AMC 10 · 2004 · #15
Easy mode Grade 4Patty has 20 coins. Some are nickels (worth 5 cents each) and the rest are dimes (worth 10 cents each). Suppose you turned every nickel into a dime and every dime into a nickel. Then she would have 70 cents more than she has now. How much is all her money worth right now?
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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