On a certain transatlantic crossing, 20% of a ship's passengers held round-trip tickets and also took their cars aboard the...

1. Translate the problem requirements: We need to clarify that "20% of all passengers have both round-trip tickets AND cars" and "60% of round-trip passengers do NOT have cars," then find what percent of all passengers have round-trip tickets.
2. Set up passenger group relationships: Define the total passenger population and establish the connection between round-trip ticket holders and those with cars.
3. Use the complement relationship: Since 60% of round-trip passengers don't have cars, then 40% of round-trip passengers do have cars, which connects to our known 20% figure.
4. Calculate using proportional reasoning: Use the fact that 40% of round-trip passengers equals 20% of total passengers to find the percentage of passengers with round-trip tickets.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what the problem is telling us in plain English:

• "20 percent of a ship's passengers held round-trip tickets and also took their cars aboard" means: 20% of ALL passengers have BOTH round-trip tickets AND cars
• "60 percent of the passengers with round-trip tickets did not take their cars" means: Among only those people who have round-trip tickets, 60% don't have cars
• We need to find: What percent of all passengers have round-trip tickets?

Think of it this way: We know a specific number about all passengers (20%), and we know a percentage breakdown within the round-trip group (60% without cars). We need to connect these two pieces of information.

Process Skill: TRANSLATE - Converting the problem's language into clear mathematical relationships

2. Set up passenger group relationships

Let's imagine we have 100 total passengers to make this concrete. We can define our groups:

• Total passengers = 100
• Passengers with round-trip tickets = Let's call this R (this is what we want to find)
• Passengers with both round-trip tickets AND cars = 20 (given as 20% of 100)

Now we can see the relationship clearly: The 20 passengers who have both round-trip tickets and cars are a subset of the R passengers who have round-trip tickets.

3. Use the complement relationship

Here's the key insight: If 60% of round-trip passengers DON'T have cars, then 40% of round-trip passengers DO have cars.

Let's think about this step by step:

• Among the R round-trip passengers, 60% don't have cars
• This means 40% of the R round-trip passengers DO have cars
• But we already know that exactly 20 passengers (from all 100) have both round-trip tickets and cars

So we can say: 40% of R passengers = 20 passengers

Process Skill: INFER - Recognizing that the complement of 60% gives us the connection we need

4. Calculate using proportional reasoning

Now we can solve this with simple proportion:

If 40% of round-trip passengers equals 20 passengers, then:

• 40% of R = 20
• \(0.4 \times \mathrm{R} = 20\)
• \(\mathrm{R} = 20 \div 0.4 = 50\)

So 50 passengers out of 100 total passengers have round-trip tickets.

Therefore: \(50/100 = 50\%\) of all passengers have round-trip tickets.

Let's verify this makes sense:

• Total passengers: 100
• Round-trip passengers: 50
• Round-trip passengers with cars: \(40\% \text{ of } 50 = 20\) ✓
• Round-trip passengers without cars: \(60\% \text{ of } 50 = 30\) ✓

Final Answer

The answer is 50%, which corresponds to choice C.

This solution demonstrates how we can use the complement relationship (40% have cars vs 60% don't have cars) to connect the given information about all passengers with the breakdown within the round-trip group.

Common Faltering Points

Errors while devising the approach

Faltering Point 1: Misinterpreting the reference group for percentages

Students often confuse which group a percentage refers to. The phrase "60 percent of the passengers with round-trip tickets did not take their cars" means 60% of ONLY the round-trip ticket holders, not 60% of ALL passengers. This is a critical distinction that affects the entire solution setup.

Faltering Point 2: Failing to recognize the subset relationship

Students may not realize that "passengers with round-trip tickets AND cars" is a subset of "passengers with round-trip tickets." They might treat these as separate, unrelated groups instead of understanding that the 20% who have both tickets and cars are part of the larger round-trip ticket group.

Faltering Point 3: Not using the complement relationship

When told that 60% of round-trip passengers don't have cars, students may struggle to recognize that this means 40% DO have cars. They might try to work directly with the 60% figure instead of using its complement, making the problem much more difficult to solve.

Errors while executing the approach

Faltering Point 1: Setting up the wrong equation

Even if students understand the relationships, they might set up incorrect equations like "60% of R = 20" instead of "40% of R = 20." This stems from not properly applying the complement relationship in their calculations.

Faltering Point 2: Arithmetic errors in percentage calculations

Students may make basic calculation errors when converting between percentages and decimals, or when dividing (20 ÷ 0.4 = 50). Some might incorrectly calculate this as 20 × 0.4 = 8 instead of dividing.

Errors while selecting the answer

Faltering Point 1: Providing the answer as a raw number instead of a percentage

After calculating that 50 passengers out of 100 have round-trip tickets, students might select "50" if it were an option, forgetting that the question asks for a percentage (50%).

Alternate Solutions

Smart Numbers Approach

We can solve this problem by choosing a convenient total number of passengers that makes our calculations clean and straightforward.

Step 1: Choose a smart number for total passengers
Let's say there are 100 passengers total. This is a logical choice because:

• Percentages are easy to work with when the total is 100
• 20% of 100 = 20 passengers (those with both round-trip tickets AND cars)
• This gives us concrete numbers to work with

Step 2: Identify what we know

• Total passengers = 100
• Passengers with both round-trip tickets AND cars = 20% of 100 = 20 passengers
• 60% of round-trip passengers do NOT have cars (so 40% of round-trip passengers DO have cars)

Step 3: Set up the relationship
Let R = number of passengers with round-trip tickets

• 40% of round-trip passengers have cars = 0.40 × R
• We know this equals 20 passengers (from Step 2)
• So: \(0.40 \times \mathrm{R} = 20\)

Step 4: Solve for round-trip passengers
\(0.40 \times \mathrm{R} = 20\)
\(\mathrm{R} = 20 \div 0.40 = 50\) passengers

Step 5: Calculate the percentage
Percent with round-trip tickets = (50 passengers ÷ 100 total passengers) × 100% = 50%

Verification:

• Round-trip passengers: 50
• Round-trip passengers with cars: \(40\% \text{ of } 50 = 20\) ✓
• Round-trip passengers without cars: \(60\% \text{ of } 50 = 30\)
• Total passengers with both round-trip tickets and cars: 20 out of 100 = 20% ✓