Question

Train A left Centerville Station, heading toward Dale City Station, at 3: 00 p.m. Train B left Dale City Station, heading toward Centerville Station, at \(3: 20 \mathrm{p} . \mathrm{m}\). on the same day. The trains rode on straight tracks that were parallel to each other. If Train A traveled at a constant speed of 30 miles per hour and Train B traveled at a constant speed of 10 miles per hour, and the distance between the Centerville Station and Dale City Station is 90 miles, when did the trains pass each other? A \(4: 45 \mathrm{p.m}\) B \(5: 00 \mathrm{p.m}\) C \(5: 20 \mathrm{p.m}\) D \(5: 35 \mathrm{p.m}\) E \(6: 00 \mathrm{p.m}\)

Short answer

5:20 p.m.

Step-by-step solution

- Understand the problem

Train A leaves Centerville Station at 3:00 p.m. traveling towards Dale City Station at 30 mph. Train B leaves Dale City Station at 3:20 p.m. traveling towards Centerville Station at 10 mph. The distance between the two stations is 90 miles. We need to determine when the two trains pass each other.

- Establish the problem in terms of time

Let us denote the time when Train A and Train B pass each other as t hours after 3:00 p.m. Important to note that Train B starts 20 minutes (or 1/3 hour) later at 3:20 p.m.

- Create distance equations for both trains

Since distance = speed × time, for Train A traveling t hours (from 3:00 p.m. to the time they pass): Distance covered by Train A = 30t For Train B which travels (t - 1/3) hours (from 3:20 p.m. to the time they pass): Distance covered by Train B = 10(t - 1/3)

- Set up the equation

The total distance covered by both trains when they pass each other is the sum of their individual distances, which equals the total distance between the stations (90 miles): 30t + 10(t - 1/3) = 90

- Simplify and solve the equation

First expand the equation: 30t + 10t - 10/3 = 90 Combine like terms: 40t - 10/3 = 90 Add 10/3 to both sides to isolate the t term: 40t = 90 + 10/3 Convert 90 to thirds: 90 = 270/3, so 40t = 270/3 + 10/3 40t = 280/3 Solve for t: t = 280 / (3 × 40) = 280 / 120 = 7/3 t = 7/3 hours Convert 7/3 hours to hours and minutes: 7/3 = 2 hours and 20 minutes.

- Calculate the passing time

Since Train A started at 3:00 p.m., add 2 hours and 20 minutes to this time: 3:00 p.m. + 2 hours 20 minutes = 5:20 p.m.

Key concepts

Train Speed Problems

Train speed problems are common in standardized tests like the GMAT. They require you to understand how two moving objects interact, often traveling at different speeds and starting at different times. The key to solving these problems lies in setting up the scenario in terms of distance, speed, and time.For example, the given problem involves two trains starting from different stations at different times. We need to determine when they will pass each other. By focusing on elements such as the distance between the stations, the speed of each train, and the time each train starts, you can systematically break down the problem.Always remember that distance is the product of speed and time. This relationship forms the basis for setting up your equations, which are essential for finding the solution.Key things to keep in mind:

• Understand the scenario described in the problem.
• Identify all given quantities (speeds, start times, distances).
• Formulate equations accordingly.

Distance-Time Relationships

Distance-time relationships form the core of problems involving moving objects. In our problem, we have two trains traveling towards each other. Train A has a speed of 30 mph and starts at 3:00 p.m., while Train B travels at 10 mph and starts at 3:20 p.m.To solve such problems, you need to express distances in terms of time and speed. For Train A, the distance traveled can be written as \(30t\), where \(t\) is the time in hours after 3:00 p.m. For Train B, which starts 20 minutes later, the expression becomes \(10(t - 1/3)\), considering \(1/3\) hour is 20 minutes.Here are a few steps to follow:

• Identify the time gap between the start times of both trains.
• Express the distance each train travels in terms of their respective speeds and times.
• Combine these expressions to form an equation representing the total distance.

Equation Solving Techniques

Solving equations accurately is crucial for cracking GMAT math problems. Once you have set up your equations based on the given information, the next step is to solve them correctly.In our problem, the equation is \(30t + 10(t - 1/3) = 90\). Let’s break it down step by step:

• Expand the equation: \(30t + 10t - 10/3 = 90\).
• Combine like terms: \(40t - 10/3 = 90\).
• Add \(10/3\) to both sides to get \(40t = 90 + 10/3\).
• Convert 90 to thirds to properly add the fractions: \(90 = 270/3\), so \(40t = 270/3 + 10/3\).
• Simplify to get \(40t = 280/3\).
• Solve for \(t\): \(t = 280 / (3 \times 40) = 7/3\).
• Convert \(7/3\) hours to hours and minutes: \(7/3\) hours is \(2\) hours and \(20\) minutes.

Word Problem Strategies

Word problems can seem daunting, but strategic approaches can make them manageable. Here’s how to tackle them effectively:1. **Read Carefully:** Understand the scenario thoroughly. In train problems, pay attention to start times, speeds, and distances.2. **Identify Variables:** Assign variables to unknown quantities, such as the time \(t\) when the trains pass each other.3. **Set Up Equations:** Use formulas like speed = distance/time to create equations. For example, here we used \(30t\) for Train A and \(10(t - 1/3)\) for Train B.4. **Combine Information:** Integrate the equations to represent the total distance or other given quantities.5. **Solve and Verify:** Solve the equations step by step, ensuring no mistake in calculations. Finally, verify your answers to be sure they make sense in the context of the problem.By following these structured steps, you can break down complex word problems into simpler parts, making them easier to solve with confidence.