Time and Distance is among the most predictable scorers in AFCAT Numerical Ability. Almost every sum springs from a single relation, Distance = Speed × Time. Once you lock the unit conversion, average speed and relative speed ideas, you can clear two to three questions in barely a minute and bank easy marks under exam pressure.
The core relation you cannot forget
Every Time and Distance problem is built on one identity:
- Distance = Speed × Time
- Speed = Distance ÷ Time
- Time = Distance ÷ Speed
Think of it as a triangle: cover the quantity you want and the other two tell you whether to multiply or divide. Speed is how much distance is covered per unit time, so its units are always a distance over a time, such as km/h or m/s. In AFCAT, the examiner rarely asks the formula directly. Instead, two of the three quantities are buried inside a story about a train, a cyclist, a runner or two friends meeting, and your job is to read the words and pull out which quantity is unknown.
A useful habit is to write down what you know in a tiny table before touching the numbers. List the speed, the time and the distance for each stage of the journey, leave a blank for the unknown, and the equation almost writes itself. This discipline stops you from panicking when a question gives three or four numbers and expects you to combine them in a particular order.
The three formulas are the same equation rearranged. If you remember D = S × T, you never have to memorise the other two separately. Distance and speed are directly linked, distance and time are directly linked, but for a fixed distance speed and time pull against each other.
Unit conversion: km/h and m/s
AFCAT mixes units deliberately. A train's speed may be given in km/h while its length is in metres, so you must convert first.
- To go from km/h to m/s, multiply by 5/18.
- To go from m/s to km/h, multiply by 18/5.
The 5/18 factor comes from 1 km = 1000 m and 1 hour = 3600 s, so 1 km/h = 1000/3600 m/s = 5/18 m/s. Notice that 1000/3600 reduces neatly to 5/18 once you divide top and bottom by 200, which is why the fraction is so clean. Going the other way, since you are undoing the conversion, you simply turn the fraction upside down to 18/5.
Why does this matter so much? Because AFCAT loves to combine a speed in km/h with a length in metres and a time in seconds. If you forget to convert, every later step is wrong even though your method is correct. Make the conversion the very first thing you do, before any subtraction or division, and write the converted value clearly so you never re-use the old unit by accident.
Convert 72 km/h into m/s.
= (72 ÷ 18) × 5
= 4 × 5
= 20 m/s
Multiples of 18 km/h convert to clean m/s values: 18→5, 36→10, 54→15, 72→20, 90→25. Memorise this strip and skip the arithmetic.
Average speed done right
Average speed is total distance ÷ total time, never a simple average of two speeds. Mixing this up is the most common error in this topic.
For a journey covering equal distances at speeds x and y, the average speed is the harmonic mean:
- Average speed = 2xy ÷ (x + y)
For three equal stretches at x, y and z, it becomes 3xyz ÷ (xy + yz + zx). The deeper reason is that the slower leg of a journey eats up more time, so it drags the average down towards the smaller speed. That is why the harmonic mean always sits below the plain arithmetic mean of the two speeds, and the gap between them is exactly the trap the examiner builds the wrong option around.
A different situation appears when the two stretches are travelled for equal times rather than equal distances. There the plain arithmetic mean (x + y)/2 is correct, because each speed acts for the same duration. So the first thing to check is whether the question fixes equal distance or equal time, and only then pick the matching formula.
If a car goes 60 km/h one way and 40 km/h back, the average is not 50. It is 2×60×40/(60+40) = 4800/100 = 48 km/h, because more time is spent at the slower speed.
A man covers half his trip at 30 km/h and the other half at 20 km/h. Find his average speed.
= 2 × 30 × 20 / (30 + 20)
= 1200 / 50
= 24 km/h
Proportionality shortcuts
These three relationships solve a huge share of AFCAT sums without a single formula plug-in:
- When time is constant, distance is directly proportional to speed.
- When distance is constant, speed is inversely proportional to time.
- When speed is constant, distance is directly proportional to time.
The inverse-proportion idea is gold. If two bodies cover the same distance, the ratio of their speeds is the inverse of the ratio of their times.
Same distance ⇒ Speed₁ : Speed₂ = Time₂ : Time₁. The times flip. This single line cracks most ratio-based questions instantly.
Two cars cover the same route. Their speeds are in the ratio 4 : 5. If the faster car takes 4 hours, how long does the slower one take?
Faster car (speed 5) takes 4 h, that is the "4" part.
One part = 4 ÷ 4 = 1 h
Slower car time = 5 parts = 5 h
Relative speed: same and opposite directions
Relative speed is how fast one object approaches or pulls away from another.
- Opposite directions (towards or away from each other): add the speeds → relative speed = a + b.
- Same direction (one chasing the other): subtract → relative speed = a − b.
The time to meet or to catch up is the gap distance divided by the relative speed. Picture yourself sitting on one of the trains. From your seat the other train seems to move at the combined speed when it comes at you head-on, and at only the difference when it runs alongside you in the same direction. That mental picture is the whole idea of relative speed, and it removes the need to track two separate motions at once.
This single idea powers a wide family of AFCAT questions: two trains crossing each other, a thief running and a policeman chasing, two cyclists starting from opposite ends of a road, and boats moving with or against a current. In every case you replace two moving bodies with one effective speed and a fixed gap, which turns a messy scenario into a one-line division.
Opposite directions close the gap fast, so you add. Same direction closes it slowly, so you subtract. Speeds must be in the same unit before you combine them.
Two trains 300 km apart move towards each other at 40 km/h and 60 km/h. When do they meet?
Time = Distance ÷ Relative speed
= 300 ÷ 100
= 3 hours
Early and late arrival problems
A staple AFCAT type: a person travelling at one speed is late, at another speed is early, and you find the distance or the scheduled time.
Set the distance D as the common unknown. Convert "late by t₁" and "early by t₂" into a time difference and write D/S₁ − D/S₂ equal to that total difference (in hours).
Walking at 5 km/h a student reaches school 6 minutes late; at 6 km/h he reaches 2 minutes early. Find the distance.
D/5 − D/6 = 8/60
(6D − 5D)/30 = 8/60
D/30 = 8/60
D = 30 × 8/60 = 4 km
Always convert minutes to hours when speeds are in km/h. A stray unit mismatch is the single biggest score-killer in this topic.
Trains crossing objects
Train questions are Time and Distance with a length twist. The distance a train travels while crossing something equals its own length plus the length of that object.
- Crossing a pole or man (a point): distance = length of train.
- Crossing a platform or bridge: distance = length of train + length of platform.
Convert the speed to m/s with 5/18 because lengths are in metres. When two trains cross each other, treat the total distance as the sum of both train lengths and use relative speed: add the speeds if they move in opposite directions, subtract if they move in the same direction. The logic is identical to the earlier relative-speed section, only now the gap that must be covered is the two lengths added together.
A 150 m long train running at 54 km/h crosses a platform 100 m long. Find the time taken.
Total distance = 150 + 100 = 250 m
Time = 250 ÷ 15 = 16.67 s ≈ 16.7 s
A quick word on boats and streams
Boats and streams is just relative speed wearing a different costume, and AFCAT slips it in regularly. Let the boat's speed in still water be b and the current (stream) speed be s.
- Downstream (with the current): effective speed = b + s.
- Upstream (against the current): effective speed = b − s.
Two handy back-calculations follow at once. The boat's still-water speed is the average of the two, b = (downstream + upstream)/2, and the stream speed is half their difference, s = (downstream − upstream)/2. These two lines answer most boat questions in seconds.
A boat goes 30 km downstream in 2 hours and 30 km upstream in 3 hours. Find the speed of the boat in still water.
Upstream speed = 30 ÷ 3 = 10 km/h
Still-water speed = (15 + 10)/2 = 12.5 km/h
Stream speed = (15 − 10)/2 = 2.5 km/h
Downstream is always faster because the current helps you. If your downstream figure ever comes out smaller than upstream, you have swapped them.
Change-in-speed effect on time
When the same distance is covered at a different speed, use the inverse-proportion ratio to find the new time fast, instead of computing distance separately.
If speed becomes k times the original, time becomes 1/k times. A speed increased to 6/5 of the old value cuts time to 5/6 of the old time.
A man covers a distance in 6 hours. If he increases his speed by 1/3, how long will the same journey take?
Time becomes inverse = 3/4 of old
New time = 6 × 3/4 = 4.5 hours
Increase speed by a fraction ⇒ flip the fraction (as 1 + fraction) and invert to scale the time. No need to find the actual distance.
Mistakes that cost marks
- Averaging two speeds arithmetically instead of using 2xy/(x+y) for equal distances.
- Forgetting to convert km/h to m/s in train and length-based sums.
- Adding speeds for same-direction problems or subtracting for opposite-direction ones.
- Mixing minutes and hours in early/late arrival sums.
- Ignoring the object's length when a train crosses a platform.
Reading "crosses a pole" and adding a length anyway. A pole is a point, so the distance is only the train's own length.
Two-minute AFCAT strategy
Time and Distance rewards drilled reflexes. Build this habit:
- Underline the units the moment you read the question.
- Decide which of the three quantities is unknown and apply D = S × T.
- Whenever a distance repeats, reach for the inverse-proportion ratio before the formula.
- Convert with 5/18 or 18/5 only once, at the start.
Carry 5/18, 18/5 and the harmonic-mean formula 2xy/(x+y) on your mental cheat-sheet. These three cover the majority of AFCAT Time and Distance questions.
- D = S × T is the master formula.
- km/h to m/s: × 5/18; m/s to km/h: × 18/5.
- Average speed for equal distance = 2xy/(x+y), never (x+y)/2.
- Same distance ⇒ speed ratio is the inverse of time ratio.
- Opposite directions add speeds; same direction subtract.
- Train crossing a platform: distance = train length + platform length.
Practice the AFCAT pattern
Solve the question below the way you would in the hall, timing yourself.
Q. A man travels from A to B at 40 km/h and returns from B to A at 60 km/h. What is his average speed for the whole journey?
Answer: The distances are equal, so use the harmonic mean: 2 × 40 × 60 / (40 + 60) = 4800 / 100 = 48 km/h. The arithmetic mean of 50 km/h is the trap option.
Whenever a body goes and returns over the same route, equal-distance average speed (2xy/(x+y)) is almost always being tested. Spot it and answer in seconds.
Frequently asked questions
How many Time and Distance questions appear in AFCAT?
AFCAT Numerical Ability typically carries two to four questions that draw on Time and Distance ideas, including direct sums, average speed, relative speed and train-crossing problems. Because the topic is formula-light and predictable, it is one of the safest scoring areas in the section.
What is the fastest way to convert km/h to m/s?
Multiply the km/h value by 5/18. For example, 90 km/h becomes 90 × 5/18 = 25 m/s. To reverse it, multiply m/s by 18/5. Memorising the strip 18→5, 36→10, 54→15, 72→20, 90→25 lets you convert common speeds instantly.
Why is average speed not just the average of the two speeds?
Average speed is total distance divided by total time. When equal distances are covered at different speeds, more time is spent at the slower speed, so the true average is lower than the arithmetic mean. The correct formula for two equal stretches is 2xy/(x+y), the harmonic mean.
When do I add speeds and when do I subtract them?
Add the speeds when two objects move in opposite directions, since they close or open the gap quickly. Subtract them when they move in the same direction, as in a chase, because the gap changes slowly. Always bring both speeds to the same unit first.
What is the best shortcut for change-of-speed problems?
Use inverse proportionality. If speed becomes k times the original over the same distance, time becomes 1/k times. So a speed raised to 4/3 of its value reduces the journey time to 3/4, letting you scale the answer without recomputing the distance.
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