2004 AMC 10A Problem 14

P11-15~4 minPrint

Ratios, Percents & Averages Diophantine Equations Organized listing / direct enumeration Set up the equation/formula and compute; no special trick needed

The 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?

Show solution without answering

Key insight

Adding one 25-cent coin raises the average by exactly 1 cent only if there were 4 coins, worth 80 cents in all. That forces three quarters and a nickel.

Solution

Step A: how many coins?

1. Say Paula has $n$ coins. An average of $20$ cents means the total value is $20n$ cents.
2. With one more quarter she has $n + 1$ coins worth $20n + 25$ cents, and the average is $21$ . So
$20n + 25 = 21(n + 1) = 21n + 21.$
3. Subtract $20n + 21$ from both sides: $4 = n$ . Paula has $4$ coins, worth $20 \times 4 = 80$ cents.

Step B: which coins?

We need four coins, each worth $1$ , $5$ , $10$ or $25$ cents, adding to $80$ .
Try using at most two quarters. The most you could get is $25 + 25 + 10 + 10 = 70$ , which is short of $80$ . So she needs at least three quarters.
Four quarters is $100$ , too much. So she has exactly three quarters, worth $75$ cents, and the fourth coin is worth $80 - 75 = 5$ cents: a nickel.

Paula has three quarters and a nickel, and no dimes.

The answer is $\boxed{\textbf{(A)}\ 0}$ .

Why this works

An average is a total divided by a count, so "the average changes when one item is added" becomes an equation for the count. A quick way to see it: the new quarter is $5$ cents above the old average, and those $5$ extra cents are shared among the new count of coins to raise the average by $1$ . So there are $5$ coins after adding it, which means $4$ before. After that, a quick check of the biggest possible totals pins down the coins.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Let Paula have $n$ coins. An average of $20$ cents means their total value is $20n$ cents. Adding a quarter gives $n + 1$ coins worth $20n + 25$ cents with average $21$ :
$20n + 25 = 21(n+1) \quad\Longrightarrow\quad 20n + 25 = 21n + 21 \quad\Longrightarrow\quad n = 4.$
So Paula has $4$ coins worth $80$ cents in total.

Now find four coins from $\{1, 5, 10, 25\}$ that sum to $80$ . The largest possible total with at most two quarters is $25 + 25 + 10 + 10 = 70 < 80$ , so at least three quarters are needed; three quarters make $75$ , and the fourth coin must be worth $5$ : a nickel. (Four quarters is $100$ , too much.) The purse holds three quarters and one nickel.

Paula has no dimes. The answer is $\boxed{\textbf{(A)}\ 0}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 6: Ratios and Percents
Introduction to Number Theory, Chapter 7: Algebra With Integers
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.2)
Introduction to Algebra, Chapter 7: Proportion (§7.1, §7.2, §7.3)