Helen and Ivan have the same number of coins. Helen has 64 20-cent coins, and Ivan has 104 20-cent coins. Helen has 40 more 50-cent coins than Ivan. Who has more money in coins, and by how much?

Let $N_{H50}$ be the number of 50-cent coins Helen has.
Helen has $N_{H50}$ 50-cent coins and 64 20-cent coins.
Total number of coins for Helen: $N_H = N_{H50} + 64$.

Let $N_{I50}$ be the number of 50-cent coins Ivan has.
Ivan has $N_{I50}$ 50-cent coins and 104 20-cent coins.
Total number of coins for Ivan: $N_I = N_{I50} + 104$.

This means Helen has 40 more 50-cent coins than Ivan. Conversely, Ivan has 40 fewer 50-cent coins than Helen.
Also, Helen has $104 - 64 = 40$ fewer 20-cent coins than Ivan.

The difference in their coin composition is that Helen has 40 more 50-cent coins and 40 fewer 20-cent coins compared to Ivan. This means Helen effectively exchanged 40 20-cent coins for 40 50-cent coins.
The value difference per coin exchange is $50 \text{ cents} - 20 \text{ cents} = 30 \text{ cents}$.
Since Helen has 40 such "exchanges" in her favor, the total difference in money is:

Converting to dollars: $\text{Difference} = \frac{1200}{100} \text{ dollars} = \$12.00$ Since Helen has the higher value coins in this exchange, Helen has more money.

Helen has more money in coins, by \boxed{\12.00}$.

Let $m_{50