2004 AMC 10A Problem 14

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?

1. We need four coins, each worth $1$, $5$, $10$ or $25$ cents, adding to $80$.
2. 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.
3. 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 $\boxed{\textbf{(A)}\ 0}$.

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.

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 $\boxed{\textbf{(A)}\ 0}$.