Time and Work Tricks for SSC CGL (LCM Method)
The fastest time and work trick for SSC CGL is the LCM (total work) method: take the LCM of the given days as the total work, divide it by each person's days to get their one-day work (efficiency), then add or subtract efficiencies and divide the total work by the result. For example, if A takes 12 days and B takes 18 days, total work = 36 units, A does 3 a day, B does 2, so together they finish in 36 ÷ 5 = 7.2 days.
🔥 Key Exam Takeaway: Together time = xy ÷ (x + y) for two people. M₁D₁H₁ ÷ W₁ = M₂D₂H₂ ÷ W₂ for men, days, hours and work. Efficiency and time are inversely proportional: if A is 25% more efficient than B, A takes 1/1.25 = 4/5 of B's time. Wages are shared in the ratio of work done, not time taken.
Time and work (with pipes and cisterns) gives 1 to 2 questions in SSC CGL Tier 1 and more in Tier 2, plus regular questions in IBPS and SBI arithmetic. All answers below were computed in Python.
Time and Work Formulas for SSC CGL
| # | Formula | When to use |
|---|---|---|
| 1 | Work = Efficiency × Time | Basic relation |
| 2 | One-day work = 1 ÷ (days to finish alone) | Fraction method |
| 3 | A and B together = xy ÷ (x + y) days | Two people, x and y days alone |
| 4 | A, B, C together = xyz ÷ (xy + yz + zx) days | Three people |
| 5 | If A + B take T days and A alone takes x, B alone = xT ÷ (x − T) | Find one person's time |
| 6 | M₁D₁H₁ ÷ W₁ = M₂D₂H₂ ÷ W₂ | Men, days, hours, work (chain rule) |
| 7 | Efficiency ratio a : b → time ratio b : a | Comparing workers |
| 8 | Wages ratio = work done ratio = efficiency ratio (if they work equal days) | Sharing wages |
| 9 | A + B, B + C, C + A given → A + B + C = half the sum of the three efficiencies | Pair questions |
| 10 | Negative worker (leak, destroyer): subtract its efficiency | Pipes, leaks, damage |
LCM Method in Time and Work (Step-by-Step)
- Take the LCM of all the "days" given. This is the total work in units.
- Efficiency of each person = total work ÷ their days.
- Combine efficiencies (add for working together, subtract for a destroyer).
- Time = work left ÷ combined efficiency.
Why it is faster: you work with whole numbers instead of fractions like 1/12 + 1/18.
| Person | Days alone | Efficiency (LCM 36) |
|---|---|---|
| A | 12 | 36 ÷ 12 = 3 units/day |
| B | 18 | 36 ÷ 18 = 2 units/day |
| A + B | 36 ÷ 5 = 7.2 days | 5 units/day |
Efficiency Method in Time and Work
When the question gives a comparison ("A is twice as efficient as B", "A is 25% more efficient"), turn it into an efficiency ratio and take total work = sum of efficiencies × combined time.
- A is twice as efficient as B → A : B = 2 : 1.
- A is 25% more efficient than B → A : B = 125 : 100 = 5 : 4.
- A takes 40% less time than B → time ratio 3 : 5 → efficiency ratio 5 : 3.
Time and Work Questions for SSC CGL with Solutions
Q1 (Basic). A can do a work in 12 days and B in 18 days. In how many days can they finish it together? Solution: LCM = 36; A = 3, B = 2 units/day. Time = 36 ÷ 5 = 7.2 days (7 1/5 days). Formula check: 12 × 18 ÷ 30 = 7.2. ✔
Q2 (Pairs). A and B can do a work in 10 days, B and C in 15 days, C and A in 12 days. How long will A, B and C take together? How long will A take alone? Solution: LCM = 60. A + B = 6, B + C = 4, C + A = 5. Sum = 15 = 2(A + B + C), so A + B + C = 7.5 units/day → 8 days. A = 7.5 − (B + C) = 7.5 − 4 = 3.5 → A alone = 60 ÷ 3.5 = 120/7 days (17 1/7 days).
Q3 (M₁D₁H₁ = M₂D₂H₂). 12 men working 8 hours a day finish a work in 10 days. In how many days will 16 men working 6 hours a day finish it? Solution: 12 × 8 × 10 = 16 × 6 × D → D = 960 ÷ 96 = 10 days.
Q4 (Efficiency ratio). A is twice as efficient as B. Together they finish a work in 14 days. How long will each take alone? Solution: A : B = 2 : 1, total work = (2 + 1) × 14 = 42 units. A alone = 42 ÷ 2 = 21 days; B alone = 42 ÷ 1 = 42 days.
Q5 (One leaves). A can do a work in 20 days and B in 30 days. A works alone for 4 days and then leaves. In how many days will B finish the rest? What is the total time? Solution: LCM = 60; A = 3, B = 2. A does 4 × 3 = 12 units; 48 left. B takes 48 ÷ 2 = 24 days. Total = 4 + 24 = 28 days.
Q6 (Alternate days). A can do a work in 10 days and B in 15 days. They work on alternate days, starting with A. In how many days is the work finished? Solution: LCM = 30; A = 3, B = 2. A 2-day cycle = 5 units. After 5 cycles (10 days), 25 units are done; 5 left. Day 11: A does 3 → 2 left. Day 12: B does 2 → done. Answer = 12 days.
Q7 (Men and women). 4 men and 6 women can finish a work in 8 days, while 3 men and 7 women can finish it in 10 days. In how many days will 10 women finish it? Solution: (4M + 6W) × 8 = (3M + 7W) × 10 → 32M + 48W = 30M + 70W → 2M = 22W → 1 man = 11 women. Work = (4 × 11 + 6) × 8 = 50 × 8 = 400 woman-days. 10 women take 400 ÷ 10 = 40 days.
Q8 (Wages). A can do a work in 10 days and B in 15 days. Working together, they earn ₹1,500. What is A's share? Solution: Efficiency ratio A : B = 15 : 10 = 3 : 2 (they work the same number of days). A's share = 3/5 × 1500 = ₹900; B's share = ₹600.
Q9 (Third person). A and B can finish a work in 12 days. With C's help they finish it in 8 days. How long will C take alone? Solution: C's one-day work = 1/8 − 1/12 = 1/24 → 24 days. (LCM method: total 24; A + B = 2, A + B + C = 3, so C = 1 → 24 days.)
Q10 (Part of work). A can finish 1/3 of a work in 5 days. How long for the whole work? Solution: 5 × 3 = 15 days.
Q11 (Percentage efficiency). A is 25% more efficient than B. B alone takes 20 days. How long does A take? Solution: Time is inverse to efficiency: 20 ÷ 1.25 = 16 days.
Q12 (Negative worker). A can build a wall in 10 days and B in 15 days, but C can demolish it in 30 days. If all three work together, how long will the wall take? Solution: LCM = 30; A = 3, B = 2, C = −1. Net = 4 units/day → 30 ÷ 4 = 7.5 days.
Common Mistakes in Time and Work Questions
| Mistake | Fix |
|---|---|
| Adding days instead of efficiencies (12 + 18 = 30 days) | Add one-day works, never days |
| "25% more efficient" means 25% less time | It means time × 100/125 = 20% less time |
| Sharing wages by days taken | Share by work done (efficiency × days worked) |
| Forgetting hours per day in men-days questions | Use M × D × H on both sides |
| Leaving a fraction of a day un-converted in alternate-day questions | Finish the last day with the leftover units ÷ that person's efficiency |
Time and Work Short Tricks: Quick Revision Table
| Given | Shortcut |
|---|---|
| A: x days, B: y days | Together = xy/(x + y) |
| A: x, B: y, C: z days | Together = xyz/(xy + yz + zx) |
| A + B: T days, A: x days | B = xT/(x − T) |
| A is k times as efficient as B, together T days | A alone = T(k + 1)/k, B alone = T(k + 1) |
| A is p% more efficient than B, B takes t days | A takes t × 100/(100 + p) |
| Men, days, hours | M₁D₁H₁ = M₂D₂H₂ (same work) |
Check the k-times row with Q4: T = 14, k = 2 → A = 14 × 3/2 = 21, B = 14 × 3 = 42. ✔
Time and Work Practice Questions (with Answers)
- A can do a job in 15 days and B in 10 days. Together? → 6 days
- A and B together finish in 6 days; A alone takes 10 days. B alone? → 15 days (10 × 6 ÷ 4)
- 20 men working 8 hours a day finish a job in 6 days. How many days will 15 men working 8 hours a day take? → 8 days
- A is 50% more efficient than B, who takes 20 days. A alone? → 40/3 days (13 1/3 days)
- A takes 20 days and B 30 days. They work together for 6 days, then A leaves. How many more days does B take? → 15 days
- A can do a work in 24 days and B in 36 days. In how many days can they finish it together? → 14.4 days
- 6 men can do a work in 10 days. How many men are needed to finish it in 4 days? → 15 men
- A, B and C take 12, 15 and 20 days. Together? → 5 days (LCM 60; 5 + 4 + 3 = 12 units/day)
Time and Work FAQs
What is the best method for time and work in SSC CGL?
The LCM (total work) method. It turns fractions into whole-number efficiencies, so most questions take under a minute.
What is the M1D1H1 = M2D2H2 formula?
Men × days × hours per day is constant for the same work. If the work changes, divide each side by its work: M₁D₁H₁/W₁ = M₂D₂H₂/W₂.
How is efficiency related to time?
They are inversely proportional. Double the efficiency means half the time.
How are wages divided in time and work?
In the ratio of work done. If all work the same number of days, that equals the efficiency ratio.
Are pipes and cisterns part of time and work?
Yes. Inlet pipes are workers, outlet pipes and leaks are negative workers, and the same LCM method applies.
Related topics: revise percentage tricks for SSC CGL, mixture and alligation questions for bank exams and RRB NTPC maths formulas next.
Practice Next
- New SSC CGL practice: SSC CGL Time and Work Mock Test 02.
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- New topic-wise test: SSC CGL Time and Work Mock Test 01: Pipes and Cisterns, Efficiency and Wages (15 Questions).
- New SSC topic test: SSC CHSL Time and Work Mock Test 01: Work Together, Leaving Midway and Pipes (15 Questions).
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