Time & Work

Time-and-work (and pipes-and-cisterns) is a reliable CDS topic. The key is to think in RATES: if a job takes n days, the one-day work is 1/n, and rates simply add.

Core method

• If A finishes in a days, A's one-day work = 1/a. Together, A and B do 1/a + 1/b per day, so combined time = ab/(a + b).
• LCM method: let the total work = LCM of the given days; each worker's rate = total/own days, in whole-number units (fast and error-free).

Work-rate equation

• M1 x D1 x H1 / W1 = M2 x D2 x H2 / W2, linking men, days, hours and work.
• Efficiency is inversely proportional to time: if A is twice as efficient as B, A takes half the time.

Pipes and cisterns

• An inlet pipe filling in x hours has rate +1/x; an outlet (leak) emptying in y hours has rate -1/y. Net rate = sum.

Exam Tricks & Tips

• 🎯 Use the LCM-of-days method: set total work = LCM so every rate is a whole number.
• 🎯 Combined time for A and B together = ab/(a + b) (product over sum).
• 🎯 Efficiency and time are inversely proportional — double efficiency means half the time.
• 🎯 For pipes, treat inlets as positive and leaks/outlets as negative rates, then add.
• 🎯 Use M1D1H1/W1 = M2D2H2/W2 when men, hours or the amount of work changes together.
• ❌ Common mistake: adding the DAYS taken by two workers instead of adding their one-day RATES.

Expected exam pattern

Combined-work, efficiency-comparison, wages-share and pipes-and-cisterns questions; 6-10 marks, quick with the rate method.

Quick recap

Work in rates (1/time), add rates for teamwork, use total = LCM of days, combined time = ab/(a+b), and treat leaks as negative rates.

Cover the answer, recall, then check. 11 cards on time and work.

Q1. If A finishes a job in n days, A's one-day work is?
A1. 1/n.

Q2. Combined time for A (a days) and B (b days) together?
A2. ab/(a + b).

Q3. A does a job in 10 days, B in 15 days. Together?
A3. (10 x 15)/(10 + 15) = 150/25 = 6 days.

Q4. The men-days-hours-work relation?
A4. M1 D1 H1 / W1 = M2 D2 H2 / W2.

Q5. How are efficiency and time related?
A5. Inversely — higher efficiency means less time.

Q6. Rate of an inlet pipe filling in x hours?
A6. +1/x per hour.

Q7. Rate of a leak emptying in y hours?
A7. -1/y per hour.

Q8. What is the LCM method for time-and-work?
A8. Set total work = LCM of the given days so each rate is a whole number.

Q9. If A is twice as efficient as B and B takes 12 days, A takes?
A9. 6 days.

Q10. A pipe fills in 6 h, a leak empties in 12 h. Net time to fill?
A10. Net rate = 1/6 - 1/12 = 1/12, so 12 hours.

Q11. In what ratio are wages shared for a joint job?
A11. In the ratio of the work done (i.e. the ratio of the rates).

Time and Work — including pipes and cisterns — is a dependable CDS arithmetic topic solved almost entirely by the "work rate" idea. Once you think in terms of work done per day, the problems become straightforward.

Core idea / what this tests

This tests the relationship between the number of workers, the time taken, and the work completed, using the concept of work rate (fraction of work per unit time). Pipes-and-cisterns is the same idea with filling and emptying rates.

Deep explanation

Beginner — the unitary (per-day) method

If a person finishes a job in n days, their one-day work = 1/n of the job. When several people work together, add their rates: if A does 1/a per day and B does 1/b per day, together they do (1/a + 1/b) per day, so together they finish in ab/(a+b) days.

Intermediate — the LCM method

Rather than juggling fractions, set total work = LCM of the given days. Then each worker's daily output is a whole number. Example: A finishes in 12 days, B in 18 days → let total work = LCM(12, 18) = 36 units. A does 36/12 = 3 units/day, B does 36/18 = 2 units/day; together 5 units/day, so time = 36/5 = 7.2 days. This avoids fraction arithmetic and is far less error-prone.

Advanced — pipes, cisterns and efficiency

• Pipes and cisterns: a filling pipe has a positive rate, an emptying (leak) pipe a negative rate; combine by adding signed rates.
• Efficiency and wages: work done is shared in proportion to (rate × time worked); wages split in the ratio of work done.
• Men-days: total work = men × days, so M₁D₁ = M₂D₂ for the same job; extend to M₁D₁H₁/W₁ = M₂D₂H₂/W₂ when hours per day and amount of work vary.

Worked example

A can do a piece of work in 12 days and B in 18 days. Working together, how long will they take?

Step 1 — use the LCM method: total work = LCM(12, 18) = 36 units.
Step 2 — daily outputs: A = 36 ÷ 12 = 3 units/day, B = 36 ÷ 18 = 2 units/day.
Step 3 — combined rate = 3 + 2 = 5 units/day.
Step 4 — time = total work ÷ combined rate = 36 ÷ 5 = 7.2 days (7 days and 1/5 of a day).
Cross-check with the formula ab/(a+b) = (12 × 18)/(12 + 18) = 216/30 = 7.2 ✓.
Answer: 7.2 days. The LCM method kept the arithmetic in whole numbers.

CDS relevance

Time-and-work questions are frequent and quick with the LCM method, which sidesteps messy fractions. Pipes-and-cisterns and men-days variations recur in CDS, and treating a leak as a negative rate handles the trickier versions cleanly.

Exam tricks & shortcuts

• LCM method: set total work = LCM of the times, so rates are whole numbers.
• Two workers together: time = ab/(a+b).
• Leak/emptying pipe = negative rate; add signed rates.
• Men-days constant: M₁D₁ = M₂D₂ (same job); with hours/work, M₁D₁H₁/W₁ = M₂D₂H₂/W₂.
• Wages split in the ratio of work actually done.

Key formulas

• If A finishes in a days, one day's work = 1/a; n workers combine rates additively.
• Two together: time = ab/(a+b); three: 1/(1/a + 1/b + 1/c).
• Work = rate × time; total work often taken as LCM of the days.
• M₁D₁H₁/W₁ = M₂D₂H₂/W₂ (men × days × hours / work).
• Efficiency ∝ 1/time; if A is twice as efficient as B, A takes half the time.
• Wages divided in ratio of work done (i.e. ratio of efficiencies × days).

Usage: take total work as the LCM of the given days to avoid fractions.

Worked Example

Problem: A can complete a piece of work in 12 days and B can complete the same work in 18 days. If they work together, in how many days will the work be finished?

Solution:
Step 1 — Express each person's one-day work as a fraction of the whole:
A's one-day work = 1/12.
B's one-day work = 1/18.

Step 2 — Add to find their combined one-day work. Use a common denominator of 36:
1/12 + 1/18 = 3/36 + 2/36 = 5/36.
So together they finish 5/36 of the work in one day.

Step 3 — Time = total work / work per day = 1 ÷ (5/36) = 36/5 = 7.2 days.

Convert the fraction: 7.2 days = 7 days and 0.2 × 24 = 4.8 hours ≈ 7 days 4.8 hours.

Answer: Working together, they finish the work in 36/5 = 7.2 days (7 days 4.8 hours).