimes sit $5$ cents apart, so swapping any one coin nudges the total by that same $5$ cents.
3.OA.C.7 Step 2 Turn the 70-cent jump into a count
- The swap raises the total by $70$ cents, so the ups beat the downs by $70$ cents.
- Each nickel-to-dime is a $+5$ and each dime-to-nickel is a $-5$, so the leftover $+70$ cents means there are $70 \div 5 = 14$ more nickels than dimes.
- In short, Patty has $14$ more nickels than dimes.
💡 A net rise of $70$ cents in $5$-cent pieces can only come from $14$ extra nickels over dimes.
4.OA.A.3 Step 3 Split the 20 coins
- Patty has $20$ coins, and $14$ more of them are nickels than dimes.
- Set the $14$ extra nickels aside; that leaves $20 - 14 = 6$ coins that pair up evenly, $3$ nickels with $3$ dimes.
- Adding the $14$ back gives $3 + 14 = 17$ nickels and $3$ dimes.
💡 After pulling out the $14$ extra nickels, the rest splits down the middle into equal counts.
4.MD.A.2 Step 4 Add up the value
- Now total the money.
- The $17$ nickels are worth $17 \times 5 = 85$ cents, and the $3$ dimes are worth $3 \times 10 = 30$ cents.
- Together that is $85 + 30 = 115$ cents, which is $\textdollar 1.15$.
- The answer is (A).
💡 Counting each pile of coins in cents and adding gives the money on the table.
4.MD.A.2 A nickel is $5$ cents and a dime is $10$ cents, a gap of $5$ cents. When a nicke 3.OA.C.7 The swap raises the total by $70$ cents, so the ups beat the downs by $70$ cents 4.OA.A.3 Patty has $20$ coins, and $14$ more of them are nickels than dimes. Set the $14$ 4.MD.A.2 Now total the money. The $17$ nickels are worth $17 \times 5 = 85$ cents, and th Review
Reasonableness: Test the swap on the answer. With $17$ nickels and $3$ dimes the total is $115$ cents. After the swap Patty would have $3$ nickels and $17$ dimes, worth $3\times 5 + 17\times 10 = 15 + 170 = 185$ cents. The difference is $185 - 115 = 70$ cents, exactly as the problem says, and $17+3 = 20$ coins checks out too. The value also lands right on a listed choice, $\textdollar 1.15$.
Alternative: Name the counts: let $n$ be nickels and $d$ dimes, so $n + d = 20$. The current value is $5n + 10d$ and the swapped value is $10n + 5d$; their difference is $(10n+5d)-(5n+10d) = 5n - 5d = 70$, giving $n - d = 14$. Solving $n+d=20$ with $n-d=14$ gives $n = 17$ and $d = 3$, so the value is $5(17)+10(3) = 115$ cents $= \textdollar 1.15$, the same answer.
CCSS standards used (min grade 4)
4.MD.A.2Solve word problems involving distances, time, liquid volumes, and money (Reading the $5$-cent gap between a nickel and a dime and adding the coin piles ($17\times 5 + 3\times 10 = 115$ cents) to get the money.)3.OA.C.7Fluently multiply and divide within 100 (Dividing the $70$-cent net change by the $5$-cent step, $70\div 5 = 14$, and multiplying the coin counts by their values.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Combining '$20$ coins total' with '$14$ more nickels than dimes' to split the coins into $17$ nickels and $3$ dimes.)
⭐ Watch what one swap changes instead of the whole pile: since each swapped coin shifts the total by $5$ cents, a $70$-cent jump means $14$ more nickels than dimes, which forces $17$ nickels and $3$ dimes worth $\textdollar 1.15$.
⭐ Watch what one swap changes instead of the whole pile: since each swapped coin shifts the total by $5$ cents, a $70$-cent jump means $14$ more nickels than dimes, which forces $17$ nickels and $3$ dimes worth $\textdollar 1.15$.
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