Time and Work is one of the most scoring and predictable arithmetic topics in the CDS / OTA written exam. Almost every paper carries two to four questions from it, and they reward a clear method rather than memory. In this Cavalier lesson you will learn the one-day work idea, efficiency ratios, and the pipes-and-cisterns twist — with solved examples throughout.
Why Time and Work matters in CDS
In every CDS Elementary Mathematics paper, the arithmetic block reliably throws up questions on Time and Work, often clubbed with its cousin, Pipes and Cisterns. These sums look wordy but rest on a single, repeatable idea, so once you train the method you can finish each in under a minute.
The examiner usually tests three things: combining the work rates of two or more people, adjusting when someone joins or leaves partway, and the inverse relationship between the number of workers and the time taken. Master these three and you cover the entire topic.
Treat the total job as 1 unit of work. Then a person who finishes the job in n days does 1/n of the work each day. Almost the whole chapter flows from this one line.
The core idea: one-day work
If a worker A completes a piece of work in 10 days, then in a single day A finishes 1/10 of it. This fraction is called A's one-day work or rate of work.
Work done in one day = 1 ÷ (days taken)
Days taken = 1 ÷ (work done in one day)
Work = Rate × Time
When two people work together, simply add their one-day works. If A does 1/10 per day and B does 1/15 per day, then together in one day they do:
1/10 + 1/15 = 3/30 + 2/30 = 5/30 = 1/6
So together they finish the whole job in 6 days — the reciprocal of their combined one-day work.
The two-worker shortcut
The add-the-fractions method always works, but a handy shortcut saves time when only two workers are involved.
If A alone takes x days and B alone takes y days, then working together they take:
Time = (x × y) ÷ (x + y) days
Using x = 10 and y = 15: (10 × 15) ÷ (10 + 15) = 150 ÷ 25 = 6 days, matching our earlier answer.
This product-over-sum shortcut is only for two workers. For three or more, fall back on adding one-day works — it never fails.
Efficiency and work ratios
Efficiency simply means how much work a person does per unit time. If A is twice as efficient as B, then A finishes the same work in half the time. Efficiency and time taken are inversely proportional.
Ratio of efficiencies = inverse of ratio of time taken.
If times are in the ratio a : b, efficiencies are in the ratio b : a.
Example: A finishes a job in 12 days and B in 18 days. Ratio of times = 12 : 18 = 2 : 3, so ratio of efficiencies = 3 : 2. A is the faster worker, contributing 3 parts of work for every 2 parts B contributes.
The LCM method avoids fractions: take the total work as the LCM of the days. If A takes 12 and B takes 18 days, let total work = LCM(12, 18) = 36 units. Then A does 36/12 = 3 units/day and B does 36/18 = 2 units/day. Whole numbers are far easier to juggle than fractions.
Men, days and the work constant
When a group of workers does a job, the product (men × days) measures the total work, assuming each worker is equally efficient. If the amount of work stays the same, this product stays constant.
M1 × D1 = M2 × D2
With hours and work added: (M1 × D1 × H1) ÷ W1 = (M2 × D2 × H2) ÷ W2
This single relation answers most "if more men are added" questions. More men means fewer days (inverse proportion); more work means more days (direct proportion).
Example: if 15 men build a wall in 20 days, how long will 25 men take for the same wall? Using M1D1 = M2D2: 15 × 20 = 25 × D2, so D2 = 300 ÷ 25 = 12 days.
Worked example: someone leaves midway
These mixed problems are the examiner's favourite. Read slowly, track the work done, and the rest is arithmetic.
A can do a piece of work in 20 days and B in 30 days. They begin together, but A leaves 5 days before the work is finished. In how many total days is the work completed?
So the work is completed in 15 days. A check: B works 15 days = 15/30 = 1/2; A works 10 days = 10/20 = 1/2; together = 1 whole job. Correct.
Pipes and cisterns — the same idea
Pipes and Cisterns is Time and Work in disguise. A pipe that fills a tank does positive work; a pipe that empties (a leak or outlet) does negative work. Add the rates with the correct signs.
Filling pipe (takes a hours): rate = +1/a per hour
Emptying pipe (takes b hours): rate = −1/b per hour
Net rate = sum of all rates; time to fill = 1 ÷ net rate
Example: pipe P fills a tank in 6 hours, pipe Q empties it in 8 hours. With both open, net rate = 1/6 − 1/8 = 4/24 − 3/24 = 1/24 per hour. The tank fills in 24 hours.
If the net rate turns out negative, the tank can never fill while both pipes run — a full tank would instead empty. Always check the sign before quoting a time.
Common mistakes to avoid
Adding the number of days instead of the rates. If A takes 10 days and B takes 15 days, the answer is never 25 days or 12.5 days — you must add 1/10 and 1/15, then take the reciprocal.
Two more traps that cost easy marks:
- Mixing up direct and inverse proportion. More men means fewer days, but more work means more days. Decide the direction before plugging into the formula.
- Ignoring working hours per day. If a question switches from 8 hours/day to 6 hours/day, include the hours in the M×D×H relation or your days will be wrong.
When the numbers are awkward fractions, switch to the LCM (unit-work) method. Turning the job into, say, 60 units of work removes denominators and slashes calculation errors under exam pressure.
Previous-year style question
This is the exact flavour of sum CDS sets year after year — three workers, a combined rate, and a missing person's time.
Q. A and B together can complete a work in 8 days. B and C together can do it in 12 days, and A and C together in 24 days. In how many days can A, B and C together finish the work?
Answer: Add the three pair-rates: (A+B) + (B+C) + (A+C) = 1/8 + 1/12 + 1/24 = 3/24 + 2/24 + 1/24 = 6/24 = 1/4. This sum equals 2(A+B+C), so A+B+C = 1/8 per day. Therefore A, B and C together finish the work in 8 days.
Quick revision
- One-day work of someone who takes n days = 1/n; add rates to combine workers.
- Two-worker shortcut: time = xy/(x+y) days.
- Efficiency is inverse of time; time ratio a:b gives efficiency ratio b:a.
- Work constant: M1D1H1/W1 = M2D2H2/W2.
- Pipes: filling is +, emptying is −; net rate gives the fill time.
- Wages split in the ratio of work actually done, not equally.
- Prefer the LCM/unit-work method to avoid fractions.
Practise five mixed sums a day from the CDS PYQ set and this topic becomes a guaranteed source of marks.
Frequently asked questions
How many Time and Work questions appear in the CDS Maths paper?
Typically two to four questions per paper, often including a Pipes and Cisterns variant. It is a high-yield, scoring topic, so it deserves focused practice.
Should I use the formula method or the LCM method?
Use the LCM (unit-work) method whenever the days lead to awkward fractions, since whole numbers reduce errors. The xy/(x+y) shortcut is fine for two clean numbers.
How do I handle a question where a worker joins or leaves midway?
Let the total time be T, write the work each person does as (their rate) times (days they actually worked), set the sum equal to 1, and solve for T.
Why is efficiency inversely proportional to time?
Because a more efficient worker does more work per day, so the same fixed job is finished in fewer days. Doubling efficiency halves the time taken.
How are wages divided in Time and Work problems?
In proportion to the work each person actually completes, which equals their one-day rate multiplied by the days they worked, not by an equal split.
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