Two Trains, Different Speeds: Solve This Classic Rate Problem

Skills This Problem Builds

• Setting up time equations using the distance-rate-time relationship
• Working with rational expressions where the variable appears in denominators
• Converting time units consistently (minutes to hours) before solving
• Solving quadratic equations that arise from clearing fractions
• Interpreting solutions in context and rejecting physically impossible answers

Solution: The Time Difference Approach

This is a classic distance-rate-time problem where we're comparing two journeys. The key insight is that we know the distance is the same for both trains, but their speeds and travel times differ in a specific way.

Step 1 — Define the variables

Let v be the speed of the slower train in km/h. Then the faster train has speed v + 5 km/h, since it travels 5 km/h faster.

Step 2 — Express the travel times

Using the relationship time = distance ÷ speed:

Step 3 — Convert the time difference to hours

The problem states the faster train takes 30 minutes less time. Since our speed is in km/h, we need time in hours: 30 minutes = 0.5 hours.

Step 4 — Set up the time difference equation

The difference between the travel times equals 0.5 hours:

Step 5 — Clear the fractions

Multiply every term by v(v + 5) to eliminate the denominators:

Step 6 — Rearrange into standard quadratic form

Multiply by 2 to clear the decimal coefficients:

Step 7 — Solve using the quadratic formula

With a = 1, b = 5, and c = -3300:

This gives us v = 55 or v = -60. Since speed must be positive, v = 55 km/h.

Step 8 — Find both speeds

Solution: Method 2 — The Distance Comparison Method

Here's an alternative approach that focuses on how much farther the faster train would travel in the same amount of time.

Step 1 — Set up using the slower train's time

Let t be the time (in hours) that the slower train takes to complete the 330 km journey. Then the faster train takes t - 0.5 hours.

Step 2 — Express speeds in terms of time

Step 3 — Use the 5 km/h speed difference

The faster train is exactly 5 km/h faster:

Step 4 — Solve for t

Multiply by t(t - 0.5):

Dividing by 5: t² - 0.5t - 33 = 0

Using the quadratic formula: t = (0.5 ± √(0.25 + 132))/2 = (0.5 ± 11.5)/2

So t = 6 hours (taking the positive solution).

Step 5 — Calculate the speeds

Verification

Let's confirm our answer by checking both the speed difference and the time difference:

Speed difference: 60 - 55 = 5 km/h ✓

Travel times:

Time difference: 6 - 5.5 = 0.5 hours = 30 minutes ✓

Both conditions are satisfied, so our solution is correct.

Watch Out For These

The Math Beneath the Surface

This problem exemplifies a fundamental pattern in rate problems: when you have a fixed distance and two different rates, comparing their times leads to a rational equation that becomes quadratic when you clear denominators.

The general form for this type of problem is:

where D is the distance, v₁ and v₂ are the speeds, and Δt is the time difference.

If the speeds are related by v₂ = v₁ + k (where k is a constant difference), then substituting gives:

This will always produce a quadratic equation in v₁. One solution will typically be positive (the physical answer), while the other will be negative and must be discarded.

The key insight is recognizing that the "30 minutes less" translates to a subtraction of times, not speeds or distances. This temporal relationship is what creates the rational equation structure.

How to Spot This Problem Type

• "travels X faster" or "X km/h faster" — indicates a speed relationship
• "takes Y minutes/hours less" — signals a time difference comparison
• "same distance" or "both travel Z km" — means you can use distance-rate-time for both journeys
• "find the speeds" — confirms this is asking for the rates, not times or distances

The combination of equal distances, different speeds, and a time comparison is the signature of this problem type. You'll see similar setups with planes, cars, runners, or any moving objects where speed differences create time differences over the same route.

Real Applications

• Transportation logistics: Companies need to calculate delivery times when vehicles have different speeds due to load capacity, fuel efficiency, or route conditions.
• Race analysis: Understanding how pace differences translate to finish time gaps in marathons, cycling, or other endurance events.
• Network performance: Comparing data transfer rates between different connection types over the same file size to predict download time differences.

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