An automobile rental agency purchases new cars from the manufacturer in lots of 5 and releases the oldest cars to...

1. Translate the problem requirements: The agency buys cars in groups of 5, sells cars in groups of 3, and ended up with 14 more cars than they started with. We need to find the minimum number of buy/sell transactions.
2. Set up the net change equation: Express the net increase of 14 cars in terms of buying transactions (B) and selling transactions (S).
3. Identify the constraint for minimum transactions: Recognize that we want to minimize the total number of transactions (B + S) while satisfying the net change requirement.
4. Test systematic combinations: Try different values of buying and selling transactions, starting from scenarios that could yield small total transactions, until we find a valid solution.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what's happening at this car rental agency in simple terms:

• The agency buys cars from the manufacturer, but they must buy exactly 5 cars at a time (this counts as 1 transaction)
• The agency sells old cars to an auctioneer, but they must sell exactly 3 cars at a time (this also counts as 1 transaction)
• At the end of last year, they had 14 more cars than they started with
• We need to find the minimum total number of buying and selling transactions

Think of it like this: every time they buy, they gain 5 cars. Every time they sell, they lose 3 cars. The net result was gaining 14 cars.

Process Skill: TRANSLATE - Converting the business scenario into a mathematical relationship

2. Set up the net change equation

Now let's express this mathematically. If we call:

• B = number of buying transactions (each adds 5 cars)
• S = number of selling transactions (each removes 3 cars)

Then the net change in cars is:
Cars gained from buying - Cars lost from selling = Net increase
\(5\mathrm{B} - 3\mathrm{S} = 14\)

This equation tells us exactly how the transactions relate to the final result. We need to find values of B and S that satisfy this equation while minimizing the total transactions \((\mathrm{B} + \mathrm{S})\).

3. Identify the constraint for minimum transactions

Our goal is to minimize the total number of transactions: \(\mathrm{B} + \mathrm{S}\)

Here's the key insight: we want the smallest possible value of \((\mathrm{B} + \mathrm{S})\) such that \(5\mathrm{B} - 3\mathrm{S} = 14\).

Since we're looking for the minimum, we should start testing small values of total transactions and see which combinations work.

Process Skill: INFER - Recognizing that we need to systematically test small values first

4. Test systematic combinations

Let's test different scenarios starting with small numbers of total transactions:

If total transactions = 3: We could have (B=3, S=0), (B=2, S=1), (B=1, S=2), or (B=0, S=3)

• B=3, S=0: \(5(3) - 3(0) = 15 ≠ 14\)
• B=2, S=1: \(5(2) - 3(1) = 10 - 3 = 7 ≠ 14\)
• B=1, S=2: \(5(1) - 3(2) = 5 - 6 = -1 ≠ 14\)
• B=0, S=3: \(5(0) - 3(3) = 0 - 9 = -9 ≠ 14\)

None work.

If total transactions = 4: We could have (B=4, S=0), (B=3, S=1), (B=2, S=2), (B=1, S=3), or (B=0, S=4)

• B=4, S=0: \(5(4) - 3(0) = 20 ≠ 14\)
• B=3, S=1: \(5(3) - 3(1) = 15 - 3 = 12 ≠ 14\)
• B=2, S=2: \(5(2) - 3(2) = 10 - 6 = 4 ≠ 14\)
• B=1, S=3: \(5(1) - 3(3) = 5 - 9 = -4 ≠ 14\)
• B=0, S=4: \(5(0) - 3(4) = 0 - 12 = -12 ≠ 14\)

None work.

If total transactions = 5: Testing (B=5, S=0), (B=4, S=1), (B=3, S=2), (B=2, S=3), (B=1, S=4), (B=0, S=5)

• B=5, S=0: \(5(5) - 3(0) = 25 ≠ 14\)
• B=4, S=1: \(5(4) - 3(1) = 20 - 3 = 17 ≠ 14\)
• B=3, S=2: \(5(3) - 3(2) = 15 - 6 = 9 ≠ 14\)
• B=2, S=3: \(5(2) - 3(3) = 10 - 9 = 1 ≠ 14\)
• Other combinations give negative results

None work.

If total transactions = 6: Testing key combinations:

• B=5, S=1: \(5(5) - 3(1) = 25 - 3 = 22 ≠ 14\)
• B=4, S=2: \(5(4) - 3(2) = 20 - 6 = 14\) ✓

Success! With B=4 and S=2, we get exactly 14 more cars, and the total transactions = 4 + 2 = 6.

4. Final Answer

The minimum number of transactions is 6, achieved with 4 buying transactions (adding 20 cars) and 2 selling transactions (removing 6 cars), for a net increase of 14 cars.

The answer is D: Six.

Common Faltering Points

Errors while devising the approach

Faltering Point 1: Misinterpreting what constitutes a transaction
Students may think that buying 5 cars counts as 5 transactions instead of 1 transaction, or selling 3 cars counts as 3 transactions instead of 1 transaction. The problem clearly states "a purchase of a lot and a release of a lot each count as one transaction," but students might focus on individual cars rather than lots.

Faltering Point 2: Setting up the equation incorrectly
Students may confuse the direction of the net change and write \(3\mathrm{S} - 5\mathrm{B} = 14\) instead of \(5\mathrm{B} - 3\mathrm{S} = 14\). They might think that since cars are being "released" (sold), this represents an increase rather than understanding that releasing cars reduces the agency's inventory.

Faltering Point 3: Misunderstanding the optimization objective
Students may try to minimize only buying transactions or only selling transactions, rather than understanding that they need to minimize the total number of transactions \((\mathrm{B} + \mathrm{S})\). This leads them to look for solutions that minimize just one variable instead of the sum.

Errors while executing the approach

Faltering Point 1: Incomplete systematic testing
Students may test only a few combinations within each total transaction count rather than exhaustively checking all possible combinations. For example, when testing 6 total transactions, they might only check (B=6, S=0) and (B=0, S=6) and miss the correct combination (B=4, S=2).

Faltering Point 2: Arithmetic errors in calculations
Students may make calculation mistakes when computing \(5\mathrm{B} - 3\mathrm{S}\) for different values. Common errors include \(5(4) - 3(2) = 20 - 6 = 15\) instead of 14, or similar computational mistakes that lead them to reject the correct answer or accept an incorrect one.

Faltering Point 3: Stopping too early in the search
Students may find a solution that works (like B=4, S=2 giving 6 transactions) but continue searching unnecessarily, or conversely, they may give up too early and conclude no solution exists when they don't find one in their first few attempts.

Errors while selecting the answer

Faltering Point 1: Confusing the number of transactions with the number of cars
Students may calculate correctly that B=4 and S=2, but then mistakenly report 14 (the net increase in cars) or 20-6 (gross cars bought minus sold) as their final answer instead of 6 (the total number of transactions).

Faltering Point 2: Selecting the first working solution without verifying it's minimal
Students may find a combination that satisfies \(5\mathrm{B} - 3\mathrm{S} = 14\) (such as B=8, S=7 which gives \(40-21=19≠14\), or a correct but non-minimal solution) and immediately select that answer without confirming they've found the minimum number of total transactions.