Time and Work problems are essentially problems about rates. When someone "does work," they consume time at a certain rate. The entire topic collapses into one fundamental relationship:
Work = Rate × Time
Or flipped: if a person completes a job in n days, their daily rate is 1/n of the total job.
Here is the analogy that makes this stick. Think of a tank being filled with water. Each worker is a pipe pouring water in at a fixed rate. Multiple workers together pour simultaneously — rates add up. The tank is full when total water poured = 1 (the whole job). Time and work is just this, dressed up in words like "A can do a work in 18 days."
Most students get tangled in fraction arithmetic. The cure is the LCM method (also called the "total work units" approach). Instead of working with fractions like 1/18 and 1/27, you pick the LCM of all the given days as the total work, then work in whole numbers.
Example: A takes 18 days, B takes 27 days. LCM(18, 27) = 54. Total work = 54 units. A's rate = 54/18 = 3 units/day. B's rate = 54/27 = 2 units/day. Together: 5 units/day. Done. The rest is arithmetic on whole numbers.
The second concept you must internalize: efficiency and time are inversely proportional. If A is twice as efficient as B, A takes half the time. If A is 60% more efficient than B (i.e., efficiency ratio = 1.6 : 1), their time ratio = 1 : 1.6. This direct inverse relationship handles the "efficiency-type" questions cleanly.
The third sub-type is the partial work problem — A works for some days, then leaves or is joined by someone else. Structure these as: work done in Phase 1 + work done in Phase 2 = 1.
Grasp these three ideas — LCM method, efficiency-time inverse, and phase-based work accounting — and you have covered 95% of what SSC CHSL throws at you.
The fraction-based approach (where you add 1/A + 1/B) works but invites errors. The LCM method removes fractions entirely.
Setup: Assign Total Work = LCM of all given time values. Each person's daily output becomes a whole number.
Example: A in 15 days, B in 25 days, C unknown.
No fractions touched until the very final step (if any).
These are problems where the composition of workers changes partway through. Structure them in two phases:
Phase 1: Identify who works and for how long → calculate units completed. Phase 2: Remaining work = Total − Phase 1 work → find time for Phase 2 at the new rate.
The critical habit: always compute "work remaining" as an absolute number of units (not a fraction) when using LCM method.
Worked structure:
4(5/7) days."X is k% more efficient than Y" translates to:
So if Shama is 60% more efficient than Annu (22 days):
13(3/4) daysNo LCM needed here — it is a pure ratio operation.
When problems involve men, hours per day, and number of days together (MDH problems), use:
M₁ × D₁ × H₁ = M₂ × D₂ × H₂ (for equal amounts of work)
For unequal work: (M₁D₁H₁)/W₁ = (M₂D₂H₂)/W₂
This formula is the only tool you need for these. Plug and chug.
When A + B work together, then A + C finish the rest, find C alone:
Keep everything in LCM units and this is just subtraction.
A filling pipe adds work (positive rate). A draining pipe removes work (negative rate). If a pipe fills a tank in f days and another drains it in d days, net rate = 1/f − 1/d. Use LCM on f and d. When the net rate is negative, the tank empties. Treat "tank full" as 1 unit of work completed.
Convert all given days to a single LCM total. Every subsequent calculation is whole-number arithmetic. For A (18 days) and B (27 days): LCM = 54, rates = 3 and 2. Adding fractions 1/18 + 1/27 = 5/54 takes ~30 seconds including error-checking. LCM method: write 3 + 2 = 5, divide 54 by 5 = 10.8 days — done in 10 seconds. Standard fraction method: ~30s. LCM method: ~10s. That is a 3× speed gain on every combined-work question.
When "X is k% more efficient than Y," time ratio = 100 : (100+k). Write the multiplier immediately: Time(X) = Time(Y) × 100/(100+k). Example: 60% more efficient → multiply Y's days by 100/160 = 5/8. Annu takes 22 days → Shama takes 22 × 5/8 = 110/8 = 13.75 days. Standard method (set up equations for daily rates, equate): 4 steps. This ratio flip: 1 step.
In phase-based problems, before computing the final time, do a sanity check: "Is remaining work less than total?" If Phase 1 uses more than the total units, you have made an error. With LCM units, this check is instant because you are comparing whole numbers. Example: Total = 24 units, Phase 1 = 12 units → remaining = 12. Quick mental check confirms you have not overshot. This catches the #1 error (wrong LCM) before you finish the problem. Saves the ~45 seconds of rewriting when you get an answer not in the options.
For Man-Days-Hours problems, draw a 2×3 grid: M, D, H across the top; Scenario 1 and 2 in rows. Multiply across each row to get man-hours. Set equal (or adjust for work ratio). This visual format stops you from multiplying the wrong variables. Setting up the equation: standard approach takes ~40s. Grid approach: ~15s because you fill 6 cells and read off one unknown.
In any problem where workers split up partway, always compute "work done in the joint phase" first, before touching the solo phase. Compute joint rate (LCM method), multiply by days → units done. Subtract from total. Then apply solo rate to remaining. Students who try to set up a single equation for the whole problem invariably get algebraic errors. Sequential chunking: 3 steps, each error-free. Single-equation attempt: 5-6 algebraic steps with fraction multiplication risk. Save ~60 seconds on each such problem.
Read the question and classify it in 5 seconds:
Type 1 — Simple Combined Work: "A in X days, B in Y days, working together, find days." → LCM method. Total work = LCM(X, Y). Rates = LCM/X and LCM/Y. Add rates, divide total work.
Type 2 — Phase Work (someone leaves or joins): → Phase 1: joint rate × days = units done. Phase 2: remaining units / new rate. Add phases.
Type 3 — Efficiency Ratio: → Identify the ratio (e.g., 1.6x more efficient). Apply inverse to given days. One multiplication.
Type 4 — Find One Person's Time Given Others: → Work backwards: combined rate − known rates = unknown rate. Invert for days.
Type 5 — MDH (Men-Days-Hours or multiple units of work):
→ MDH formula. Write as ratio: (M₁×D₁×H₁)/W₁ = (M₂×D₂×H₂)/W₂.
Decision rule: if you see fractions accumulating, stop and restart with LCM. If the answer you compute is not among the options, check whether you used total work correctly (LCM vs. the number 1).
Why this question: Riya-Sangeeta is the canonical "join later" phase problem. It tests whether you can handle partial work neatly.
Solving path:
4(5/7) days.Why this question: This three-person problem (A + B first, then A + C) is a structure that recurs almost every year. It tests sequential work accounting.
Solving path:
Why this question: The "fraction of work remaining" type. Extremely common and solvable in under 30 seconds with LCM.
Solving path:
Why this question: Efficiency problems trip up students who try to use rates instead of the direct ratio shortcut. This one is clean and fast if you know the inverse rule.
Solving path:
13(3/4) days.Why this question: Another "fraction remaining" question but with different numbers — tests whether you have the pattern locked in, not just memorized one example.
Solving path:
Using fractions instead of LCM units. Adding 1/12 + 1/18 manually under time pressure leads to arithmetic errors. Commit to LCM units on every problem where at least two people's days are given.
Forgetting to subtract Phase 1 work before computing Phase 2 time. Students sometimes divide the total work by Phase 2's rate instead of the remaining work. Always write: Remaining = Total − Done so far, as a separate line.
Inverting the efficiency ratio in the wrong direction. "A is 60% more efficient" means A takes fewer days, not more. The time ratio is 100 : 160, so A's days = given days × 100/160. Many students multiply by 160/100 and get a number larger than the original — that would mean A is less efficient.
In MDH problems, forgetting to scale work. If the second scenario involves two ponds instead of one, the right side of the MDH equation is 2W, not W. A straight M₁D₁H₁ = M₂D₂H₂ equation without adjusting for work quantity gives the wrong answer.
Treating "A leaves after k days" differently from "B joins after k days." Both are phase problems with the same structure. The first phase is always the group that starts; the second phase is whoever remains or joins. Do not let the wording change your framework.
Wrong LCM calculation. If LCM is computed incorrectly (e.g., taking the product instead of LCM), every subsequent number is wrong. Quick check: LCM(a, b) must be divisible by both a and b. Verify with a 2-second mental check before proceeding.