The RTD Framework
Strategy: Always build an RTD table. Label rows for each traveler/leg. Columns: Rate, Time, Distance. The relationship D = R × T always holds in every row.
Problem Types & RTD Tables
| Problem Type | Key Relationship | Example |
|---|---|---|
| Same direction, catch up | $R_1 T = R_2 T + \text{gap}$ | A starts 2 hr before B; when does B catch A? |
| Opposite directions, meet | $(R_1+R_2)T = D$ | Two trains approaching each other |
| Round trip | $D_1 = D_2$ (same distance) | Drive to city and back |
| Multiple legs | $D_{total} = D_1 + D_2 + ...$ | Trip with different speeds each segment |
Relative Speed
Average Speed — The Harmonic Mean Trap
10 Rate & Distance Traps
1. Average speed ≠ arithmetic mean
For round trips with equal distances: use $\frac{2r_1r_2}{r_1+r_2}$, not $\frac{r_1+r_2}{2}$.
2. Units mismatch
Rate in mph but time in minutes — convert to same unit before multiplying.
3. Head start in chase problems
If A starts 2 hours early at speed $r$, A's head start = $2r$ miles, not 2 hours.
4. Opposite direction distance
Two trains moving opposite: combined distance = $(R_1+R_2) \times T$. This is their COMBINED travel.
5. Round trip: distance is same, not time
Going 60 mph and returning 40 mph means different times but SAME distance each way.
6. Speed of river current vs boat
Upstream: effective speed = boat − current. Downstream: boat + current.
7. Overtaking: find when, not where
Catch-up problems ask "when" — set distances equal and solve for time.
8. Total distance vs each leg
The total distance in a multi-leg problem is the sum of all legs — don't count segments twice.
9. Rate table setup
Always clearly label which row is which traveler — mixed-up rows create wrong equations.
10. Speed and distance are proportional if time is constant
If two travelers travel for the same time, the faster one goes farther by a proportional amount.
10 GMAT Practice Questions
A car travels 240 miles at 60 mph. How many hours does the trip take?
(B) 4 hours. $T = D/R = 240/60 = 4$ hours.
Two trains start from cities A and B, which are 400 miles apart, traveling toward each other. Train 1 travels at 80 mph and Train 2 at 120 mph. How many hours until they meet?
(C) 2 hours. Combined rate = 80+120 = 200 mph. Time = 400/200 = 2 hours.
A cyclist rides to a destination at 12 mph and returns at 8 mph. What is the average speed for the round trip?
(A) 9.6 mph. Average speed (equal distances) = $\frac{2(12)(8)}{12+8} = \frac{192}{20} = 9.6$ mph.
Tom drove from City X to City Y in 3 hours at 60 mph. He returned at 45 mph. How many total hours did the round trip take?
(B) 7 hours. Distance X to Y = 60 × 3 = 180 miles. Return time = 180/45 = 4 hours. Total = 3+4 = 7 hours.
Runner A starts at noon and runs at 6 mph. Runner B starts at 1 PM and runs the same route at 9 mph. At what time does B catch A?
(B) 3:00 PM. At 1 PM, A has gone 6 miles. Relative speed = 9−6 = 3 mph. Time to close 6-mile gap = 6/3 = 2 hours after 1 PM = 3:00 PM.
A boat travels 30 miles upstream in 3 hours and the same 30 miles downstream in 2 hours. What is the speed of the current?
(B) 2.5 mph. Upstream speed = 10 mph; downstream = 15 mph. Boat speed = (10+15)/2 = 12.5 mph. Current = (15−10)/2 = 2.5 mph.
Is the distance from Town A to Town B greater than 200 miles?
(1) A car traveling at 50 mph takes more than 4 hours to travel from A to B.
(2) A car traveling at 60 mph takes less than 4 hours to travel from A to B.
(A) Statement (1) alone sufficient. (1): Distance > 50×4 = 200. Yes, distance > 200. Sufficient. (2): Distance < 60×4 = 240. But distance could be 201 or 150 — we can't determine if > 200. NOT sufficient. Answer: (A).
A plane flies 1,200 miles with a tailwind in 3 hours and returns against the same wind in 4 hours. What is the plane's speed in still air?
(B) 350 mph. With wind: $p+w = 400$. Against wind: $p-w = 300$. Adding: $2p = 700$ → $p = 350$ mph.
At 9 AM, Person X leaves home and drives at 40 mph. At 10 AM, Person Y leaves the same home and drives in the same direction at 60 mph. At what time will Y catch X?
(C) 1 PM. At 10 AM, X has 40 miles lead. Relative speed = 60−40 = 20 mph. Time = 40/20 = 2 hours after 10 AM = 12 PM. Wait — let's verify: at 12 PM, X has been driving 3 hours → 120 miles. Y has driven 2 hours → 120 miles. ✓ Answer is (B) 12 PM. (B) 12 PM is correct.
A car travels a total distance of 300 miles. The first 100 miles are at 50 mph, the next 100 at 60 mph, and the last 100 at 75 mph. What is the total time for the trip?
(B) 5 hours. $T_1 = 100/50 = 2$ hr. $T_2 = 100/60 = 5/3$ hr. $T_3 = 100/75 = 4/3$ hr. Total = $2 + 5/3 + 4/3 = 2 + 3 = 5$ hours.
Lesson Summary — Key Takeaways
D = R × T — the master equation
Build a table: one row per traveler/leg. Three columns: R, T, D. Every row satisfies D = R × T.
Opposite: add speeds. Same: subtract
Two objects approaching: add rates. One chasing: subtract to get the closing rate.
Average speed = Total D / Total T
Never add speeds and divide by 2 for average speed unless time on each leg is equal.
Head start in chase: convert to distance
If A starts 2 hours before B at rate r, the head start = 2r miles — then use relative speed to catch up.