Distance problems are the most predictable marks in the quantitative section. The question families are few — a single traveller, two travellers in opposite directions, a chase, a train crossing a platform, a boat on a river — and each has one standard setup. Learn the five setups and a whole block of questions stops requiring thought.
They are also where unit mistakes are most expensive. An answer 18 times too big comes from mixing km/h with minutes, and it will always be one of the options.
One equation governs everything in this family:
Write down the three quantities you know, in consistent units, then solve for the fourth. Ninety percent of mistakes here come from skipping the unit step, so make it a habit to write the unit beside every number before computing anything.
The second idea is relative motion: when two objects move, only their relative speed matters, and it is the sum when they approach each other and the difference when one chases the other.
| Convert | Multiply by | Example |
|---|---|---|
| km/h to m/s | 5 ÷ 18 | 54 km/h = 15 m/s |
| m/s to km/h | 18 ÷ 5 | 20 m/s = 72 km/h |
| minutes to hours | ÷ 60 | 150 min = 2.5 h |
| km to m | × 1000 | 1.5 km = 1,500 m |
Convert first: 54 km/h = 54 × 5 ÷ 18 = 15 m/s. The distance covered from the moment the engine reaches the platform to the moment the last carriage leaves is 120 + 180 = 300 m. Time = 300 ÷ 15 = 20 seconds.
Approaching, so speeds add: 90 km/h. Time = 180 ÷ 90 = 2 hours, so they meet 80 km from the bus's town.
Downstream speed = 12 km/h and upstream = 6 km/h. Still water = (12 + 6) ÷ 2 = 9 km/h; current = (12 − 6) ÷ 2 = 3 km/h. Check: 9 + 3 = 12 downstream and 9 − 3 = 6 upstream.
The time difference is 15 minutes, or 0.25 hours. Writing d for the distance: d ÷ 4 − d ÷ 5 = 0.25, so d ÷ 20 = 0.25 and d = 5 km. Check: 5 km at 4 km/h takes 75 minutes and at 5 km/h takes 60 minutes, a difference of 15 minutes.
A car travels 150 km in 2 hours 30 minutes. What is its speed?
A thief runs at 10 km/h, and a policeman starts 1 km behind at 12 km/h. How long until he catches up?
A train 200 m long crosses a pole in 10 seconds. What is its speed in km/h?
A cyclist covers the first half of a journey at 10 km/h and the second half at 15 km/h. What is the average speed?
Without looking back: how do you convert km/h to m/s, and how do you combine the speeds of two objects moving towards each other?
The answer: multiply by 5 and divide by 18; and add the speeds, because the gap closes at the sum of the rates.
To convert km/h into m/s, multiply by and divide by 18. Two objects approaching each other close the gap at the of their speeds, while a chase closes at the . A train crossing a pole covers its own .
A question says: "A man walks at 5 km/h for 2 hours, then cycles at 15 km/h for 40 minutes. What is his average speed for the whole journey?" Solve it, and name the wrong answer that the options will contain.
Distance on foot = 5 × 2 = 10 km. Distance cycled = 15 × (40 ÷ 60) = 15 × 2/3 = 10 km. Total distance 20 km; total time 2 hours 40 minutes = 8/3 hours. Average speed = 20 ÷ (8/3) = 20 × 3 ÷ 8 = 7.5 km/h.
The expected distractor is 10 km/h, which is the plain mean of 5 and 15 — the trap for candidates who average the two speeds without weighting them by time. Because the two legs happen to cover equal distances here, the harmonic mean also gives 7.5, but the totals method is the one that works for every version of this question.
The equation distance = rate × time has a twin: work = rate × time. The next lesson applies exactly the same algebra to workers and pipes, where the trick is to convert every contributor into a fraction of the job per hour.