The toll for crossing a certain bridge is $0.75 each crossing. Drivers who frequently use the bridge may instead purchase...

1. Translate the problem requirements: We need to find the minimum number of days (x) where buying a monthly sticker (\(\$13\)) plus reduced tolls (\(\$0.30\) per crossing) costs less than paying regular tolls (\(\$0.75\) per crossing). The driver crosses twice per day on x days.
2. Set up the cost comparison: Express the total monthly cost for both options in terms of x days, accounting for 2 crossings per day.
3. Create and solve the inequality: Find when the sticker option becomes cheaper than the regular toll option.
4. Interpret the mathematical result: Since x must be a whole number of days, determine the smallest integer value that satisfies our condition.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what's happening here in everyday terms. A driver has two payment options for crossing a bridge:

Option 1 (Regular tolls): Pay \(\$0.75\) every single time you cross
Option 2 (Monthly sticker): Pay \(\$13.00\) upfront for the month, then only \(\$0.30\) per crossing

The driver will cross the bridge exactly twice per day on x days next month, and won't cross on any other days. So if x = 10 days, that means 20 total crossings for the month.

We need to find the smallest value of x where Option 2 (the sticker) saves money compared to Option 1 (regular tolls).

Process Skill: TRANSLATE - Converting the problem setup into clear mathematical relationships

2. Set up the cost comparison

Now let's express the total monthly cost for each option in terms of x (the number of days):

Option 1 - Regular tolls:
• 2 crossings per day × x days = 2x total crossings
• Cost per crossing = \(\$0.75\)
• Total monthly cost = \(\$0.75 \times 2\mathrm{x} = \$1.50\mathrm{x}\)

Option 2 - Monthly sticker:
• Monthly sticker cost = \(\$13.00\)
• 2 crossings per day × x days = 2x total crossings
• Reduced cost per crossing = \(\$0.30\)
• Total crossing costs = \(\$0.30 \times 2\mathrm{x} = \$0.60\mathrm{x}\)
• Total monthly cost = \(\$13.00 + \$0.60\mathrm{x}\)

So we're comparing:
• Regular option: \(\$1.50\mathrm{x}\)
• Sticker option: \(\$13.00 + \$0.60\mathrm{x}\)

3. Create and solve the inequality

For the sticker to save money, its total cost must be less than the regular toll cost:

Sticker cost < Regular cost
\(\$13.00 + \$0.60\mathrm{x} < \$1.50\mathrm{x}\)

Let's solve this step by step:
\(\$13.00 + \$0.60\mathrm{x} < \$1.50\mathrm{x}\)
\(\$13.00 < \$1.50\mathrm{x} - \$0.60\mathrm{x}\)
\(\$13.00 < \$0.90\mathrm{x}\)

Dividing both sides by \(\$0.90\):
\(\mathrm{x} > \$13.00 \div \$0.90\)
\(\mathrm{x} > 14.44...\)

Process Skill: MANIPULATE - Systematically isolating the variable to find the threshold

4. Interpret the mathematical result

Our calculation shows that \(\mathrm{x} > 14.44...\) But since x represents the number of days, it must be a whole number.

The smallest integer greater than 14.44 is 15.

Let's verify this makes sense:

If x = 14 days:
• Regular cost: \(\$1.50 \times 14 = \$21.00\)
• Sticker cost: \(\$13.00 + \$0.60 \times 14 = \$13.00 + \$8.40 = \$21.40\)
• Regular is cheaper (\(\$21.00 < \$21.40\)), so no savings with sticker

If x = 15 days:
• Regular cost: \(\$1.50 \times 15 = \$22.50\)
• Sticker cost: \(\$13.00 + \$0.60 \times 15 = \$13.00 + \$9.00 = \$22.00\)
• Sticker is cheaper (\(\$22.00 < \$22.50\)), so sticker saves \(\$0.50\)

Final Answer

The least value of x for which the driver can save money by using the sticker is 15 days.

The answer is B. 15

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the crossing pattern: Students may incorrectly interpret "twice on each of x days" as meaning x total crossings instead of 2x total crossings. This leads to setting up the cost equations incorrectly from the start.

2. Setting up the wrong inequality direction: Students may confuse which option should be cheaper and write the inequality as "Regular cost < Sticker cost" instead of "Sticker cost < Regular cost," leading them to find when the sticker is more expensive rather than when it saves money.

3. Forgetting the monthly sticker fee: Students may only compare the per-crossing costs (\(\$0.75\) vs \(\$0.30\)) without accounting for the upfront \(\$13.00\) monthly fee, leading to an incomplete cost analysis.

Errors while executing the approach

1. Arithmetic errors in division: When calculating \(\$13.00 \div \$0.90\), students may make computational mistakes and get an incorrect decimal result, such as 14.4 instead of 14.44..., which could lead to selecting the wrong answer.

2. Sign errors when manipulating the inequality: Students may make algebraic mistakes when moving terms across the inequality, such as incorrectly handling \(\$13.00 + \$0.60\mathrm{x} < \$1.50\mathrm{x}\), potentially changing signs incorrectly or making substitution errors.

Errors while selecting the answer

1. Selecting the boundary value instead of the next integer: Since \(\mathrm{x} > 14.44...\), students may incorrectly choose 14 (thinking it's "close enough") instead of recognizing that they need the smallest integer that satisfies the strict inequality, which is 15.

2. Not verifying the boundary cases: Students may arrive at x = 15 but fail to check that x = 14 doesn't work and x = 15 does work, potentially second-guessing their correct answer and switching to an incorrect choice.