Time and Work Shortcuts: LCM Method, Efficiency Ratios and Pipes & Cisterns Tricks for SSC & Bank Exams

Quick Answer: The LCM Method in a Nutshell

Quick Answer: The LCM method is the fastest Time and Work shortcut for SSC and Bank exams. Instead of writing a worker’s rate as 1/x and juggling fractions, assume the total work equals the LCM of the days given. Every worker’s per-day output becomes a whole number, so the whole question collapses into simple addition, subtraction and division — solvable in under 30 seconds without a single fraction.

Why Fraction-Based Methods Waste Time in SSC and Bank Exams

The traditional textbook approach writes A’s one-day work as 1/A, B’s as 1/B, and then asks you to add fractions like 1/12 + 1/15 + 1/20 under exam pressure. Finding common denominators mid-question is where seconds leak away — and where calculation slips happen.

Compare the two approaches for the same question: “A does a job in 12 days, B in 15 days. Together?”

Same answer, but the LCM column is pure whole-number arithmetic — no fraction addition, no LCM-of-denominators step hidden inside the working. Across a 25-question Quant section, this saves 20–40 seconds per question.

Step-by-Step LCM Method for Time and Work Questions

• Step 1 — Take the LCM: Assume total work = LCM of all the “days” figures in the question.
• Step 2 — Compute per-day work: Each worker’s daily efficiency = Total work ÷ their days.
• Step 3 — Add/subtract efficiencies: People working together add; someone leaving subtracts; outlet pipes subtract.
• Step 4 — Divide: Time = Total work ÷ combined efficiency. Remaining work ÷ efficiency gives remaining days.

Solved Examples: Two Workers, Three Workers and Leaving-Joining Cases

Example 1 (A and B together): A finishes a work in 10 days, B in 15 days. LCM = 30 units. A = 3/day, B = 2/day. Together = 5/day → 30 ÷ 5 = 6 days.

Example 2 (Alternate days): Same A (3/day) and B (2/day) work on alternate days, A starting. In 2 days they complete 5 units. After 10 days (5 cycles) = 25 units done. Day 11: A adds 3 → 28. Day 12: B adds 2 → 30. Total = 12 days.

Example 3 (Worker leaves midway): A (4/day) and B (6/day) start a 48-unit work together. After 4 days (10/day × 4 = 40 units), A leaves. Remaining 8 units at B’s 6/day = 1 1/3 days. Total = 5 1/3 days.

Efficiency Ratios: The Second Core Shortcut

Efficiency and time are inversely proportional. If A is twice as efficient as B, A takes half the time. So if the efficiency ratio is 2 : 3, the time ratio is 3 : 2 — never 2 : 3. This single inversion is the most common trap in the topic.

Example: A is thrice as efficient as B and finishes a work 8 days sooner. Time ratio B : A = 3 : 1, so difference = 2 parts = 8 days → 1 part = 4 days. A takes 4 days, B takes 12 days.

Wages and Money-Sharing Problems Using Efficiency Ratios

Wages are divided in the ratio of work done = efficiency × days worked. If A works at 3 units/day for 5 days and B at 2 units/day for 6 days, work ratio = 15 : 12 = 5 : 4 — so ₹1,800 total wages split as ₹1,000 and ₹800.

Example: A and B together earn ₹870. A does a job in 6 days, B in 4 days. LCM = 12; A = 2/day, B = 3/day. Working together, wage ratio = 2 : 3, so A gets ₹348 and B gets ₹522.

Pipes and Cisterns: Same Logic, Different Labels

Pipes and Cisterns is Time and Work wearing a disguise:

• Inlet pipe = a worker (positive efficiency)
• Outlet/leak pipe = negative efficiency (subtract it)
• Tank capacity = LCM of individual filling times, often read in litres

Nothing else changes — Steps 1 to 4 apply exactly as before.

Solved Pipes and Cisterns Problems for SSC and Bank Prelims

Problem 1 (Two inlets + one outlet): Pipes A and B fill a tank in 12 and 15 hours; outlet C empties it in 20 hours. LCM = 60 units. A = +5, B = +4, C = −3. Net = 6/hour → tank fills in 60 ÷ 6 = 10 hours.

Problem 2 (Partial filling): A fills in 10 hours (6/hr), B in 15 hours (4/hr) — total 60 units. Both open for 3 hours: 30 units filled. B is closed; A fills remaining 30 at 6/hr = 5 hours. Total = 3 + 5 = 8 hours.

Negative and Alternate Work Patterns (Demolition, Leakage, Alternating Days)

Destructive work: A builder constructs 10 units/day; a demolition crew destroys 4 units/day. Net = +6/day. A 60-unit wall takes 60 ÷ 6 = 10 days. If the destroyers can outpace builders, the job never finishes — check for that trap option.

Leakage: A pump fills a tank in 2 hours (efficiency 6, tank = 12 units); with a leak it takes 2.4 hours → actual rate = 5. Leak = 6 − 5 = 1 unit/hr, so the leak alone empties the full tank in 12 hours.

Alternate days with unequal workers: Always compute per-cycle (2-day) work, run full cycles, then handle the tail end day-by-day — as shown in the solved section above.

Common Mistakes and Traps in Time and Work Questions

• Inverse-ratio confusion: Efficiency 2 : 3 means time 3 : 2, not 2 : 3.
• Mixing efficiency with time: Never add days directly; always convert to per-day units first.
• Unit mismatch: If one figure is in hours and another in days, convert everything to one unit before taking the LCM.
• Alternate-day tail errors: After full cycles, check whether the last day’s worker overshoots the total — that day still counts as one day.
• Wages vs days: Wages follow work done (efficiency × days), not days alone, unless efficiencies are equal.

Practice Set: 10 Exam-Style Questions with Answer Key

1. A does a work in 8 days, B in 12. Together? (Work 24; 3+2=5 → 4 4/5 days)
2. A is twice as efficient as B and finishes 10 days earlier. B’s time? (Time ratio 1:2 → 1 part = 10 → B = 20 days)
3. A and B fill a tank in 6 and 8 hours; drain C empties it in 24. All open? (LCM 24; 4+3−1=6 → 4 hours)
4. A, B, C take 12, 15, 20 days. Together? (LCM 60; 5+4+3=12 → 5 days)
5. Men-hours: 12 men finish a job in 16 days; 8 men? (192 man-days ÷ 8 = 24 days)
6. A works 5 days then leaves; B finishes in 7 more. A alone takes 12 days. B alone? (Work 60; A does 25, B does 35 in 7 days → 10 days)
7. A and B alternate, B starts, A = 6/day, B = 4/day, work = 60. Days? (2-day cycle 10; 5 cycles = 50, B-day 6th cycle adds 4, A finishes last 6 → 12 days)
8. Wages ₹1,400; A works 5 days at 3/day, B 5 days at 4/day. A’s share? (15:20 → ₹600)
9. Two pipes fill in 20 and 30 min; leak empties in 60. Time to fill? (LCM 60; 3+2−1=4 → 15 min)
10. A builds 8/day, wrecking crew undoes 3/day. A 72-unit job? (Net 5/day → 14.4 days)

Answer key: 1) 4 4/5 days 2) 20 days 3) 4 hours 4) 5 days 5) 24 days 6) 10 days 7) 12 days 8) ₹600 9) 15 minutes 10) 14.4 days

Frequently Asked Questions

Q: What is the LCM method in Time and Work?

Assume total work = LCM of the days given by each worker, so each worker’s daily work becomes a whole number. All further arithmetic — combining, subtracting, dividing — stays in whole numbers.

Q: How do you convert efficiency ratio into time ratio?

Efficiency and time are inversely proportional. If the efficiency ratio is 2 : 3, the time ratio is 3 : 2. Flip the ratio; never use it as-is.

Q: How do you solve Pipes and Cisterns with one outlet pipe?

Treat the outlet as negative efficiency. Take the LCM of filling/emptying times as tank capacity, add inlet rates, subtract the outlet rate, then divide capacity by the net rate.

Q: How are wages divided when workers work different days?

Divide wages in the ratio of work done = efficiency × days worked by each person. Equal efficiency reduces this to a simple days ratio.

Q: Is the LCM method faster than the fraction method in exams?

Yes. It avoids fraction addition entirely and gives whole-number arithmetic, typically saving 20–40 seconds per question — a meaningful edge in SSC CGL and IBPS prelims time limits.

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