AMC 10 · 2004 · #14
Grade 8 arithmeticThe average value of all the pennies, nickels, dimes, and quarters in Paula's purse is 20 cents. If she had one more quarter, the average value would be 21 cents. How many dimes does she have in her purse?
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: Paula's purse holds only pennies, nickels, dimes, and quarters, and the average value of all these coins is $20$ cents. If one more quarter were added, the average would become $21$ cents. Find how many dimes are in the purse.
Givens: The purse contains only pennies ($1$¢), nickels ($5$¢), dimes ($10$¢), and quarters ($25$¢).; The average value of all the coins is $20$ cents.; Adding one more quarter ($25$¢) would make the average $21$ cents.; Answer choices: (A) $0$, (B) $1$, (C) $2$, (D) $3$, (E) $4$.
Unknowns: The number of dimes in the purse.
Understand
Restated: Paula's purse holds only pennies, nickels, dimes, and quarters, and the average value of all these coins is $20$ cents. If one more quarter were added, the average would become $21$ cents. Find how many dimes are in the purse.
Givens: The purse contains only pennies ($1$¢), nickels ($5$¢), dimes ($10$¢), and quarters ($25$¢).; The average value of all the coins is $20$ cents.; Adding one more quarter ($25$¢) would make the average $21$ cents.; Answer choices: (A) $0$, (B) $1$, (C) $2$, (D) $3$, (E) $4$.
Plan
Primary tool: #4 Introduce a Variable
Secondary: #13 Convert to Algebra, #2 Make a Systematic List
The number of dimes is hidden behind two facts about averages, and an average is total divided by count, so name the two hidden quantities — the coin count and the total value — with variables (Tool #4). Turn each averaging fact into an equation and solve the pair (Tool #13); this pins down exactly how many coins there are and how much they are worth together. Then, because only a few coins can make that exact total, list the coin combinations that fit (Tool #2) and see how many dimes are forced.
Execute — Answer: A
6.SP.B.5 Step 1 Turn the first average into an equation
- An average is the total divided by how many things there are.
- Let $n$ be the number of coins in the purse and let $T$ be their total value in cents.
- The average value is $20$ cents, so $T \div n = 20$.
- Multiplying both sides by $n$ gives $T = 20n$.
- This one equation ties the total value to the coin count.
💡 Average value is just the total value shared equally over every coin, so total equals average times the number of coins.
8.EE.C.7 Step 2 Add the quarter and solve for the count
- Adding one quarter raises the value by $25$ cents and the count by $1$, and the new average is $21$ cents, so $\dfrac{T+25}{n+1} = 21$.
- Clearing the fraction gives $T + 25 = 21(n+1) = 21n + 21$.
- Replace $T$ with $20n$ from the first equation: $20n + 25 = 21n + 21$.
- Subtracting $20n$ and $21$ from both sides leaves $4 = n$.
- So there are $n = 4$ coins, worth $T = 20 \times 4 = 80$ cents in total.
💡 Two facts about the same coins give two equations, and substituting one into the other leaves a single unknown to solve.
2.MD.C.8 Step 3 List the coins that make 80 cents
- Now find $4$ coins whose values add to $80$ cents.
- To reach $80$ with only $4$ coins you need mostly quarters.
- Three quarters give $75$ cents, leaving $5$ cents for the last coin — a nickel — which works: $25+25+25+5 = 80$.
- Fewer quarters fail: two quarters leave $30$ cents for two coins, but the largest two non-quarter coins are two dimes worth only $20$ cents, not enough.
- So the coins must be three quarters and one nickel.
- That leaves no room for any dimes, so the number of dimes is $0$, which is choice (A).
💡 Building a large total from very few coins forces the biggest coins, which crowds the dimes out completely.
6.SP.B.5 An average is the total divided by how many things there are. Let $n$ be the num 8.EE.C.7 Adding one quarter raises the value by $25$ cents and the count by $1$, and the 2.MD.C.8 Now find $4$ coins whose values add to $80$ cents. To reach $80$ with only $4$ c Review
Reasonableness: Check the found coins directly: three quarters and one nickel are $4$ coins worth $75+5 = 80$ cents, and $80 \div 4 = 20$ cents average, matching the first fact. Adding one more quarter makes $5$ coins worth $105$ cents, and $105 \div 5 = 21$ cents, matching the second fact. Both conditions hold with zero dimes, so (A) is consistent. It also makes sense that the average rose: a $25$-cent quarter is above the old $20$-cent average, so it pulls the average up.
Alternative: Reason about the average directly instead of solving equations. A coin worth exactly the current average of $20$ cents would leave the average unchanged. The added quarter is $25$ cents, which is $5$ cents above that average, and those extra $5$ cents get shared out over the new pile of coins. They raise the average by exactly $1$ cent only when the new pile has $5$ coins, so the purse started with $4$ coins totaling $80$ cents. Listing the coins then forces three quarters and one nickel, giving $0$ dimes.
CCSS standards used (min grade 8)
6.SP.B.5Summarize numerical data sets by reporting number of observations and measures (Reading the average value as total value divided by the number of coins to write $T = 20n$.)8.EE.C.7Solve linear equations in one variable (Solving $20n + 25 = 21n + 21$ (variable on both sides) to find the coin count $n = 4$ and total $T = 80$.)2.MD.C.8Solve word problems involving dollar bills, quarters, dimes, nickels, and pennies (Finding which $4$ coins add to $80$ cents — three quarters and one nickel — so the dime count is forced to $0$.)
⭐ Average value means the total divided by how many coins, so turn each average into an equation, solve for the count and total, then check which coins can actually make that total.
⭐ Average value means the total divided by how many coins, so turn each average into an equation, solve for the count and total, then check which coins can actually make that total.
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