Where do I start for this question.

What is the most amount of money--all coins but no dollar coins--that you could have in your pocket and still not be able to give exact change for a dollar, a half dollar, a quarter, a dime or a nickel?

$0.49

never mind $0.99

half-dollar, quarter, dime, nickel, 4 pennies

To solve this problem, we need to find the largest amount of money in coins (excluding dollar coins) that cannot be used to give exact change for a dollar, half dollar, quarter, dime, or nickel.

Let's break it down step by step:

1. Start by listing all the possible coin denominations that can be used: penny (1 cent), nickel (5 cents), dime (10 cents), quarter (25 cents), half dollar (50 cents), and dollar (100 cents).

2. We are not allowed to use dollar coins, so we exclude them from our list. We are left with penny, nickel, dime, quarter, and half dollar.

3. Now, let's consider the largest coin, which is the half dollar. If we have more than 1 half dollar, we can already give exact change for a dollar, because 2 half dollars make a dollar. So, we can only have a maximum of 1 half dollar.

4. Next, let's consider the remaining coins. Since we don't want to be able to give exact change for a dollar, we need to find a combination of the remaining coins that cannot add up to 50 cents. We need to avoid any combination that would equal 50 cents. For instance, we can't have 2 quarters because they add up to 50 cents.

5. By considering the remaining coins (penny, nickel, dime, and quarter), we can deduce that the highest possible amount we can have without being able to give exact change for the given denominations is 49 cents. This can be achieved by having 2 quarters, 4 dimes, and 9 pennies, which totals to 49 cents.

So, the answer to the question is that the largest amount of money (using all coins except dollar coins) that cannot give exact change for a dollar, half dollar, quarter, dime, or nickel is 49 cents.

Remember, the approach explained here involved considering the various possibilities and combinations to find the answer.