🧮 Aptitude & Reasoning · Chapter 3

Time, speed and distance

Drive to Mysuru at 60 km/h and back at 40, and your average is 48, not 50 — because you spent longer at the slow speed, and time, not distance, is what averages are weighted by.

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The basics

Speed is distance covered per unit of time: distance = speed × time. Every question in this topic is that one equation, rearranged.

The problem it solves: working out when something arrives, how fast it must go, or where two moving things meet.

The mental model: fix whichever of the three quantities the question holds constant. Same distance? Speed and time move in opposite directions (go 25% faster, take 20% less time). Same time? Distance grows with speed. Two things moving? Treat one as standing still and give the other the relative speed — the speeds add when they approach each other and subtract when one chases the other.

Units are where most marks are lost. Trains are measured in metres and seconds, speeds in km/h. To convert, ×5/18 turns km/h into m/s (because 1 km/h = 1,000 m ÷ 3,600 s).

Explained

Bengaluru to Mysuru is about 120 km by road. You drive there at 60 km/h and come back at 40 km/h.

Average speed? Most people say 50. It is 48.

There: 2 hours. Back: 3 hours. 240 km in 5 hours = 48 km/h. You spent more time at the slow speed, so it drags the average down. That one idea — average over time, not over speeds — is the trap in half of this chapter.

Two trains at Itarsi junction

You're on a sleeper train waiting at Itarsi. On the next track, another train starts moving the same way and for a few seconds you're sure your train is moving.

Then it overtakes you so slowly you can read every coach number. Later a train going the opposite way flashes past in a blur, though it is no faster than the first.

Nothing changed but direction. Same direction, you see the difference of speeds. Opposite directions, you see the sum. That is relative speed, and it is why the examiner can make one train cross another in 10 seconds or in 90 seconds with the same two trains.

Average speed: method, shortcut, and when it breaks

Standard method (always works): total distance ÷ total time.

120 km at 60 km/h  → 2 h
120 km at 40 km/h  → 3 h
average = 240 / 5  = 48 km/h

Shortcut — equal distances only:

average = 2xy / (x + y) = 2 × 60 × 40 / 100 = 48

When the shortcut fails:

• *Equal times, not distances* — 2 h at 12 km/h and 2 h at 18 km/h averages (12 + 18)/2 = 15. Use the plain average
• Unequal times — 2 h at 12 and 1 h at 18 → (24 + 18)/3 = 14, neither 15 nor 14.4. Go back to totals
• Three legs of unequal length — only totals work

The shortcut is a special case of the method, so when in doubt, write the two totals.

Picture it: what 'crossing' means

A 240 m train crosses a 360 m platform:

start:  [=====train 240=====]|------platform 360------|
                             ^ engine reaches platform
end:                          |------platform 360------|[=====train 240=====]
                                                        ^ tail clears it
engine travels: 240 + 360 = 600 m

Two trains crossing (150 m and 100 m):

 →→ [====150====]          opposite: closing speed = 50 + 40 = 90 km/h
          [===100===] ←←   same way: closing speed = 50 − 40 = 10 km/h
 distance to clear = 150 + 100 = 250 m in both cases

The length you cover is always both lengths; only the speed you divide by changes.

Solving any speed question

Most wrong answers are right arithmetic on the wrong distance or the wrong unit.

The impossible hill

A cyclist climbs a 30 km hill road near Ooty at 30 km/h. She wants her average speed for the whole up-and-down trip to be 60 km/h.

Infosys round, 60 seconds on the clock

Two buses start towards each other on a 200 km stretch of highway. Bus A leaves at 7:00 at 50 km/h. Bus B leaves the other end at 8:00 at 25 km/h. When and where do they meet?

Step 1 — handle the head start first. By 8:00, A has covered 50 km. Gap left = 200 − 50 = 150 km.

Step 2 — closing speed (approaching, so add): 50 + 25 = 75 km/h.

Step 3 — time to close: 150 ÷ 75 = 2 hours → they meet at 10:00.

Step 4 — where: A has driven 3 hours × 50 = 150 km from its start. Check: B drove 2 × 25 = 50 km; 150 + 50 = 200 ✓.

The trap answer is 200 ÷ 75 = 2 h 40 min from 7:00, which ignores that B sat still for the first hour.

The traps examiners set

• Mixing units — a 240 m train at 54 km/h: divide 600 m by 15 m/s, not by 54
• Forgetting the platform or the other train — crossing a pole covers one length; anything with length adds its length
• Averaging the speeds — 60 and 40 over equal distances is 48, not 50
• Same-direction relative speed — subtract; a train at 50 overtaking one at 40 crosses at 10 km/h
• Head starts — remove the distance already covered before dividing by the closing speed
• Minutes in an hours equation — 10 minutes late is 1/6 hour, not 10
• Speed up x%, time down x% — no: +25% speed saves 20% of the time, because 1/1.25 = 0.8

Words to know

In two minutes

• Distance = speed × time; at a fixed distance, speed and time are inversely proportional.
• Convert km/h to m/s with ×5/18 and back with ×18/5 — trains in metres need it every time.
• Relative speed: add speeds for opposite directions, subtract for the same direction.
• A train crossing a pole covers its own length; crossing a platform or another train covers the sum of lengths.
• Average speed = total distance ÷ total time; for two equal distances it is 2xy/(x + y), always below the plain average.
• Speed up by x% and time falls by x/(100 + x)% — +25% speed saves 20% of the time.
• Boats: downstream = b + s, upstream = b − s; so b = (down + up)/2 and s = (down − up)/2.

Test yourself

There are 10 questions, 3 puzzles and 5 recall cards for this lesson in the app — the recall cards come back on later days, just before you would forget.