7.1 Uniform Motion: Rate, Time & Distance
Key Takeaways
- Uniform motion is D = R × T. Convert every quantity into matching units before multiplying or dividing; mixing mph with minutes, or km with meters, is the highest-frequency arithmetic miss on Focus Quant.
- Average speed is always total distance divided by total time. It equals the arithmetic mean of two speeds only when the traveler spends equal time at each speed — never merely because the two distances look similar.
- Equal-distance legs at speeds a and b (including a same-route round trip) have average speed 2ab/(a + b), the harmonic mean. That value is always strictly less than (a + b)/2 when a ≠ b, and it sits closer to the slower speed.
- Build a three-column D–R–T table for multi-leg trips. Fill two cells in each row, compute the third, then add the distance column or the time column — never average the rate column.
- A speed increase that saves a stated amount of time produces a rational equation in 1/v. Clear denominators, reject the negative root, and convert minutes to hours before you write the equation.
Distance, Rate, and Time
GMAT Focus Quantitative Reasoning is 21 Problem Solving questions in 45 minutes, with no calculator. Official Guide Quantitative Review 2026–2027 Math Review 3.3 (Rates, Ratios, and Percents) treats a rate as a ratio of two quantities. Uniform motion is the cleanest case of that idea: an object travels at a constant speed R for a time T and covers a distance D. The identity is
D = R × T, with rearrangements R = D / T and T = D / R.
One equation, three letters. Every "how far," "how fast," or "how long" stem on this exam is that identity plus unit discipline. You never need a calculator to multiply a rate by a time if the units already match. You do need a conversion whenever they do not.
A rate carries two units at once — miles per hour, meters per second, kilometers per minute. Distance must use the same length unit as the numerator of the rate. Time must use the same time unit as the denominator. If a train travels at 90 km/h for 40 minutes, writing 90 × 40 = 3,600 is not a computation error; it is a unit error. Forty minutes is 40/60 = 2/3 of an hour, so D = 90 × (2/3) = 60 km.
Memorize the minute-to-hour fractions that appear constantly on Quant. They replace decimal arithmetic you cannot afford to fumble without a calculator:
| Minutes | Hours |
|---|---|
| 10 | 1/6 |
| 12 | 1/5 |
| 15 | 1/4 |
| 20 | 1/3 |
| 30 | 1/2 |
| 40 | 2/3 |
| 45 | 3/4 |
| 50 | 5/6 |
Two less common conversions still show up when a stem mixes metric length with seconds, or customary length with seconds:
- 1 km/h = 1,000 m / 3,600 s = 5/18 m/s, so 1 m/s = 18/5 = 3.6 km/h
- 1 mph = 5,280 ft / 3,600 s = 22/15 ft/s
Worked example (unit trap). A cyclist rides at 24 km/h for 15 minutes, then at 18 km/h for 20 minutes. First-leg time is 1/4 hour, so D₁ = 24 × 1/4 = 6 km. Second-leg time is 1/3 hour, so D₂ = 18 × 1/3 = 6 km. Total distance is 12 km, not 24×15 + 18×20.
When a stem gives mixed units, convert everything to the unit the question asks for before you write D = R × T. If the answer is requested in minutes, you can keep rates in miles per minute (divide mph by 60) instead of converting the final hour-answer at the end. Either path is correct; mixing paths in the middle of the algebra is not.
Average Speed Is Total Distance over Total Time
Average speed for an entire trip is defined, not derived as a convenience:
V_avg = (total distance) / (total time) = (D₁ + D₂ + … + Dₙ) / (T₁ + T₂ + … + Tₙ)
It is not the arithmetic mean of the listed speeds. That trap is one of the most reliable GMAT Quant misses in this chapter. Traveling slower takes more time, so the slower speed weights the average more heavily. The arithmetic mean (R₁ + R₂)/2 is correct only when the traveler spends equal time at each speed.
Equal-distance legs: harmonic mean
Suppose the outward distance D is covered at speed a and the return distance D is covered at speed b.
- Outward time T₁ = D/a
- Return time T₂ = D/b
- Total distance 2D
- Total time D(1/a + 1/b) = D(a + b)/(ab)
- Average speed = 2D / [D(a + b)/(ab)] = 2ab / (a + b)
That expression is the harmonic mean of a and b. By the AM–HM inequality, 2ab/(a + b) < (a + b)/2 whenever a ≠ b. The harmonic mean also sits closer to the slower speed than to the faster one — a useful estimate check when you are eliminating choices.
Worked example. Out at 30 mph, back the same 60-mile road at 60 mph. Harmonic mean: 2(30)(60)/(30 + 60) = 3,600/90 = 40 mph. Arithmetic mean would have been 45 mph. Direct check: outward time 60/30 = 2 hours, return time 60/60 = 1 hour, total 120 miles in 3 hours, 120/3 = 40 mph. More time was spent at 30 mph, so 40 is closer to 30 than to 60.
For three equal-distance legs at speeds a, b, and c, the same derivation gives V_avg = 3abc / (ab + bc + ca). Speeds 20, 30, and 60 over equal distances average 3(20)(30)(60) / (600 + 1,800 + 1,200) = 108,000 / 3,600 = 30. That is again closer to the slowest listed speed than a casual (20+30+60)/3 = 36 would suggest.
Equal-time legs: arithmetic mean
If the traveler spends the same time t at speed a and at speed b, then D₁ = a t, D₂ = b t, and V_avg = (at + bt)/(2t) = (a + b)/2. Equal time is the one setting in which averaging the speeds is legal. Equal distance is not that setting. A stem that says "for the first two hours … then for the next two hours" is equal-time. A stem that says "to the office and back along the same route" is equal-distance unless the two legs have different lengths.
| Scenario | Condition | Average-speed formula | 30 mph and 60 mph |
|---|---|---|---|
| Equal distance (incl. same-route round trip) | D₁ = D₂ | 2ab/(a + b) | 40 mph |
| Equal time | T₁ = T₂ | (a + b)/2 | 45 mph |
| Three equal distances | D₁ = D₂ = D₃ | 3abc/(ab + bc + ca) | 20, 30, 60 → 30 mph |
| General multi-leg | arbitrary Dᵢ, Tᵢ | (Σ Dᵢ) / (Σ Tᵢ) | compute from a table |
Multi-Leg Tables and Round Trips
When a journey changes speed, pauses, or turns around, do not average the listed rates in your head. Draw a three-column table and force every row to satisfy D = R × T:
| Leg | Distance D | Rate R | Time T = D/R |
|---|---|---|---|
| 1 | D₁ | R₁ | T₁ |
| 2 | D₂ | R₂ | T₂ |
| Total | D₁ + D₂ | V_avg (computed last) | T₁ + T₂ |
Protocol:
- Identify the invariant. Is total distance fixed, total time fixed, or one quantity written in terms of another?
- Fill the known cells. Convert units first.
- For each row, compute the missing cell from D = R × T.
- Add a column — distances or times — and solve. The average-rate cell of the total row is never an input; it is an output from (total D)/(total T).
Worked example (round trip with a given total time). A motorist drives from City A to City B at 40 mph and returns the same route at 60 mph. Total driving time is 5 hours. Find the one-way distance.
Harmonic-mean path: V_avg = 2(40)(60)/(40 + 60) = 48 mph. Round-trip distance = 48 × 5 = 240 miles, so one-way D = 120 miles.
Table path: T₁ = D/40, T₂ = D/60, and D/40 + D/60 = 5. Multiply through by LCM(40, 60) = 120: 3D + 2D = 600, so 5D = 600 and D = 120 miles. Same answer, no formula name required. On a no-calculator exam the LCM clear is often faster than computing 48 first.
Worked example (unequal distances). A driver covers 120 miles at 40 mph, then 160 miles at 32 mph. Times: 120/40 = 3 hours and 160/32 = 5 hours. Total 280 miles in 8 hours, so V_avg = 280/8 = 35 mph. Arithmetic mean (40+32)/2 = 36 is a trap. A distance-weighted average of the speeds, (120·40 + 160·32)/280, is also wrong — that expression weights by miles rather than by hours, which is not the definition of average speed.
Speed Changes That Save Time
A large family of Quant stems fixes the distance and changes the speed by a constant, then tells you how much time is saved (or lost). Translate directly into a difference of two time fractions.
If distance D is covered at speed v, time is D/v. At speed v + k the time is D/(v + k). Saving h hours means
D/v − D/(v + k) = h.
Convert h from minutes to hours before this line. Clearing denominators produces a quadratic. Keep the positive root; discard the negative one because speed is positive.
Worked example. An express train covers 180 miles at v mph. Raising the speed by 15 mph shortens the trip by 1 hour. Then 180/v − 180/(v + 15) = 1. Multiply through by v(v + 15): 180(v + 15) − 180v = v(v + 15), so 2,700 = v² + 15v, or v² + 15v − 2,700 = 0. Factor as (v + 60)(v − 45) = 0. The usable root is v = 45 mph. Check: 180/45 = 4 hours; 180/60 = 3 hours; the difference is 1 hour.
If the saved time is 40 minutes, write 40/60 = 2/3 hour on the right-hand side. Leaving 40 on the right-hand side while speeds are in mph is the same unit mismatch as 90 km/h × 40 minutes.
Formula sheet for this section
| Task | Formula |
|---|---|
| Uniform motion | D = R × T |
| Average speed | (total distance) / (total time) |
| Equal-distance average of a, b | 2ab / (a + b) |
| Equal-time average of a, b | (a + b) / 2 |
| Minutes to hours | divide by 60 |
| km/h to m/s | multiply by 5/18 |
A cyclist rides 90 kilometers to a town at a constant 15 kilometers per hour and returns along the same road at a constant 45 kilometers per hour. What is the cyclist's average speed, in kilometers per hour, for the entire round trip?
A truck travels 200 miles at a constant 40 miles per hour and then travels 100 miles at a constant 60 miles per hour. What is the truck's average speed, in miles per hour, for the entire 300-mile trip?
A car covers a 240-mile route at a constant speed of v miles per hour. If the car instead traveled 20 miles per hour faster, the trip would take 1 hour less. What is v?