Time & Work
🔒 Log in to trackRates, LCM units and the men-days chain solve nearly every question: combined work, efficiency ratios, pipes and cisterns, mid-work join/leave, alternate days, provisions and wages. One slot is near-guaranteed in Tier 1, with staged pipe problems and wage splits rising in Tier 2.
One page per subtopic: detailed notes, every question type, formulas, tricks and practice sets.
Every formula on one printable page, grouped by subtopic.
5 exam-level questions worked step by step.
61 questions — untimed practice or a timed test with analysis.
Track record in the exam
Questions per shift in recent SSC CGL papers.
Test difficulty mix (61 questions)
Question patterns exams keep repeating
Taken from previous-year papers. If a pattern is marked "very common", expect to see it in your exam.
Two or three people working together
very commonIndividual completion times are given and the time working together is asked — or the reverse, with the together time given and one person's time missing.
How to solve: Take the job as LCM units and add the daily rates; for two people use ab/(a+b). To find a missing worker, subtract his partners' rates from the together rate. The combined time must be less than the fastest individual time.
Example: A can finish a job in 15 days and B in 10 days. Working together, they finish it in:
Job = 30 units; A does 2/day, B does 3/day → 5/day → 6 days.
Pipes & cisterns
very commonFill pipes and an outlet (or a leak) act on a tank; the fill time is asked, or a pipe opens mid-way and the tank must be re-solved in stages.
How to solve: Mark inlets + and outlets −, set the tank as LCM units, and take the net rate. A fill–drain pair alone is ab/(b−a). For staged problems, compute the stage-1 work first, then divide the remainder by the new net rate.
Example: A pipe fills a tank in 12 hours and another empties the full tank in 24 hours. With both open, the tank fills in:
Net rate = 1/12 − 1/24 = 1/24 → 24 hours.
Efficiency comparisons
common'A is twice/thrice as efficient as B', an efficiency ratio like 3 : 2, or a percent such as '50% more efficient', with a together time or one individual time given.
How to solve: Convert everything to a rate ratio first (50% more = 3:2). Weaker worker = 1 unit/day, stronger = k units/day, so the job = (k+1) × together-days units; divide by each rate for individual times.
Example: A is twice as efficient as B and together they finish a work in 18 days. B alone takes:
Job = 3 × 18 = 54 units → B alone 54 days (A alone 27).
A leaves or joins mid-work
commonSomeone works a few days and then leaves (or joins late); the remaining or total time, or the fraction each person did, is asked.
How to solve: Work actually done = rate × days. Subtract it from the job and hand the remainder to the continuing workers at their own rate. For a late joiner, add the solo phase to the team phase.
Example: A can do a work in 12 days and B in 18 days. A works 3 days and leaves. B alone finishes the rest in:
A did 3/12 = 1/4 → B needs (3/4) × 18 = 13.5 days.
Alternate days
commonA and B work one day each in turn (someone starts); the total time or the finishing day is asked.
How to solve: Count progress per 2-day cycle: 1/T_A + 1/T_B per cycle. Complete the full cycles that fit, then walk the leftover day by day — checking whose turn each leftover day is.
Example: A can do a job in 3 days and B in 6 days. Working on alternate days starting with A, the job is finished in:
Cycle = 1/3 + 1/6 = 1/2 per 2 days → exactly 2 cycles = 4 days.
Men–days–hours chain
commonMen, days and hours per day appear in two scenarios (efficiency too, sometimes); one value of the second scenario is missing.
How to solve: Equate the products: M1D1H1E1 = M2D2H2E2 for the same work. Put everything that grows with time beside M and solve for the missing quantity; more men always means fewer days.
Example: 12 men working 8 hours a day finish a work in 15 days. 18 men working 10 hours a day will finish it in:
12 × 15 × 8 = 1440 man-hours → 1440 ÷ (18 × 10) = 8 days.
Provisions / consumption
commonA garrison, fort or hostel has food for M men for D days; after some days men join or leave and the remaining duration is asked.
How to solve: Stock = M × D man-days. Subtract M × t for the days already passed, then divide the remaining stock by the new headcount.
Example: A garrison of 400 men has food for 25 days. After 5 days, 100 more men join. The food now lasts:
Stock left = 400 × 20 = 8000 man-days → 8000 ÷ 500 = 16 days.
Wages distribution
commonA total wage for a jointly completed job is given; one worker's share is asked, sometimes with unequal days or an unknown third worker.
How to solve: Shares follow work done: rate × days for each worker, as a ratio. Equal days reduce it to the rate ratio; find an unknown member's rate as team rate minus the others.
Example: A and B earn ₹600 together. A alone does the job in 10 days, B in 15 days. A's share is:
Rates 3 : 2 → A gets (3/5) × 600 = ₹360.