A certain library assesses fines for overdue books as follows. On the first day that a book is overdue, the...

1. Translate the problem requirements: Understand that we start with \(\$0.10\) on day 1, and for each subsequent day we compare two options: (1) add \(\$0.30\) to current total, or (2) double the current total, then choose whichever gives the smaller amount
2. Track the daily progression systematically: Calculate the fine for each day from day 1 through day 4, applying the "lesser amount" rule at each step
3. Apply the comparison rule consistently: For each new day, compute both options (add \(\$0.30\) vs. double) and select the smaller result
4. Verify the final amount: Ensure our day 4 calculation matches one of the given answer choices

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what the library fine system means in everyday terms:

Think of this like a penalty that gets worse each day, but the library wants to be somewhat fair - they'll always choose whichever penalty is smaller between two options.

Here's how it works:
• Day 1 (first day overdue): The fine starts at \(\$0.10\)
• Each additional day: We calculate two possible fines and pick the smaller one:
  - Option A: Take yesterday's fine and add \(\$0.30\)
  - Option B: Take yesterday's fine and double it

We want to find the total fine on the fourth day the book is overdue.

Process Skill: TRANSLATE - Converting the problem's penalty structure into a clear day-by-day calculation method

2. Track the daily progression systematically

Now let's work through each day step by step, like following a recipe:

Day 1:
The fine starts at \(\$0.10\) (given in the problem)
Fine = \(\$0.10\)

Day 2:
Now we need to compare our two options:
• Option A (add \(\$0.30\)): \(\$0.10 + \$0.30 = \$0.40\)
• Option B (double it): \(\$0.10 \times 2 = \$0.20\)

Since \(\$0.20\) is less than \(\$0.40\), we choose \(\$0.20\)
Day 2 fine = \(\$0.20\)

Day 3:
Starting from yesterday's \(\$0.20\), we compare:
• Option A (add \(\$0.30\)): \(\$0.20 + \$0.30 = \$0.50\)
• Option B (double it): \(\$0.20 \times 2 = \$0.40\)

Since \(\$0.40\) is less than \(\$0.50\), we choose \(\$0.40\)
Day 3 fine = \(\$0.40\)

3. Apply the comparison rule consistently

Day 4:
Starting from yesterday's \(\$0.40\), let's compare our options:
• Option A (add \(\$0.30\)): \(\$0.40 + \$0.30 = \$0.70\)
• Option B (double it): \(\$0.40 \times 2 = \$0.80\)

Since \(\$0.70\) is less than \(\$0.80\), we choose \(\$0.70\)
Day 4 fine = \(\$0.70\)

Notice the pattern: early on, doubling gives the smaller amount because the fine is small. But as the fine grows, adding \(\$0.30\) becomes the better (smaller) option.

4. Verify the final amount

Our calculation shows that on the fourth day, the total fine is \(\$0.70\).

Let's double-check our work:
Day 1: \(\$0.10\) ✓
Day 2: \(\min(\$0.10 + \$0.30, \$0.10 \times 2) = \min(\$0.40, \$0.20) = \$0.20\) ✓
Day 3: \(\min(\$0.20 + \$0.30, \$0.20 \times 2) = \min(\$0.50, \$0.40) = \$0.40\) ✓
Day 4: \(\min(\$0.40 + \$0.30, \$0.40 \times 2) = \min(\$0.70, \$0.80) = \$0.70\) ✓

Looking at our answer choices, \(\$0.70\) corresponds to choice B.

Final Answer: B. \(\$0.70\)

Common Faltering Points

Errors while devising the approach

1. Misunderstanding what "total fine" means: Students might think they need to add up all the daily fines from day 1 through day 4, rather than understanding that "total fine" refers to the accumulated fine amount on each specific day. The problem is asking for the fine amount that has built up by day 4, not a sum of separate daily charges.

2. Confusion about when the comparison rule applies: Students may incorrectly apply the "add \(\$0.30\) or double, whichever is lesser" rule to day 1, when this rule only applies starting from day 2. Day 1 has a fixed fine of \(\$0.10\) with no choice involved.

3. Misinterpreting "additional day": Students might get confused about which day is considered an "additional day." They need to understand that day 1 is the first overdue day (fixed at \(\$0.10\)), and the comparison rule begins on day 2, which is the first "additional" day.

Errors while executing the approach

1. Arithmetic errors in calculations: Students may make simple calculation mistakes when adding \(\$0.30\) or doubling amounts. For example, calculating \(\$0.40 + \$0.30 = \$0.60\) instead of \(\$0.70\), or \(\$0.40 \times 2 = \$0.70\) instead of \(\$0.80\).

2. Choosing the wrong option in comparisons: Students might consistently choose the larger amount instead of the smaller amount, misremembering that the library chooses "whichever results in the lesser amount." This could lead them to pick doubling over adding \(\$0.30\) on day 4.

3. Losing track of which day they're calculating: Students may skip a day or repeat a calculation, especially when working through the multi-step progression from day 1 to day 4. They might jump from day 2 directly to day 4, or calculate day 3 twice.

Errors while selecting the answer

1. Selecting an intermediate day's result: Students might correctly calculate all the daily amounts but then select the fine for day 2 (\(\$0.20\)) or day 3 (\(\$0.40\)) instead of day 4 (\(\$0.70\)), especially if they lose track of which day the question is asking about.