Time, Speed and Distance

Why this topic is a scoring goldmine

Almost every CDS Elementary Mathematics paper carries 3 to 6 questions built directly on Time, Speed and Distance (TSD). Unlike geometry proofs, these sums follow predictable templates — once you recognise the type, the answer is a one or two-step calculation.

The chapter also feeds into other topics: ratio and proportion, percentage change in speed, and even simple equations. So the hours you invest here pay off across the paper.

What makes TSD especially friendly is that it rewards concept clarity over heavy calculation. Most sums need nothing more than the basic relation, a unit conversion, and one shortcut. Candidates who lose marks here usually do so through careless unit handling or by confusing average speed with the simple mean — both avoidable with disciplined practice. At The Cavalier we encourage students to first label the type of problem (train, boat, race, average, relative) before touching the numbers, because correct classification is half the solution.

The one relationship that runs the chapter

Everything begins with a single idea: distance equals speed multiplied by time. Rearranged, it gives all three forms you will ever need.

A handy memory device is the “DST triangle”: put D on top, S and T at the bottom. Cover the quantity you want, and the position of the other two tells you whether to multiply or divide.

Two quick proportionalities follow and are tested often:

• When time is constant, distance is directly proportional to speed.
• When distance is constant, speed is inversely proportional to time — double the speed, halve the time.

This inverse relationship is the secret behind many “late by” and “early by” questions. For example, if a person increases speed and reaches a fixed destination earlier, the distance never changes, so the ratio of speeds is simply the inverse ratio of times. Spotting that constant distance lets you set up a clean proportion instead of two separate equations, and it is exactly the kind of reasoning CDS examiners reward.

Unit conversion: km/h to m/s

CDS examiners love mixing units — the speed in km/h but the length of a train in metres and time in seconds. Get the conversion wrong and the whole sum collapses.

The factor comes from 1 km = 1000 m and 1 hour = 3600 s, so 1000÷3600 = 5÷18.

Average speed – the trap most candidates fall into

Average speed is total distance divided by total time — never the simple average of the two speeds.

The 2ab÷(a+b) shortcut applies only when the two parts of the journey cover the same distance — for example, going to a place at one speed and returning at another. If a journey is split into three equal parts at three different speeds, the same logic extends to a three-term harmonic mean, though CDS rarely goes beyond two segments.

To see why total-distance-over-total-time is the only safe definition, picture a 120 km trip done as 60 km at 30 km/h (2 hours) and 60 km at 60 km/h (1 hour). Total distance is 120 km, total time is 3 hours, so the average is 40 km/h. The simple mean of 30 and 60 would wrongly give 45 km/h. The gap exists because the slower leg eats up more time, dragging the true average down.

Relative speed: same or opposite direction

When two bodies move, we measure speed of one relative to the other.

This single idea powers all train-crossing and overtaking sums. If two objects move toward each other, the gap closes at the combined speed; if one chases another, the gap closes only at the difference.

A useful way to picture it: sit on one of the moving objects and treat yourself as stationary. From that frame, the other object appears to move at the relative speed. This “ride-along” trick turns a confusing two-body problem into a simple one-body problem and is especially handy in overtaking and circular-track sums.

Problems on trains

A train is not a point; its own length matters. The key is identifying the total distance the train must travel for the event described.

• Crossing a pole / standing man: distance = length of train.
• Crossing a platform / bridge / tunnel: distance = length of train + length of platform.
• Crossing another train: distance = sum of both lengths; use relative speed (sum if opposite, difference if same direction).

Boats and streams

Here the water current either helps or opposes the boat. Let the boat's speed in still water be b and the stream speed be s.

Downstream means moving with the current (faster); upstream means against it (slower). Many CDS questions give you the two times or two speeds and ask for b or s — the last two formulas crack those instantly.

A common variant gives the time taken to row a fixed distance downstream and upstream. Since the distance is the same, the ratio of upstream time to downstream time equals the inverse ratio of the two speeds. From that single ratio you can recover the boat and stream speeds without forming long equations. Watch for the word “still water”, which always refers to the boat's own speed b.

Races and circular tracks

In a 100 m race, “A beats B by 10 m” means when A finishes 100 m, B has run only 90 m. “A beats B by 4 seconds” means B takes 4 s more to finish.

On a circular track of length L, two runners starting together:

• Opposite directions: first meeting time = L ÷ (u + v).
• Same direction: first meeting time = L ÷ (u − v).

If the question asks when they next meet at the starting point, compute each runner's time for one full lap (L÷u and L÷v) and take the LCM of those two times. This neatly connects TSD with the HCF and LCM chapter, so a strong grip on both pays double dividends in the exam.

Worked example: average speed

Let us apply the equal-distance rule cleanly.

Notice the answer is 48 km/h, not the tempting 50 km/h. The harmonic mean always sits below the simple average because more time is spent at the slower speed. Examiners deliberately place the simple-average value (here 50) among the options to catch hasty candidates, so train yourself to pause whenever you see “to and fro” or “same road back” in a question.

Worked example: train crossing

A second illustration shows the train-plus-platform template in action.

Common mistakes to avoid

• Forgetting to convert units — mixing km/h with metres and seconds.
• Using (a+b)÷2 for average speed over equal distances.
• Adding speeds when objects move in the same direction (should subtract).
• Ignoring the train's own length when it crosses a platform or another train.
• Swapping downstream and upstream formulas in boat sums.

Previous-year style question

This question blends relative speed, unit conversion and the “sum of lengths” rule — a classic CDS combination. Practising mixed-concept sums like this is the fastest way to build exam confidence.

To extend your understanding, try the same numbers with the trains moving in the same direction: the relative speed becomes 60 − 48 = 12 km/h = 10÷3 m/s, and the crossing time jumps to 200 ÷ (10÷3) = 60 seconds. The dramatic change from under 7 seconds to a full minute shows just how much direction matters — a single word in the question can multiply the answer ninefold.

Quick revision

Drill 10–15 mixed sums daily for a week and TSD becomes one of your most reliable scoring zones in the CDS & OTA Maths paper.