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Aptitude & Reasoning · A — Quantitative Aptitude

Time & Work (with Pipes)

If you can think in 'work done per day', pipes filling and emptying a tank are the same equation with a sign flipped.

Test weight: HighTime / question: 55–75 secDifficulty: Medium

Convert every worker into a rate: if A finishes in a days, A does 1/a of the job each day. Rates simply add when people work together.

Combined work

If A takes a days and B takes b days, together they take 1 / (1/a + 1/b) days. A leak or emptying pipe is just a worker with a negative rate — subtract it from the total.

How to Approach It

  • Set the job as LCM units — let total work equal the LCM of the given days, so each worker's daily output is a whole number and combined work is a sum, not a fraction.
  • Add rates, subtract leaks — workers and inflow pipes add to the rate; leaks and outflow pipes subtract. Time = total work ÷ net rate.
  • Scale with Men × Days — for the same job, men times days stays constant, so more workers means proportionally fewer days.
  • Split pay by contribution — wages divide in the ratio of work actually done, the inverse of the time each person takes.

Techniques & Methods

  • LCM-of-days units — let the job = LCM of the days; each rate is units/day; add them. e.g. 10 and 15 days → job 30 units → 3 + 2 = 5/day → 6 days.
  • Negative rate for leaks — a leak is a worker with a minus rate. e.g. fill 6 h, leak 8 h → 1/6 − 1/8 = 1/24 → 24 h.
  • Men × Days is constant — for the same job, M1×D1 = M2×D2. e.g. 6×10 = 12×5 days.
  • Pay follows efficiency — wages split in the ratio 1/a : 1/b. e.g. 10 and 15 days → 3 : 2.
The Edge
Stop juggling fractions — take the LCM of the days as the total work units. If A=10 and B=15 days, let the job be 30 units: A does 3/day, B does 2/day, together 5/day → 30/5 = 6 days. For a leaking tank, the leak is just a worker with a negative rate — add it like any other pipe.
Worked example
Pipe A fills a tank in 12 hours; pipe B empties it in 18 hours. If both are open, how long to fill?
  1. Think in rates per hour. The filling pipe adds 1/12 of the tank each hour; the emptying pipe removes 1/18 each hour.
  2. Because one fills and one empties, combine them with a minus sign: net rate = 1/12 − 1/18.
  3. Use a common denominator of 36: 3/36 − 2/36 = 1/36 of the tank per hour.
  4. Filling 1/36 of the tank each hour means the whole tank takes 36 hours.
Worked example
12 men can finish a job in 18 days. After 6 days, 4 more men join. How many more days to finish?
  1. Measure the whole job in man-days: 12 men × 18 days = 216 man-days of work.
  2. In the first 6 days, 12 men complete 12 × 6 = 72 man-days, leaving 216 − 72 = 144 man-days.
  3. From day 7 onward there are 12 + 4 = 16 men working.
  4. Remaining time = remaining work ÷ workers = 144 ÷ 16 = 9 days.

Worked Drills

Worked example
A and B can do a job in 10 and 15 days. They start together but A leaves 5 days before completion. The work lasts:
  1. Let total time be T. A works (T−5) days at 1/10/day; B works all T days at 1/15/day.
  2. Set up: (T−5)/10 + T/15 = 1.
  3. Solving gives T = 9 days.
Worked example
A is 50% more efficient than B, who alone finishes in 30 days. A alone finishes in:
  1. B's rate = 1/30. A is 50% more efficient, so A's rate = 1.5 × 1/30 = 1/20.
  2. A alone therefore takes 20 days.
Worked example
A+B finish in 8 days, B+C in 12 days, C+A in 8 days. B alone takes:
  1. Add all three pairs: 2(A+B+C) = 1/8 + 1/12 + 1/8 = 1/3, so A+B+C = 1/6.
  2. B = (A+B+C) − (C+A) = 1/6 − 1/8 = 1/24.
  3. So B alone takes 24 days.
Worked example
12 men finish a job in 36 days. After 16 days, 6 men leave. The remaining work takes:
  1. Total work = 12 × 36 = 432 man-days. In 16 days, 12 men do 12 × 16 = 192 man-days.
  2. Work left = 432 − 192 = 240 man-days, with 6 men remaining.
  3. Remaining time = 240 ÷ 6 = 40 days.
Worked example
A tap fills a tank in 4 h, a second empties it in 6 h, a third fills it in 12 h. With all open it fills in:
  1. Net rate = 1/4 − 1/6 + 1/12 (fill, empty, fill).
  2. Common denominator 12: 3/12 − 2/12 + 1/12 = 2/12 = 1/6 per hour.
  3. Filling 1/6 per hour means the tank fills in 6 hours.
Worked example
A (20 days) and B (25 days) start together; A leaves after 5 days. B finishes the rest in:
  1. In 5 days together they do 5(1/20 + 1/25) = 5 × 9/100 = 45/100 of the job.
  2. Remaining = 55/100, done by B alone at 1/25 per day.
  3. Time = (11/20) ÷ (1/25) = 13.75 days.
Worked example
A takes 12 days and B takes 6 days. Together they finish in:
  1. Combined rate = 1/12 + 1/6 = 1/12 + 2/12 = 3/12 = 1/4 per day.
  2. Together they finish in 4 days.
Worked example
8 men or 12 women finish a job in the same time. The ratio of one man's to one woman's efficiency is:
  1. 8 men do the same work as 12 women, so 8 men = 12 women in output.
  2. man : woman = 12 : 8 = 3 : 2.
⚠ Watch out
  • You cannot average the days (A=10, B=15 does not mean 12.5) — average the rates.
  • A leak subtracts from the fill rate; watch the sign.
  • "Men × Days" is constant for the same job — use it for quick scaling.
Takeaways
  • Turn every worker, pipe and leak into a rate per unit time, then just add (subtract leaks).
  • The LCM-of-days trick replaces messy fractions with whole-number daily outputs.
  • Men × Days is constant for a fixed job — the fastest tool for scaling worker counts.
  • Pay divides in the ratio of work done, i.e. the inverse of each person's time.
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