2004 AMC 10A Problem 14

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

An average is a total divided by a count, so a statement about how the average moves when one item is added is a linear equation in the count. The quick mental version: the new quarter exceeds the old average by $5$ cents, and that surplus is spread over the new count to raise the average by $1$, so the new count is $5$. After that, a small bounding argument (at most two quarters cannot reach $80$) pins down the coins with no real casework.

• Finding n = 4 coins and 80 cents but then guessing a coin mix (like 25 + 25 + 10 + 10 + 10) without checking it uses exactly four coins.
• Setting up $\frac{20n + 25}{n} = 21$ (forgetting the count also increases), which gives $n = 25$.