Speed, Time and Distance Shortcuts | Banking Quant Mastery

Why This Chapter Matters

This chapter rewards clean unit conversion and calm setup. It covers straight-line motion, average speed, relative speed, trains, platform crossing, and time loss or gain from speed changes.

Core Ideas

Use $distance = speed \times time$ as the base relation.
Convert $m/s$ to $km/h$ by multiplying by $\frac{18}{5}$ .
When a train crosses a pole, only the train length matters. When it crosses a platform or another train, add lengths.
Relative speed handles chase and opposite-direction questions neatly.
For equal distances travelled at two different speeds, the average speed is the harmonic mean, not the simple mean.
If speed changes but distance remains fixed, time changes inversely.

High-Value Formulas

Concept	Formula / Rule
Core relation	$d = s t$
Unit conversion	$1 m/s = \frac{18}{5} km/h$
Time to cross pole	$time = \frac{train length}{speed}$
Average speed for equal distances	$\overset{s}{ˉ} = \frac{2 ab}{a + b}$
Relative speed after meeting	$\frac{speed of P}{speed of Q} = \frac{b}{a} when post-meeting times are a, b$

How To Approach Questions

Convert all speeds into one unit before computing.
Write the effective distance that must be covered.
Use relative speed for moving-object interactions.
For return journeys or equal-distance problems, choose the right average-speed formula instead of averaging directly.
If the question gives total journey time with different segment speeds, build the equation from the segment times.

Worked Examples

Example 1

Prompt: A train $180$ metres long crosses a pole in $9$ seconds. Find its speed.

Approach: Speed $= 180 \div 9 = 20 m/s = 72 km/h$ .

Example 2

Prompt: At $72 km/h$ , how much distance is covered in $25$ seconds?

Approach: Convert speed to $20 m/s$ . Distance $= 20 \times 25 = 500 m$ .

Example 3

Prompt: A traveller covers a distance at $40 km/h$ and returns over the same distance at $10 km/h$ . Find the average speed for the whole trip.

Approach: For equal distances, average speed $= \frac{2 ab}{a + b} = \frac{2 \times 40 \times 10}{40 + 10} = 16 km/h$ .

Example 4

Prompt: A car covers a distance in $10$ hours, moving at $40 km/h$ for the first half of the time and $20 km/h$ for the second half. Find the distance.

Approach: The first $5$ hours cover $200$ km and the next $5$ hours cover $100$ km, so the total distance is $300$ km.

Common Mistakes

Mixing metres with kilometres or hours with seconds.
Using train length alone when a platform is also being crossed.
Missing the sign change in relative speed questions.
Taking a simple average of speeds when the distances are equal.
Forgetting that speed gain and time saved are inverse effects on a fixed route.

Quick Revision

If the units are clean and the actual distance condition is identified correctly, speed-time-distance becomes a structured formula chapter rather than a guessing chapter.

8. Speed, Time and Distance Shortcuts