AMC 10 · 2004 · #14

Grade 8 arithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

mean-median-mode-rangesystems-of-equations convert-to-algebrasystematic-enumeration ↑ Prerequisites: linear-equations-one-var

📏 Medium solution 💡 2 insights

Problem

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.

Pick an answer.

(A)

$\ 0$

(B)

$\ 1$

(C)

$\ 2$

(D)

$\ 3$

(E)

$\ 4$

AMC 10 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Introduce a Variable

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.

1STEP 1

Turn the first average into an equation

Let $n$ be the coin count and $T$ the total value in cents; the average is 20, so $\frac{T}{n} = 20$ , giving $T = 20n$ .

T/n = 20 → T = 20n

2STEP 2

Add the quarter and solve for the count

The extra quarter gives $\frac{T+25}{n+1} = 21$ ; substituting $T = 20n$ leaves $20n + 25 = 21n + 21$ , so $n = 4$ and $T = 80$ .

20n + 25 = 21n + 21 → n = 4, T = 80

3STEP 3

List the coins that make 80 cents

Four coins worth 80 cents force $25+25+25+5$ ; two quarters would leave 30 cents for two coins, impossible — so 0 dimes, choice (A).

25+25+25+5 = 80 → 0 dimes, (A)

Answer

Check the found coins directly: three quarters and one nickel are 4 coins worth 75+5 = 80 cents, and 80 ÷ 4 = 20 cents average, matching the first fact. Adding one more quarter makes 5 coins worth 105 cents, and 105 ÷ 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.

💡Key takeaway

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.

Turn the first average into an equation
Add the quarter and solve for the count
List the coins that make 80 cents

Turn the first average into an equation

Add the quarter and solve for the count

List the coins that make 80 cents

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