Eddie spills his cash register drawer on the floor and has to sweep up the coins. He has 680 coins, which are a mixture of pennies, nickels, dimes, and quarters. There are 10 times as many pennies as dimes, and half as many quarters as dimes. If he has $\$ 28.00$, how many dimes are there?

Step 1 ：Let's denote the number of pennies as p, the number of nickels as n, the number of dimes as d, and the number of quarters as q.

Step 2 ：We have the following equations based on the problem: \(p + n + d + q = 680\) (total number of coins), \(0.01p + 0.05n + 0.10d + 0.25q = 28.00\) (total value of coins), \(p = 10d\) (number of pennies is 10 times the number of dimes), and \(q = 0.5d\) (number of quarters is half the number of dimes).

Step 3 ：We can substitute the equations \(p = 10d\) and \(q = 0.5d\) into the equations \(p + n + d + q = 680\) and \(0.01p + 0.05n + 0.10d + 0.25q = 28.00\) to solve for d (the number of dimes).

Step 4 ：This gives us the new equations: \(11.5d + n = 680\) and \(0.325d + 0.05n = 28.0\).

Step 5 ：Solving these equations, we find that \(d = 24\) and \(n = 404\).

Step 6 ：So, the number of dimes is \(\boxed{24}\).