A tourist purchased a total of $1{,}500 worth of traveler's checks in $10 and $50 denominations, During the trip the...

1. Translate the problem requirements: We need to find the minimum value of lost checks. The tourist bought $1,500 in $10 and $50 checks, cashed exactly 7 checks total, and the number of $10 checks cashed differs by exactly 1 from the number of $50 checks cashed.
2. Identify possible cashing scenarios: Since 7 checks were cashed and the $10 and $50 check counts differ by 1, we need to find valid combinations that sum to 7 checks.
3. Calculate cash value for each scenario: For each valid combination of cashed checks, determine the total dollar amount that was cashed.
4. Determine minimum loss amount: The scenario that results in the highest cashed amount will give us the minimum lost amount (since lost = $1,500 - cashed).

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we know in plain English:

• The tourist started with $1,500 total in traveler's checks
• These checks came in only two types: $10 checks and $50 checks
• During the trip, exactly 7 checks were cashed
• The remaining checks were all lost
• The number of $10 checks cashed was either one more or one less than the number of $50 checks cashed

Our goal is to find the minimum possible value of the lost checks. This means we want to find the scenario where the tourist cashed the maximum amount of money, leaving the least amount to be lost.

Process Skill: TRANSLATE - Converting the problem language into clear mathematical understanding

2. Identify possible cashing scenarios

Since exactly 7 checks were cashed, and the number of $10 checks differs by exactly 1 from the number of $50 checks, let's think about this systematically.

If we call the number of $10 checks cashed 'x' and the number of $50 checks cashed 'y', then:

• \(\mathrm{x + y = 7}\) (total checks cashed)
• Either \(\mathrm{x = y + 1}\) OR \(\mathrm{x = y - 1}\) (the difference condition)

Let's work through both cases:

Case 1: \(\mathrm{x = y + 1}\) (one more $10 check than $50 checks)
Substituting into \(\mathrm{x + y = 7}\):
\(\mathrm{(y + 1) + y = 7}\)
\(\mathrm{2y + 1 = 7}\)
\(\mathrm{2y = 6}\)
\(\mathrm{y = 3}\)

So \(\mathrm{x = 4}\), meaning 4 ten-dollar checks and 3 fifty-dollar checks were cashed.

Case 2: \(\mathrm{x = y - 1}\) (one fewer $10 check than $50 checks)
Substituting into \(\mathrm{x + y = 7}\):
\(\mathrm{(y - 1) + y = 7}\)
\(\mathrm{2y - 1 = 7}\)
\(\mathrm{2y = 8}\)
\(\mathrm{y = 4}\)

So \(\mathrm{x = 3}\), meaning 3 ten-dollar checks and 4 fifty-dollar checks were cashed.

Process Skill: CONSIDER ALL CASES - Ensuring we examine both possible relationships between the check counts

3. Calculate cash value for each scenario

Now let's calculate how much money was cashed in each scenario:

Scenario 1: 4 ten-dollar checks + 3 fifty-dollar checks
Amount cashed = \(\mathrm{(4 \times \$10) + (3 \times \$50)}\)
Amount cashed = \(\mathrm{\$40 + \$150 = \$190}\)

Scenario 2: 3 ten-dollar checks + 4 fifty-dollar checks
Amount cashed = \(\mathrm{(3 \times \$10) + (4 \times \$50)}\)
Amount cashed = \(\mathrm{\$30 + \$200 = \$230}\)

We can see that Scenario 2 results in more money being cashed ($230 vs $190).

4. Determine minimum loss amount

Since we want the minimum amount lost, we need the scenario where the maximum amount was cashed.

From our calculations above, the maximum amount cashed was $230 (Scenario 2).

Therefore, the minimum amount lost = Total money - Maximum amount cashed
Minimum amount lost = \(\mathrm{\$1,500 - \$230 = \$1,270}\)

Final Answer

The minimum possible value of the checks that were lost is $1,270.

Looking at our answer choices, this matches option (D) $1,270.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the optimization goal: Students may confuse what they're trying to minimize. The question asks for the minimum possible value of checks lost, but students might incorrectly try to minimize the amount cashed instead of maximizing it. To minimize losses, you need to maximize the amount cashed first.

2. Missing the constraint interpretation: Students often overlook that "one more or one less" creates exactly two distinct scenarios that must both be analyzed. They might only consider one case (like only "one more") and miss the complete solution space.

3. Misreading the relationship constraint: The phrase "the number of $10 checks cashed was one more or one less than the number of $50 checks cashed" might be misinterpreted. Students could incorrectly think this means the values (dollar amounts) differ by one, rather than the count of checks differing by one.

Errors while executing the approach

1. Algebraic setup errors: When setting up equations for the two cases (\(\mathrm{x = y + 1}\) and \(\mathrm{x = y - 1}\)), students frequently make substitution mistakes. For example, writing \(\mathrm{(y - 1) + y = 7}\) as \(\mathrm{2y - 1 = 7}\) but then incorrectly solving to get \(\mathrm{y = 3}\) instead of \(\mathrm{y = 4}\).

2. Arithmetic calculation mistakes: Students often make computational errors when calculating the cash values, such as computing \(\mathrm{(3 \times \$10) + (4 \times \$50)}\) as \(\mathrm{\$30 + \$200 = \$220}\) instead of $230, or similar mistakes in the other scenario.

3. Forgetting to compare scenarios: After calculating both scenarios, students might stop at the first calculation and not compare the results to determine which gives the maximum amount cashed.

Errors while selecting the answer

1. Selecting the wrong optimization result: Students may correctly calculate both scenarios ($190 and $230 cashed) but then select the scenario that minimizes the amount cashed rather than maximizes it, leading them to choose \(\mathrm{\$1,500 - \$190 = \$1,310}\) instead of \(\mathrm{\$1,500 - \$230 = \$1,270}\).

2. Reporting amount cashed instead of amount lost: After finding that the maximum amount cashed is $230, students might mistakenly select this value or an answer choice close to it, forgetting that the question asks for the amount lost, not the amount cashed.