GRE Speed, Time & Distance: A Guide to Rates, Motion & QC Strategy

Rate and motion problems on the GRE Quant section test your conceptual grasp of proportionality, relative speed, and weighted averages. By mastering core frameworks and Quantitative Comparison shortcuts, you can bypass tedious calculations and solve rate questions with speed and confidence.

Section 1: The Core Framework & Proportionality

The TSD Triangle & Unit Conversions

Master the fundamental Distance = Speed × Time relationship, explore the 3 algebraic rearrangements, and practice instant unit conversions with our interactive tool.

The TSD Triangle & Unit Conversions

The Fundamental Motion Formula

All motion questions test the relationship between three variables: Distance, Speed (Rate), and Time.

Distance
$$ D = S \times T $$

Distance = Speed × Time

Speed
$$ S = \frac{D}{T} $$

Speed = Distance / Time

Time
$$ T = \frac{D}{S} $$

Time = Distance / Speed

Golden Rule of Units: Units across all 3 variables must always be compatible. If speed is in miles per hour (mph), time must be in hours and distance in miles. If speed is in meters/sec, time must be in seconds.
1Tab 1 of 3The Core Formulas

Speed-Time Proportionality & Multipliers

Master inverse variation (constant distance) and direct variation (constant time) with step-by-step interactive problem models.

Speed, Time & Proportionality

Constant Distance: Inverse Speed-Time Ratio

When distance is constant, Speed and Time are inversely proportional.

If speed is multiplied by k, time required becomes 1/k.

Step 1The Problem

Carl and Ruth

Carl averaged 2m mph on a trip that took him h hours.
If Ruth made the same trip in (2/3)h hours, what was her average speed in mph?

Step 2Method 1: The Algebraic Approach (Formula)

Step A: Find the total distance.

$$ \text{Distance } (D) = \text{Speed} \times \text{Time} = (2m) \times h = 2mh $$

Step B: Divide distance by Ruth's time.

$$ \text{Ruth's Speed} = \frac{D}{\text{Ruth's Time}} = \frac{2mh}{\frac{2}{3}h} = 2mh \times \frac{3}{2h} = \mathbf{3m \text{ mph}} $$

Notice that variable h cancels out cleanly.

Step 3Method 2: The Ratio Shortcut (Proportionality)

Principle: For constant distance, the speed ratio is the inverse of the time ratio:

$$ \frac{S_1}{S_2} = \frac{T_2}{T_1} $$

Step A: Compare their times.

Ruth's time is 2/3 of Carl's time (ratio of Carl's time to Ruth's time is 3 : 2).

Step B: Invert the ratio for speed.

Since speed is inversely proportional to time, Ruth's speed must be 3/2 of Carl's speed:

$$ \text{Ruth's Speed} = \frac{3}{2} \times (2m) = \mathbf{3m \text{ mph}} $$

Step 4Key Takeaway

Why use the Ratio Shortcut?

The algebraic method requires setting up and simplifying compound algebraic fractions. The ratio shortcut lets you solve it in 5 seconds mentally: Time scaled by 2/3 → Speed scaled by 3/2 → (3/2) × 2m = 3m.

1Tab 1 of 2Constant Distance (Inverse Ratio)

The Average Speed Trap

Learn why simple averages fail for round trips and how to apply the true Total Distance / Total Time framework seamlessly.

The Average Speed Trap

Car travels A to B at 40 km/hr.
Returns B to A at 60 km/hr.

What is the average speed?

1Tab 1 of 3The Problem

Section 2: Relative Motion & Multi-Object Scenarios

Relative Speed Concepts

Understand the physics of opposite-direction closing speeds vs. same-direction overtaking speeds with our dynamic simulator.

Opposite Direction (Add Speeds)

A (2 m/s) and B (3 m/s) are 100m apart.

🏃‍♂️ A
🏃‍➡️ B
1Tab 1 of 2Opposite Direction

Advanced Scenarios: Trains, Boats & Escalators

Step-by-step breakdowns for train platform crossings, upstream/downstream river currents, and escalator problems.

Step 1Scenario

100m train crosses 300m platform at 20 m/s.

Step 2Total Distance

Distance = Train Length + Platform Length
$$ D = 100 + 300 = 400 \text{ m} $$

Step 3Time

$$ T = \frac{400}{20} = 20 \text{ seconds} $$
1Tab 1 of 3Trains

Section 3: GRE Quantitative Comparison Strategy

GRE Strategy: Quantitative Comparison in Motion

Bypass tedious formulas on QC questions by using speed weighting logic and speed-distance ratios to compare quantities instantly.

GRE Quantitative Comparison: Motion & Speed Strategy

QC Strategy 1: Average Speed vs. Midpoint Value

On GRE Quantitative Comparison, you can determine whether average speed is greater or less than the midpoint without computing any exact numbers.

Step 1The QC Problem

A motorist travels from Town A to Town B at 40 mph and returns along the same route from Town B to Town A at 60 mph.

Quantity AAverage speed of the car for the entire round trip
Quantity B50 mph

Step 2The Slow Way: Numerical Calculation

Assume distance between Town A and B = 120 miles (LCM of 40 and 60):

$$ \text{Time}_1 = \frac{120}{40} = 3 \text{ hrs}, \quad \text{Time}_2 = \frac{120}{60} = 2 \text{ hrs} $$
$$ \text{Average Speed} = \frac{120 + 120}{3 + 2} = \frac{240}{5} = 48 \text{ mph} $$

Since 48 mph < 50 mph → Quantity B is greater (takes 45–60 seconds).

Step 3The GRE Shortcut: Time-Weighting Logic

Why is the average speed always pulled below 50 mph?

Average speed is a time-weighted average, not a simple average of numbers:

  • To cover the exact same distance, the car takes 3 hours (60% of total trip time) at 40 mph, and only 2 hours (40% of trip time) at 60 mph.
  • Because more than half the journey time is spent driving at the slower speed, the weighted average is pulled closer to 40 mph than to 60 mph.
Since >50% of trip time is spent at 40 mph → Average Speed < 50 mph → Quantity B is strictly greater!

Correct Answer: (B)

1Tab 1 of 2QC Strategy 1: The Average Speed Comparison
Topic Drill in Preparation

Topic Drill Coming Soon

A dedicated question set for Speed, Time & Distance is currently being authored. In the meantime, test your baseline score on the full Diagnostic: