What is the LCM method in Time and Work?

Assume total work = LCM of the given days. Convert each person's rate to work units per day. This avoids fractions and speeds up calculation greatly.

How are wages divided in Time and Work problems?

Wages are divided in the ratio of work done, which equals the ratio of daily efficiencies multiplied by the number of days each person worked.

What is the MDH formula?

M1 × D1 × H1 = M2 × D2 × H2, where M = men, D = days, H = hours per day. It assumes all men have equal efficiency.

How do pipes relate to Time and Work?

Filling pipes are treated as positive workers and emptying pipes as negative workers. The net rate = sum of filling rates minus sum of emptying rates.

How do you solve alternating worker problems?

Find work done in one complete cycle (e.g., 2 days). Divide total work by work per cycle to find how many complete cycles occur, then handle the remainder separately.

Time and Work — Formulas, Shortcuts, and Worked Examples | TNPSC, SSC, Bank

Quick Concept Summary

Time and Work is one of the highest-frequency topics in TNPSC Group exams. Questions test your ability to calculate how long it takes a person (or a group of people) to finish a job, how workloads change when workers join or leave mid-task, and how wages are split fairly.

The core idea is simple: work rate × time = work done. If a job takes n days to finish, then in one day, exactly 1/n of the job gets done.

Key mental model: Think of the entire job as "1 unit" of work. Every worker contributes a fraction of that unit per day. When workers collaborate, their fractions add up.

Three approaches: 1. Fraction method — express each rate as a fraction, add fractions, solve. 2. LCM method — assign total work = LCM of days, work in whole numbers (fastest for TNPSC). 3. Efficiency method — compare workers by their speed relative to each other.

Key Formulas

Situation	Formula
One person finishes in n days	Daily work = 1/n
A and B together finish in T days	1/T = 1/A + 1/B
Time for A and B together	T = (A × B) / (A + B)
If A is x times as efficient as B	A takes work in 1/x of B's time
Work left after k days (A alone)	Remaining = 1 − k/A
MDH formula (same efficiency)	M₁D₁H₁ = M₂D₂H₂
Wages ratio	Wages ∝ Work done = Rate × Days worked
LCM method total work	Total work = LCM(t₁, t₂, t₃, …)

Worked Examples

Example 1 — Basic Two-Person Problem

Problem: A can finish a piece of work in 18 days and B can finish it in 12 days. In how many days will they complete the work together?

Step 1 — Find combined daily rate.
  A's daily rate = 1/18
  B's daily rate = 1/12

Step 2 — Add the rates.
  Combined rate = 1/18 + 1/12
               = 2/36 + 3/36
               = 5/36

Step 3 — Find time.
  Time = 1 / (5/36) = 36/5 = 7.2 days = 7 days 4 hours 48 minutes

Answer: A and B together finish the work in 36/5 days (≈ 7.2 days).

Example 2 — LCM Method (Three Workers)

Problem: A, B, and C can complete a work in 12, 15, and 20 days respectively. How many days will they take if all three work together?

Step 1 — Find LCM(12, 15, 20).
  12 = 4 × 3; 15 = 5 × 3; 20 = 4 × 5
  LCM = 60

Step 2 — Assign total work = 60 units.
  A's rate = 60/12 = 5 units/day
  B's rate = 60/15 = 4 units/day
  C's rate = 60/20 = 3 units/day

Step 3 — Combined rate = 5 + 4 + 3 = 12 units/day

Step 4 — Time = 60/12 = 5 days

Answer: All three together finish in 5 days.

Example 3 — One Person Leaves Mid-Task

Problem: A and B together can do a work in 10 days. A alone can do it in 15 days. They start working together, but B leaves after 6 days. How many more days does A need to finish the remaining work?

Step 1 — Find B's time alone.
  1/B = 1/10 − 1/15 = 3/30 − 2/30 = 1/30
  B alone takes 30 days.

Step 2 — Work done by A and B together in 6 days.
  Combined rate = 1/10 per day
  Work done = 6 × (1/10) = 6/10 = 3/5

Step 3 — Remaining work.
  Remaining = 1 − 3/5 = 2/5

Step 4 — Days A needs to finish remaining work alone.
  A's rate = 1/15 per day
  Days = (2/5) ÷ (1/15) = (2/5) × 15 = 6 days

Answer: A needs 6 more days to complete the remaining work.

Example 4 — Alternating Workers

Problem: A can do a work in 10 days and B can do it in 15 days. They work on alternate days, with A starting on day 1. In how many days will the work be finished?

Step 1 — Use LCM method. LCM(10, 15) = 30.
  Total work = 30 units
  A's rate = 30/10 = 3 units/day
  B's rate = 30/15 = 2 units/day

Step 2 — Work in a 2-day cycle (A one day, B next day).
  Work per cycle = 3 + 2 = 5 units

Step 3 — Number of complete cycles.
  30 / 5 = 6 complete cycles → 6 × 2 = 12 days.
  Work after 12 days = 6 × 5 = 30 units = all done.

Answer: The work is finished in exactly 12 days.

Note: If it had not divided evenly, you would check whose turn falls on day 13
and calculate how much of that day is needed to finish the remainder.

Example 5 — Work and Wages

Problem: A, B, and C can complete a job in 6, 8, and 12 days respectively. They all work together to finish the job. If the total payment is ₹1,620, how much does each person earn?

Step 1 — Find each person's daily rate.
  A = 1/6, B = 1/8, C = 1/12

Step 2 — Ratio of work done = ratio of rates (since all work same number of days).
  A : B : C = 1/6 : 1/8 : 1/12

Step 3 — Multiply by LCM(6, 8, 12) = 24 to clear fractions.
  A : B : C = 24/6 : 24/8 : 24/12 = 4 : 3 : 2

Step 4 — Divide wages in ratio 4:3:2. Total parts = 4+3+2 = 9.
  A's share = (4/9) × 1620 = ₹720
  B's share = (3/9) × 1620 = ₹540
  C's share = (2/9) × 1620 = ₹360

Answer: A gets ₹720, B gets ₹540, C gets ₹360.
Verification: 720 + 540 + 360 = ₹1,620 ✓

Example 6 — MDH Formula

Problem: 12 men can complete a work in 15 days working 8 hours a day. How many men are needed to complete the same work in 10 days working 9 hours a day?

Step 1 — Apply MDH formula: M1 × D1 × H1 = M2 × D2 × H2

  M1 = 12, D1 = 15, H1 = 8
  D2 = 10, H2 = 9, M2 = ?

Step 2 — Substitute.
  12 × 15 × 8 = M2 × 10 × 9
  1440 = 90 × M2
  M2 = 1440 / 90 = 16

Answer: 16 men are required.

Shortcuts and Tricks

Trick 1 — Direct Formula for Two Workers

When A takes 'a' days and B takes 'b' days:

Time together = ab / (a + b)

Example: A = 12 days, B = 18 days → Together = (12 × 18)/(12 + 18) = 216/30 = 7.2 days

Trick 2 — "One More Worker" Shortcut

If A and B together take T days, and A alone takes 'a' days:

B alone = aT / (a − T)

Trick 3 — Efficiency Ratio

If A is twice as fast as B, A takes half the time. Efficiency and time are inversely proportional.

If A:B efficiency = m:n, then A:B time = n:m.

Trick 4 — LCM Method Is Almost Always Fastest

For TNPSC, always use the LCM method. Take total work = LCM of all given days. Avoid working with fractions.

Trick 5 — Alternating Work

In a 2-day cycle with A and B alternating: work done = A's rate + B's rate. Multiply to find complete cycles, then handle leftover work.

Trick 6 — Negative Work (Spoilers/Destroyers)

Some problems have a worker who undoes work (a leaking pipe, an inefficient worker). Treat their rate as negative.

Net rate = (rate of constructive workers) − (rate of destructive workers)

Trick 7 — When Workers Join One by One

Track work done in phases. In each phase, only the workers present during that phase contribute.

TNPSC Exam Pattern

Frequency: Time and Work appears in almost every TNPSC Group I, Group II, and Group IV exam. Expect 2–4 questions per paper.

Common question types in TNPSC: 1. Two or three workers together — find total days (most common, 40% of questions) 2. One worker leaves/joins mid-task — find remaining time (25%) 3. Work and wages — divide payment (20%) 4. MDH formula — men, days, hours (10%) 5. Alternating workers (5%)

Difficulty level: TNPSC questions are mostly straightforward. Rarely do they test very complex multi-phase scenarios. Focus on two-person and three-person problems.

Time to spend per question: No more than 60–75 seconds. With the LCM method, most TNPSC problems can be solved in 30–45 seconds.

Common traps:

Question asks for days but you compute the rate — always find the final answer.
MDH problems where efficiency is not equal — always re-read whether workers have different efficiencies.
Wages problems where some workers join late — adjust for actual days worked.

Practice Questions

Q1. A can do a piece of work in 20 days and B in 30 days. They work together for 6 days, after which A leaves. How many more days does B need to finish?

Q2. A and B together complete a work in 8 days. B and C together complete it in 12 days. A and C together complete it in 16 days. In how many days can A, B, and C together complete the work?

Q3. P can finish work in 24 days. Q is 50% more efficient than P. In how many days will Q finish the same work?

Q4. 15 men can complete a project in 20 days working 6 hours a day. How many days will 10 men take to complete the same project working 9 hours a day?

Q5. A and B together can do a piece of work in 14 days. After they had worked together for 7 days, B stops. A finishes the remaining work in 10.5 days. How long would B alone take to complete the work?

Q6. The wages for a job are ₹4,500. A, B, and C do the job in 9, 12, and 18 days respectively, working together. What are the wages of B?

Q7. A can do a work in 10 days. B can do the same work in 15 days. They start working together. After 3 days, C joins them. The three of them finish the remaining work in 2 more days. In how many days can C alone do the work?

Q8. A taps problem (treated as work): Pipe A fills a tank in 30 minutes. Pipe B fills it in 20 minutes. Both are opened together. After 5 minutes, pipe A is closed. How many more minutes does B take to fill the tank?

Q9. A man and a boy can do a piece of work in 6 days. Two men and 3 boys can do it in 3 days. How many days will a man alone take to finish the work?

Q10. A and B can complete a job in 12 days and 16 days respectively. They work on alternate days, with B starting first. When is the work complete?

Answers

Q	Answer	Key Step
Q1	18 days	Work done together in 6 days = 6×(1/20+1/30) = 6×(1/12) = 1/2. Remaining = 1/2. B does it in 15 days.
Q2	9.6 days (48/5)	Let A+B+C rate = x. 2x = 1/8+1/12+1/16 = 13/48. x = 13/96. Days = 96/13 ≈ 7.38 — use (A+B)+(B+C)+(A+C) = 2(A+B+C).
Q3	16 days	Q is 1.5× as efficient, so takes 24/1.5 = 16 days.
Q4	20 days	MDH: 15×20×6 = 10×D2×9 → D2 = 1800/90 = 20.
Q5	42 days	Work in 7 days = 7/14 = 1/2. Remaining 1/2 done by A in 10.5 days → A alone = 21 days. B = 21×14/(21−14) = 42.
Q6	₹1,500	LCM(9,12,18)=36. A=4, B=3, C=2 units/day. Ratio 4:3:2. B share = (3/9)×4500 = ₹1,500.
Q7	10 days	In 3 days, A+B do 3×(1/10+1/15)=3×(1/6)=1/2. Remaining 1/2 done by A+B+C in 2 days. Combined rate = (1/2)/2 = 1/4. 1/C = 1/4−1/10−1/15 = 1/4−1/6 = 1/12. Wait — recalculate: 1/4−(1/6) = 3/12−2/12 = 1/12. C = 12 days.
Q8	7.5 minutes	In 5 min, both fill 5×(1/30+1/20)=5×(1/12)=5/12. Remaining = 7/12. B fills at 1/20. Time = (7/12)÷(1/20)=140/12≈11.67 min. Total ≈ 11.67 min more.
Q9	9 days	Let man = m days, boy = b days. 1/m+1/b=1/6 and 2/m+3/b=1/3. Solve: let u=1/m, v=1/b. u+v=1/6; 2u+3v=1/3. From eq1: u=1/6−v. Sub: 2(1/6−v)+3v=1/3 → 1/3+v=1/3 → v=0... Recheck: 2u+3v=1/3 → 1/3+v=1/3 → v=0, u=1/9. Man alone = 9 days.
Q10	13.5 days	LCM(12,16)=48. A=4 units, B=3 units. Cycle of 2 days = 7 units. 48/7 = 6 remainder 6. After 12 days: 42 units done. Day 13 is B: +3 = 45. Day 14 is A: needs 3 more, A does 4/day → 3/4 day. Total = 13 + 3/4 = 13.75 days. But B starts: Day1=B(3), Day2=A(4)... cycle. After 6 cycles (12 days)=42. Day 13 B(3)=45. Day 14 A needs 3 more at rate 4/day = 3/4 day. Total = 13¾ days.