Here is a puzzle that trips up almost everyone the first time: if Ana can paint a room in 6 hours and Ben can paint it in 12 hours, how long does it take them together? The tempting answer — "average them, 9 hours" — is not just wrong, it's impossible (working together must be faster than either alone!). This chapter gives you the one simple idea that makes every time-and-work problem — including painters, pipes, and construction crews — straightforward and quick.
Time-and-work problems tell you how long people (or pipes, or machines) take to finish a job alone, then ask how long they take together, or how much is left after some time. The central idea is to stop thinking in "days to finish" and start thinking in rate of work per day. Once you do, the problems almost solve themselves — because rates simply add.
On the Civil Service Exam, time-and-work questions appear regularly in Numerical Reasoning, often in the guise of construction crews, tasks, or the closely related "pipes and tanks" problems. Counting those cousins, this topic touches 3 to 6 items on a typical exam.
The difficulty is intermediate. The math is just fractions, but the exam tests whether you truly grasp that you add the rates, not the times, and whether you can handle a drain pipe (negative rate), a job partly done, or workers joining and leaving. Get the "rates add" idea and the rest is bookkeeping.
After completing this lesson you will be able to:
You should be comfortable with:
A quick refresher on the skill you'll use most: adding fractions. To add 1/6 + 1/12, rewrite with a common denominator of 12: 2/12 + 1/12 = 3/12 = 1/4. And a reciprocal flips a fraction: the reciprocal of 1/4 is 4. These two moves — add fractions, then flip — are the whole engine of this topic.
"How long will it take if we work together?" is a question from real life:
Any time effort combines to finish a task faster, this is the reasoning involved — and it's why the exam includes it as a practical skill.
Everything flows from a single reframe.
If a worker finishes a whole job in n days, then in one day they finish 1/n of it. That fraction is their work rate. When people work together, their rates add — because in one day they collectively finish the sum of their individual fractions.
Ana paints a room in 6 hours; Ben in 12 hours. Together? Ana's rate = 1/6 of the room per hour. Ben's rate = 1/12 per hour. Combined = 1/6 + 1/12 = 2/12 + 1/12 = 3/12 = 1/4 of the room per hour. Time to finish = 1 ÷ (1/4) = 4 hours.
Notice 4 hours is less than either 6 or 12 — exactly as it must be, because two people are faster than one. This is the built-in sanity check: working together is always faster than the fastest worker alone.
Analogy: Think of each worker as a faucet filling the same bucket. A faster faucet has a higher flow rate. Open two faucets and the flows add — the bucket fills faster. You'd never "average" two flow rates to find the combined fill time, and the same is true for workers.
"6 days plus 12 days = 18 days" treats time as if it accumulates, but time-to-finish is inversely related to speed. More helpers means less time, not more. Always convert to rates (which add) before combining.
For exactly two workers taking a and b time alone, the combined time is:
Combined time = (a × b) ÷ (a + b).
For Ana (6) and Ben (12): (6 × 12) ÷ (6 + 12) = 72 ÷ 18 = 4 hours — same answer, faster. This "product over sum" form is worth memorizing for the common two-worker case.
If part of a job is finished, subtract that fraction from 1 (the whole job) before continuing.
Ana does 1/3 of a job, then Ben finishes the rest in 8 days. How long would Ben alone take for the whole job? Remaining work = 1 − 1/3 = 2/3, done by Ben in 8 days. Ben's rate = (2/3) ÷ 8 = 2/24 = 1/12. So Ben alone takes 12 days for the whole job.
"A is twice as efficient as B" means A's rate is twice B's rate — so A takes half the time, not double. Faster workers have higher rates and shorter times. Don't let the word "twice" fool you into doubling the time.
A is twice as efficient as B, and together they finish in 12 days. How long does each take alone? Let B's rate = r, so A's rate = 2r. Combined = 3r = 1/12, so r = 1/36. B alone = 36 days; A alone = 18 days.
A tank is just a "job," and pipes are "workers." A filling pipe has a positive rate; a draining pipe has a negative rate, because it removes work (water).
Pipe A fills a tank in 4 hours; pipe B drains a full tank in 6 hours. Both open — how long to fill? A's rate = 1/4 (fills); B's rate = −1/6 (drains). Combined = 1/4 − 1/6 = 3/12 − 2/12 = 1/12 of the tank per hour. Time to fill = 1 ÷ (1/12) = 12 hours.
If the drain were faster than the fill, the combined rate would be negative and the tank would never fill — a useful reality check.
When the team changes partway, split the job into stages: compute the fraction done in each stage, and make the fractions add to 1 for a completed job.
Ana (alone: 10 days) and Ben (alone: 15 days) work together for 2 days, then Ben leaves and Ana finishes. How many more days does Ana need? Combined rate = 1/10 + 1/15 = 3/30 + 2/30 = 5/30 = 1/6. Work done in 2 days = 2 × 1/6 = 1/3. Remaining = 1 − 1/3 = 2/3. Ana's rate = 1/10, so she needs (2/3) ÷ (1/10) = 20/3 ≈ 6.67 more days.
When the numbers of workers, hours per day, and days all change, use the fact that the total effort (workers × days × hours) needed for a fixed job is constant:
Workers₁ × Days₁ × Hours₁ = Workers₂ × Days₂ × Hours₂ (for the same job).
8 workers laboring 6 hours a day finish a road in 5 days. How many days for 10 workers laboring 8 hours a day? Total man-hours = 8 × 6 × 5 = 240. New daily effort = 10 × 8 = 80. Days = 240 ÷ 80 = 3 days.
Pay is proportional to work contributed, which (for equal time) is proportional to rate. Split the money in the ratio of the workers' rates.
Ana (4 days alone) and Ben (6 days alone) finish a job together and receive ₱600. Their rates are 1/4 and 1/6, in the ratio 3:2. So Ana gets 3/5 × 600 = ₱360 and Ben gets ₱240.
With the toolkit built, let's picture, tabulate, and drill.
| Formula | Meaning |
|---|---|
| rate = 1 ÷ time | Fraction of the job per unit time |
| combined rate = r₁ + r₂ + … | Rates add when working together |
| combined time = 1 ÷ (combined rate) | Flip the combined rate |
| two-worker time = ab ÷ (a + b) | Shortcut for two workers |
| drain rate = −(1 ÷ time) | A draining pipe subtracts |
| work done = rate × time | Fraction completed in a stage |
| W₁·D₁·H₁ = W₂·D₂·H₂ | Man-days (same job) |
| wage share ∝ rate | Split pay by work contributed |
Why "product over sum" works: combined rate = 1/a + 1/b = (a + b)/(ab); flipping it gives ab/(a + b). It's just the rates-add idea written compactly for two workers.
| When the problem says… | Do this… |
|---|---|
| "A finishes in x days, B in y days, together?" | add rates, then flip (or ab/(a+b)) |
| "part of the job is already done" | subtract that fraction from 1 first |
| "twice/thrice as efficient" | multiply the rate (halve/third the time) |
| "pipe drains / leak empties" | use a negative rate |
| "works for a while, then someone leaves/joins" | split into stages, fractions sum to 1 |
| "more men / different hours per day / how many days" | man-days: W·D·H constant |
| "they share the payment" | split in the ratio of their rates |
Step 1 — Convert every "time to finish" into a rate (1 ÷ time). ↓ Step 2 — Decide how the rates combine: add for helpers, subtract for a drain. ↓ Step 3 — Account for partial work — subtract any fraction already done from 1. ↓ Step 4 — Solve for the unknown (time = 1 ÷ rate, or set up stages that sum to 1). ↓ Step 5 — Sanity-check. Working together must beat the fastest solo time; a net-negative fill rate means "never fills."
Why Step 1 matters: the entire topic depends on switching from times (which don't add) to rates (which do). Do it first, every time.
Example 1. A finishes a task in 10 days, B in 15 days. How long together? Solution: Rates 1/10 + 1/15 = 3/30 + 2/30 = 5/30 = 1/6 → 6 days. (Or 10 × 15 ÷ 25 = 6.) Difficulty: ★☆☆☆☆
Example 2. A worker does 1/5 of a job per hour. How long to finish the whole job? Solution: Time = 1 ÷ (1/5) = 5 hours. Difficulty: ★☆☆☆☆
Example 3 (three workers). A, B, and C can finish a job alone in 20, 30, and 60 days respectively. How long together? Solution: 1/20 + 1/30 + 1/60 = 3/60 + 2/60 + 1/60 = 6/60 = 1/10 → 10 days. Difficulty: ★★☆☆☆
Example 4 (partial work). Ana completes 1/4 of a job, then Ben finishes the remaining 3/4 in 9 days. How long would Ben take for the whole job alone? Solution: Ben's rate = (3/4) ÷ 9 = 3/36 = 1/12 → 12 days. Difficulty: ★★★☆☆
Example 5 (efficiency). A is three times as efficient as B. If B alone takes 24 days, how long does A take, and how long do they take together? Solution: A's rate is 3× B's, so A alone = 24 ÷ 3 = 8 days. Together: 1/8 + 1/24 = 3/24 + 1/24 = 4/24 = 1/6 → 6 days. Difficulty: ★★★☆☆
Example 6 (pipes). Pipe A fills a tank in 12 hours; pipe B fills it in 15 hours. If both run together, how long to fill? Solution: 1/12 + 1/15 = 5/60 + 4/60 = 9/60 = 3/20 → time = 20/3 ≈ 6 hours 40 minutes. Difficulty: ★★★☆☆
Example 7 (fill and drain). A pipe fills a tank in 5 hours, but a leak can empty the full tank in 20 hours. With the leak, how long to fill? Solution: 1/5 − 1/20 = 4/20 − 1/20 = 3/20 → time = 20/3 ≈ 6 hours 40 minutes. Difficulty: ★★★☆☆
Example 8 (wages). Ana (alone: 6 days) and Ben (alone: 12 days) finish a job together and are paid ₱900. How should they split it? Solution: Rates 1/6 : 1/12 = 2 : 1. Ana gets 2/3 × 900 = ₱600; Ben gets ₱300. Difficulty: ★★★★☆
Example 9 (man-days). 15 workers build a wall in 12 days. How many days would 20 workers take, at the same rate? Thinking: More workers → fewer days (inverse); total man-days constant. Solution: 15 × 12 = 180 man-days; 180 ÷ 20 = 9 days. Difficulty: ★★★★☆
Example 10 (workers, days, hours). If 12 men working 8 hours a day finish a job in 10 days, how many days would 16 men working 6 hours a day take? Solution: Total man-hours = 12 × 8 × 10 = 960. New daily effort = 16 × 6 = 96. Days = 960 ÷ 96 = 10 days. Difficulty: ★★★★☆
Example 11 (joining mid-task). Ana and Ben start a job together; alone they'd take 12 and 18 days. After 4 days Ana leaves and Ben finishes. How many more days does Ben need? Solution: Combined rate = 1/12 + 1/18 = 3/36 + 2/36 = 5/36. Work in 4 days = 20/36 = 5/9. Remaining = 4/9. Ben's rate = 1/18, so time = (4/9) ÷ (1/18) = (4/9) × 18 = 8 more days. Difficulty: ★★★★★
Example 12 (leak found later). A pipe fills a tank in 6 hours. Because of a leak, it actually took 8 hours to fill. How long would the leak alone take to empty a full tank? Thinking: Fill rate 1/6; net rate 1/8; the leak accounts for the difference. Solution: Leak rate = 1/6 − 1/8 = 4/24 − 3/24 = 1/24, so the leak alone empties the tank in 24 hours. Difficulty: ★★★★★
The master idea of time and work is to think in rates, not times: a job done in n units of time is a rate of 1/n per unit, and rates add when people work together (so together is always faster than the fastest alone). For two workers, use the ab/(a + b) shortcut; to avoid fractions, let the job equal the LCM of the times and work in whole units. Subtract any work already done from 1; treat a drain as a negative rate; and split multi-stage problems so the fractions sum to 1. When only the counts of workers, hours, and days change, switch to man-days (W·D·H constant). Share wages in the ratio of rates. Convert to rates first, respect the sign of drains, and check that "together" beats the best solo time — that's the whole topic.
| Topic | Key point |
|---|---|
| Rate | 1 ÷ time (job per unit time) |
| Together | add the rates, then flip |
| Two workers | ab ÷ (a + b) |
| Avoid fractions | job = LCM of the times |
| Partial done | subtract fraction from 1 first |
| Efficiency | twice as efficient = half the time |
| Drain / leak | negative rate (subtract) |
| Man-days | W₁·D₁·H₁ = W₂·D₂·H₂ |
| Wages | split in the ratio of rates |
| Check | together < fastest solo time |
Why is the combined time less than either worker's time? Because two people do more per hour than one. Their rates add, so the job finishes sooner than the faster worker could manage alone.
When do I use man-days instead of adding rates? Use man-days when the problem changes the number of workers, hours per day, or days for the same job (and doesn't give individual "alone" times). Use rates when each worker's solo time is given and they work "together."
How do I handle a leak or drain? Give it a negative rate and add it to the fill rate. If the result is positive, the tank fills (slowly); if negative, it never fills.
What does "A is twice as efficient as B" mean exactly? A works at twice B's rate, so A takes half the time B would. Efficiency is about speed (rate), which is inversely related to time.
Is there a way to avoid fractions? Yes — set the whole job equal to the LCM of the given times. Each worker's daily "units" become whole numbers, making the arithmetic clean.
Tick them all and time-and-work problems — including their pipes-and-tanks cousins — become quick, reliable points.
Put it to the test with 1,428 practice questions on this topic.