Time and Work is fundamentally about one thing: rate of work. If a person can finish a job in n days, their daily work rate is 1/n of the total job per day. That's the entire foundation — everything else is just combining and manipulating rates.
Here's a useful analogy. Think of a water tank being filled by taps. Each tap fills at its own rate. Open two taps together, and the combined rate fills the tank faster. A person doing work is exactly like a tap filling a tank. Their "speed" is their daily fraction of work. Two people working together simply add their fractions.
Why does this matter for SSC GD? Because every single Time & Work question — whether it involves two workers, partial completion, or a man-and-son pair — reduces to the same operation: add rates, subtract rates, or scale them. There is no question in this chapter that cannot be solved once you internalize this one principle.
The two methods you need to know are:
Fraction Method — Express each person's daily work as a fraction, add them to get combined daily work, then take the reciprocal to get total days.
LCM Method (faster) — Assign the total work as the LCM of the given days, compute each person's daily work in units, add units for combined work, and divide total by combined rate.
The LCM method is faster for virtually every SSC GD question because it eliminates fractions entirely. You will see why once you compare both approaches on the same problem.
One more thing to understand before the Deep Dive: efficiency is just another word for daily work rate. If A is "twice as fast" as B, A's rate is 2× B's rate. If A does "3/4 of a work in 12 days," A's rate is (3/4) ÷ 12 = 1/16 of the work per day. These are not new concepts — they are the same fraction-based thinking applied differently.
If A completes work in a days and B in b days:
1/a1/b1/a + 1/b = (a+b)/(ab)ab/(a+b)This formula ab/(a+b) works only when both start and finish together. The moment the problem says "A leaves after 5 days" or "B joins after 3 days," you cannot plug into this formula blindly. You have to track the work done piece by piece.
Look — the LCM method turns fractions into whole numbers, which is faster to compute under exam pressure.
Steps:
Example: A finishes in 10 days, B in 15 days.
No fraction arithmetic at all. That's why you should default to this method every time.
This is tested frequently: "A and B together finish in X days, B alone takes Y days. How long does A alone take?"
When a worker leaves early or joins late, break the problem into phases:
Phase 1: Work done during the period both work together. Phase 2: Remaining work done by the person who stays.
Formula for remaining work after d days of combined work:
Remaining = Total Work − (Combined daily units × d)
Then: Days for remaining = Remaining ÷ Remaining person's daily units
Sometimes: "A does 3/4 of work in 12 days. In how many days does A finish 1/8 of the work?"
Don't overcomplicate this. Find A's per-day rate:
(3/4)/12 = 1/16 per day.(1/8) ÷ (1/16) = (1/8) × 16 = 2 days.Or more intuitively: if 3/4 takes 12 days, then 1/4 takes 4 days, and 1/8 takes 2 days. Direct proportion.
"A is twice as fast as B." Translate: A's rate = 2 × B's rate.
If their combined rate gives them T days together:
x per day, A's rate = 2x per day.x + 2x = 1/T → x = 1/(3T) → B alone takes 3T days.2x = 2/(3T) → A alone takes 3T/2 days.For T = 8: B alone = 24 days, A alone = 12 days.
When total work doubles or changes, the combined rate stays the same — only the total work changes. If copying 50 pages takes T days together, copying 100 pages takes 2T days. Scale linearly.
Always assign total work = LCM of all given days. This converts every fraction into a whole number. For A in 10 days, B in 15 days: LCM = 30, so A = 3 units/day, B = 2 units/day, together = 5 units/day, answer = 30/5 = 6. Compare: fraction method requires computing 1/10 + 1/15 = 3/30 + 2/30 = 5/30, then inverting to 30/5 = 6. Same answer, but with LCM you never write a fraction. Standard method: 4 steps with fraction arithmetic. LCM method: 4 steps, all integers. Saves roughly 20 seconds per question.
When you know combined time and one person's solo time, find the other's solo time using: 1/A = 1/Together − 1/B. With LCM: Total = LCM(Together, B_days). Combined daily units − B daily units = A daily units. Then A_days = Total ÷ A_units. Example: Together = 12 days, B = 28 days. LCM(12,28) = 84. Together = 7 units/day, B = 3 units/day, A = 4 units/day. A alone = 84/4 = 21 days. No fraction subtraction needed. Steps: 3 vs 5 in fraction method.
When partial work fractions are given, scale using direct proportion before computing rates. "3/4 work in 12 days → full work in 16 days." Then "1/8 work takes 16/8 = 2 days." You never need to write the rate explicitly. The rule: full work days = given days × (1 ÷ given fraction). Then scale to the asked fraction. Standard method: compute rate then divide. This method: two multiplications, no division. Saves 15 seconds.
If A is k times faster than B, then A takes 1/k times the days B takes. Equivalently: days are in the ratio 1:k (A:B). Combined, if they finish in T days, then B alone = T × (k+1)/k × ... no — just use this: combined rate = (k+1) × B's rate, so B's days = T × (k+1). A's days = T × (k+1)/k. For k=2, T=8: B = 8 × 3 = 24, A = 8 × 3/2 = 12. You go from T to both solo days in two multiplications. Standard algebra: 4 steps. This pattern: 2 steps.
For "A and B work together for d days, then A leaves, B finishes alone" problems — compute total work in LCM units, subtract (combined rate × d), divide remainder by B's rate. Example: A=25 days, B=20 days, LCM=100. A=4, B=5, together=9 units/day. 5 days together = 45 units done. Remaining = 55 units. B finishes in 55/5 = 11 days. The entire calculation uses only integers. Compare fraction method: track (1/25+1/20)×5, subtract from 1, divide by 1/20 — three fraction operations vs three integer operations. Saves 25 seconds.
In the exam hall, read the question and classify it in 3 seconds:
Type 1 — Both work together start to finish? Use ab/(a+b) directly or LCM. Answer in under 30 seconds.
Type 2 — Find one person's solo time given combined and other's solo? 1/A = 1/Together − 1/B. Use LCM subtraction. 30 seconds.
Type 3 — Partial work given (X/Y fraction in Z days)? Scale to full work, then scale to asked fraction. Two multiplications.
Type 4 — One person leaves early or joins late? Phase 1: compute work done together. Phase 2: divide remaining work by solo rate. LCM units throughout.
Type 5 — Efficiency ratio given (A is k× faster)? B's solo days = T × (k+1). A's solo days = T × (k+1)/k.
Type 6 — Total work scales (e.g., 50 pages vs 100 pages)? Find combined rate for original work, scale time proportionally.
Always default to LCM. If the LCM is large or ugly (e.g., LCM > 200), switch to fraction method — it will be cleaner in that case. But for 90% of SSC GD questions, LCM stays below 100.
Why this question: The most basic template — A and B given individually, find combined. Every SSC GD paper has one like this.
Solving path: LCM(10, 15) = 30. A does 3 units/day, B does 2 units/day. Together = 5 units/day. Days = 30/5 = 6 days. Answer: Option B.
Why this question: Reverse-engineering one person's solo time from combined and partner's solo. This tests whether you can reverse the rate-addition.
Solving path: LCM(12, 28) = 84. Together (A+B) = 84/12 = 7 units/day. B alone = 84/28 = 3 units/day. A alone = 7 − 3 = 4 units/day. A's days = 84/4 = 21 days. Answer: Option A.
Why this question: Father-son variant — combined and father's solo known, find son's solo. Tests the subtraction of rates.
Solving path: LCM(10, 8) = 40. Man's rate = 40/10 = 4 units/day. Together = 40/8 = 5 units/day. Son's rate = 5 − 4 = 1 unit/day. Son alone = 40/1 = 40 days. Answer: Option D.
Why this question: Fractional work — A does part of work in known days, find time for a different fraction. Appears regularly to trap students who compute rates unnecessarily.
Solving path: 3/4 work takes 12 days → full work takes 12 × (4/3) = 16 days. Now, 1/8 work takes 16 × (1/8) = 2 days. Answer: Option D.
Why this question: Half-time variant where B takes half the time A takes — this is an efficiency ratio problem phrased differently.
Solving path: A takes 18 days. B takes half = 9 days. A's daily work = 1/18. B's daily work = 1/9. Together = 1/18 + 2/18 = 3/18 = 1/6 of the work per day. Answer: Option A.
Why this question: Scaled total work — original manuscript is 50 pages but they need 100 pages. Tests whether you can proportionally scale time.
Solving path: For 50 pages: George takes 8 hours, Sonia takes 6 hours. Combined rate = 1/8 + 1/6 = 3/24 + 4/24 = 7/24 per hour. Time for 50 pages = 24/7 hours. For 100 pages (double the work): time = 2 × 24/7 = 48/7 = 6 and 6/7 hours. Answer: Option A.
Why this question: Partial completion — they work together for 5 days, then one leaves. The most common "advanced" variant in SSC GD.
Solving path: LCM(25, 20) = 100. A = 4 units/day, B = 5 units/day. Together = 9 units/day. In 5 days: 9 × 5 = 45 units done. Remaining = 100 − 45 = 55 units. B finishes alone: 55 ÷ 5 = 11 days. Answer: Option B.
Why this question: Efficiency ratio — man is twice as fast as woman. Tests the "k times faster" translation into rates and solo days.
Solving path: Let woman's rate = x per day. Man's rate = 2x per day. Combined = 3x per day. Together they finish in 8 days: 3x = 1/8 → x = 1/24. Woman alone = 24 days. Man alone = 1/(2x) = 12 days. Man: 12 days, Woman: 24 days. Answer: Option B.
Using ab/(a+b) when work phases differ. This formula only holds when both people work the entire duration from start to finish. If one person leaves early, use phase-by-phase calculation with LCM units — never plug into the combined formula.
Forgetting to scale when total work changes. In the 100-page vs 50-page type, students compute the combined time for the original job and forget to multiply by the scaling factor. Combined rate does not change — only the total work does.
Misreading "B takes half the time" as "B is half as efficient." Half the time means twice the efficiency — B's daily rate is double. Re-read: "B does the same work in half the time taken by A" means B is faster, not slower.
Computing LCM of the wrong numbers. In partial-completion problems, you only need LCM of the individual solo days — not the "together" days (which is a derived value, not an input). Using all three numbers for LCM will give a needlessly large number.
Not simplifying fractional work before computing rate. When the question says "A does 3/4 of work in 12 days," first find full work days (16), then work with that. Students who jump to rate = (3/4)/12 without simplifying often make arithmetic errors under pressure.
Assuming "together" always means both work for all days. Questions like "A goes away after 5 days" mean A worked only 5 days, not the full duration. Re-read the question stem once specifically for who-leaves-when before setting up the calculation.