Time-Speed-Distance & Time-Work

Snapshot

• Time, speed and distance and time and work are the same idea twice: a rate acting for some time produces an output, which is distance in one case and finished work in the other.
• One set of habits covers both: convert units first, think in rates, add rates for things that happen together, and use the LCM of the days to avoid fractions.
• This page explains why each rule works, gives a LaTeX formula sheet, shortcuts with the cases where they fail, 15 worked examples (trains, boats, pipes, wages, men-days), a traps table, 16 rapid-fire questions with answers and a one-week routine.
• Exam reality: +5 for a correct answer, −1 for a wrong one. These are short questions, and the marks go to whoever converts units and sets the rate correctly.

Part 1 — The core idea: rate times time equals output

Written and reviewed by Devendra Dubey, Mathematics & Commerce faculty at UniDrill (M.Sc. Mathematics, Osmania University · 10 years teaching). Prepared for the CUET UG General Test.

One equation, two costumes

Distance = speed × time is the same equation as Work = rate × time. A car with a speed of 60 km/h covers 60 km every hour. A worker who completes a job in 10 days does one-tenth of the job every day, so the rate is 1/10 of the job per day. When you see a "train" or a "tap" or a "worker", ask what the rate is, and how long it runs.

Why speed and time are inverse when distance is fixed

If the distance is fixed, going twice as fast takes half as long. In general, speed × time is constant. So a speed rise of 25% (a factor of 5/4) makes time 4/5 of the original, which is a 20% fall. The same logic gives "workers × days is constant" for a fixed job.

Why rates add

Two workers in the same room each do their own share per day, and the shares add. So A and B working together have a rate $\dfrac{1}{a} + \dfrac{1}{b}$ and take the reciprocal to get the time. Days do not add or average, because time is the inverse of rate. A leaking tank is just a worker with a negative rate.

Why relative speed adds or subtracts

If two trains approach each other, the gap between them closes at the sum of their speeds, because each one eats into the gap. If one chases the other in the same direction, the gap closes only by the difference. Imagine you are sitting in one train: you see the other approaching at the relative speed.

Why 5/18 appears

1 km is 1000 m and 1 hour is 3600 s, so 1 km/h equals $\dfrac{1000}{3600} = \dfrac{5}{18}$ m/s. Multiply by 5/18 to go from km/h to m/s. Multiply by 18/5 to go the other way.

Pause and predict: A can do a job in 6 days and B in 12 days. Working together, will they take more than 9 days, between 6 and 9, or less than 6?

Check your answer: less than 6. Together they must beat the faster worker alone, and adding any worker can only speed things up. The exact time is $\dfrac{6 \times 12}{6 + 12} = 4$ days.

Part 2 — Formula sheet

Speed, time and distance:

$$\text{Distance} = \text{Speed} \times \text{Time} \qquad 1 \text{ km/h} = \dfrac{5}{18}\ \text{m/s}$$

$$\text{Average speed} = \dfrac{\text{Total distance}}{\text{Total time}} \qquad \text{Equal distances at } a, b: \dfrac{2ab}{a + b}$$

Work:

$$\text{Rate of A} = \dfrac{1}{a} \text{ per day} \qquad \text{A and B together: } \dfrac{ab}{a + b} \text{ days}$$

$$\dfrac{M_1 D_1 H_1}{W_1} = \dfrac{M_2 D_2 H_2}{W_2}$$

Part 3 — Shortcut methods and when they fail

Unit habit. Before any calculation, write the unit next to each number. A length in metres divided by a speed in km/h is meaningless until one of them is converted.

Stages for incomplete work. When a worker joins or leaves partway, do not use the "together" formula for the whole job. Compute the work done in each stage as rate × time, add the pieces and keep going until the total reaches 1 (or the LCM units).

Part 4 — Worked examples: speed, distance and trains

Worked example 1: conversion. Convert 72 km/h to m/s and 15 m/s to km/h.

72 km/h $= 72 \times \dfrac{5}{18} = 20$ m/s. 15 m/s $= 15 \times \dfrac{18}{5} = 54$ km/h.

Worked example 2: late arrival. A man walking at 30 km/h instead of his usual 40 km/h reaches 30 minutes late. How far is his destination?

Common wrong step: writing the time difference as 30 without converting minutes to hours. A speed in km/h needs a time in hours.

Worked example 3: average speed. A car goes to a town at 40 km/h and returns by the same road at 60 km/h. Find the average speed.

$\dfrac{2 \times 40 \times 60}{40 + 60} = \dfrac{4800}{100} = 48$ km/h. The car spends longer at the lower speed, so the answer sits below 50.

Worked example 4: two trains, opposite directions. Two trains of lengths 120 m and 130 m run towards each other at 40 km/h and 50 km/h. How long do they take to cross?

Worked example 5: finding the length of a train. A train passes a pole in 10 seconds and a platform 250 m long in 20 seconds. Find its length and speed.

Let the length be L. The speed is $\dfrac{L}{10}$. For the platform, $\dfrac{L + 250}{20} = \dfrac{L}{10}$, so $L + 250 = 2L$ and $L = 250$ m. Speed $= \dfrac{250}{10} = 25$ m/s $= 90$ km/h.

Common wrong step: using 250 m as the distance for the pole. A pole has no length, so the train travels only its own length.

Worked example 6: overtaking. A 150 m train runs at 60 km/h. How long does it take to overtake a man walking at 6 km/h in the same direction?

Relative speed $= 60 - 6 = 54$ km/h $= 15$ m/s. Distance to clear is 150 m (the man has no length). Time $= \dfrac{150}{15} = 10$ s.

Worked example 7: meeting point. Two towns are 300 km apart. A car leaves the first at 40 km/h and another leaves the second at 60 km/h at the same time, towards each other. When and where do they meet?

Relative speed 100 km/h, so the time is $\dfrac{300}{100} = 3$ hours. The first car has travelled $40 \times 3 = 120$ km, so they meet 120 km from the first town (and 180 km from the second).

Part 5 — Worked examples: boats, work and pipes

Worked example 8: finding boat and stream speed. A boat goes 16 km downstream in 2 hours and returns 8 km upstream in 2 hours. Find the boat's speed in still water and the stream's speed.

Downstream speed $= \dfrac{16}{2} = 8$ km/h and upstream speed $= \dfrac{8}{2} = 4$ km/h. Boat $= \dfrac{8 + 4}{2} = 6$ km/h and stream $= \dfrac{8 - 4}{2} = 2$ km/h.

Worked example 9: round trip on a river. A boat's still-water speed is 12 km/h and the stream flows at 3 km/h. How long does it take to go 45 km downstream and come back?

Downstream: $\dfrac{45}{15} = 3$ h. Upstream: $\dfrac{45}{9} = 5$ h. Total = 8 hours.

Common wrong step: using the speed 12 km/h for both legs to get 7.5 h. The current helps in one direction and hurts in the other, and the loss is larger than the gain.

Worked example 10: two workers. A can do a job in 12 days and B in 18 days. How long do they take together?

$$\dfrac{12 \times 18}{12 + 18} = \dfrac{216}{30} = 7.2 \text{ days}$$

Check: the answer must be below 12 and above half of 12, which it is.

Worked example 11: LCM method. A does a job in 10 days and B in 15 days. How long together?

Take the job as LCM(10, 15) = 30 units. A does 3 units a day and B does 2, together 5 a day, so the time is $\dfrac{30}{5} = 6$ days.

Worked example 12: a worker leaves. A (10 days alone) and B (15 days alone) start a job together. A leaves after 4 days. How many days does the whole job take?

Worked example 13: pipes and a leak. (a) A pipe fills a tank in 6 hours but a leak empties it in 9 hours. How long to fill it with the leak? (b) Pipes A and B fill a tank in 10 and 15 hours, and outlet C empties it in 30 hours. How long with all three open?

(a) $\dfrac{6 \times 9}{9 - 6} = 18$ hours. (b) Take the tank as 30 units. A fills 3 an hour, B fills 2, C empties 1. Net $= 3 + 2 - 1 = 4$ a hour, so the time is $\dfrac{30}{4} = 7.5$ hours.

Worked example 14: wages. A and B can finish a job in 12 days and 18 days. They finish it together and are paid ₹1,500. How should it be shared?

Work per day is in the ratio $\dfrac{1}{12} : \dfrac{1}{18} = 3 : 2$. Shares are $\dfrac{3}{5} \times 1500 = 900$ and $\dfrac{2}{5} \times 1500 = 600$, that is ₹900 and ₹600.

Worked example 15: men, days and hours. 12 men build a wall in 20 days working 8 hours a day. How many days will 16 men take if they work 6 hours a day?

$$12 \times 20 \times 8 = 16 \times D \times 6 \Rightarrow D = \dfrac{1920}{96} = 20 \text{ days}$$

More men shorten the time, fewer hours lengthen it, and the two effects cancel exactly.

Part 6 — Common traps

Part 7 — Rapid-fire practice with answers

Part 8 — A timed-practice routine

In the exam. For a work question, write the job as an LCM of the days on the first line. For a speed question, write the units on the first line. If you cannot decide between direct and inverse, test with an extreme: doubling the workers should halve the days.

Part 9 — Where this connects

Inverse proportion and the percentage flip come from Percentage, Ratio and Alligation, and the equal-distance harmonic mean is explained in Averages and Ages. Questions that give speed against time as a graph are handled in Data Interpretation. The idea of a rate of change, which is what speed really is, becomes a derivative in Application of Derivatives. Distance and direction puzzles continue in Direction Sense.

One-line revision: distance = speed × time, convert with 5/18, average speed for equal distances is 2ab/(a + b), relative speed adds for opposite movement and subtracts for same-way movement, and for work add the rates (or use the LCM of the days).