Time and Work, Pipes and Cisterns
CSAT · Quantitative Aptitude
Introduction
Time-and-work questions look varied but share one structure: work = rate × time. The LCM (total units) method avoids fractions and is the fastest approach under exam pressure.
Key concepts
The LCM method
Take the total work as the LCM of the individual times. If A takes 12 days and B 18 days, total = 36 units; A does 3 units a day, B 2, together 5, so they need 36/5 = 7.2 days.
Efficiency
If A is twice as efficient as B, A does 2 units for every 1 unit of B and takes half the time.
Someone leaves or joins
Split the job into phases: work done together for x days, then the remainder by whoever is left. Add the phase times.
Wages
Wages are shared in the ratio of work done (not days worked): rate × days for each person.
Pipes and cisterns
Pipes are workers: a filling pipe adds units, an emptying pipe (leak) removes them. Net rate = sum of filling rates − sum of emptying rates. If the net is negative, the tank empties.
Men–days–hours
M₁D₁H₁/W₁ = M₂D₂H₂/W₂: more men, days or hours mean more work; use it for 'how many men are needed' questions.
- Time together = 1/(1/a + 1/b) = ab/(a + b)
- Total work = LCM of times; rate = total ÷ time
- Net pipe rate = Σ filling − Σ emptying
- M₁D₁H₁/W₁ = M₂D₂H₂/W₂
Solved examples
Example 1
A can finish a piece of work in 6 days and B in 15 days. In how many days can they finish it working together?
(a) 21 days (b) 10 1/2 days (c) 5 2/7 days (d) 4 2/7 days
- Take total work = LCM(6, 15) = 30 units.
- A does 5 units/day, B does 2 units/day; together 7.
- Time = 30 ÷ 7 = 4 2/7 days.
Answer: (d) 4 2/7 days
Example 2
Pipes A and B can fill a tank in 6 and 15 hours. Pipe C can empty the full tank in 12 hours. If all three are opened together, how long will it take to fill the empty tank?
(a) 3 3/19 hours (b) 8 2/3 hours (c) 6 2/3 hours (d) 4 2/7 hours
- Tank = LCM(6, 15, 12) = 60 units.
- A fills 10, B fills 4, C empties 5 units/hour.
- Net = 9 units/hour → time = 60 ÷ 9 = 6 2/3 hours.
Answer: (c) 6 2/3 hours
Example 3
Pipes A and B can fill a tank in 20 and 6 hours. Pipe C can empty the full tank in 20 hours. If all three are opened together, how long will it take to fill the empty tank?
(a) 6 hours (b) 4 8/13 hours (c) 8 hours (d) 3 3/4 hours
- Tank = LCM(20, 6, 20) = 60 units.
- A fills 3, B fills 10, C empties 3 units/hour.
- Net = 10 units/hour → time = 60 ÷ 10 = 6 hours.
Answer: (a) 6 hours
Example 4
Pipes A and B can fill a tank in 8 and 10 hours. Pipe C can empty the full tank in 30 hours. If all three are opened together, how long will it take to fill the empty tank?
(a) 7 5/23 hours (b) 5 5/23 hours (c) 4 4/9 hours (d) 3 27/31 hours
- Tank = LCM(8, 10, 30) = 120 units.
- A fills 15, B fills 12, C empties 4 units/hour.
- Net = 23 units/hour → time = 120 ÷ 23 = 5 5/23 hours.
Answer: (b) 5 5/23 hours
Example 5
A can finish a piece of work in 40 days and B in 36 days. In how many days can they finish it working together?
(a) 38 days (b) 18 18/19 days (c) 19 18/19 days (d) 76 days
- Take total work = LCM(40, 36) = 360 units.
- A does 9 units/day, B does 10 units/day; together 19.
- Time = 360 ÷ 19 = 18 18/19 days.
Answer: (b) 18 18/19 days
Example 6
Pipes A and B can fill a tank in 20 and 15 hours. Pipe C can empty the full tank in 12 hours. If all three are opened together, how long will it take to fill the empty tank?
(a) 30 hours (b) 5 hours (c) 32 hours (d) 8 4/7 hours
- Tank = LCM(20, 15, 12) = 60 units.
- A fills 3, B fills 4, C empties 5 units/hour.
- Net = 2 units/hour → time = 60 ÷ 2 = 30 hours.
Answer: (a) 30 hours
Example 7
Pipes A and B can fill a tank in 6 and 12 hours. Pipe C can empty the full tank in 30 hours. If all three are opened together, how long will it take to fill the empty tank?
(a) 6 8/13 hours (b) 4 8/13 hours (c) 4 hours (d) 3 9/17 hours
- Tank = LCM(6, 12, 30) = 60 units.
- A fills 10, B fills 5, C empties 2 units/hour.
- Net = 13 units/hour → time = 60 ÷ 13 = 4 8/13 hours.
Answer: (b) 4 8/13 hours
Example 8
A and B can do a job in 10 and 20 days respectively. They start together, but A leaves after 2 days. In how many days in total is the job finished?
(a) 16 days (b) 14 days (c) 6 2/3 days (d) 17 days
- Total work = 20 units; A = 2/day, B = 1/day.
- In 2 days together: 2 × 3 = 6 units.
- Remaining 14 units by B alone: 14 ÷ 1 = 14 days.
- Total = 2 + 14 = 16 days.
Answer: (a) 16 days
- Always convert to units per day – adding rates is easy, adding times is wrong.
- ab/(a + b) for two people is quickest when the numbers are small.
- If A, B, C together take x days and A, B take y days, C alone takes xy/(y − x).
- Adding times (12 + 18 days) instead of rates.
- Forgetting that a leak works against the filling pipes.
- Sharing wages by days instead of work done.
Practice set
Attempt all 12 questions (about 90 seconds each), then check the worked solutions below.
Q1. Pipes A and B can fill a tank in 6 and 8 hours. Pipe C can empty the full tank in 20 hours. If all three are opened together, how long will it take to fill the empty tank?
(a) 6 4/29 hours (b) 4 4/29 hours (c) 2 38/41 hours (d) 3 3/7 hours
Q2. A and B can do a job in 40 and 10 days respectively. They start together, but A leaves after 2 days. In how many days in total is the job finished?
(a) 8 days (b) 7 1/2 days (c) 9 1/2 days (d) 10 1/2 days
Q3. A is three times as efficient as B. B alone can finish a job in 36 days. How long will they take together?
(a) 7 1/5 days (b) 12 days (c) 18 days (d) 9 days
Q4. A can finish a piece of work in 20 days and B in 36 days. In how many days can they finish it working together?
(a) 28 days (b) 13 6/7 days (c) 56 days (d) 12 6/7 days
Q5. Pipes A and B can fill a tank in 20 and 12 hours. Pipe C can empty the full tank in 12 hours. If all three are opened together, how long will it take to fill the empty tank?
(a) 22 hours (b) 20 hours (c) 4 8/13 hours (d) 7 1/2 hours
Q6. Pipes A and B can fill a tank in 20 and 8 hours. Pipe C can empty the full tank in 40 hours. If all three are opened together, how long will it take to fill the empty tank?
(a) 6 2/3 hours (b) 5 5/7 hours (c) 5 hours (d) 8 2/3 hours
Q7. Pipes A and B can fill a tank in 6 and 10 hours. Pipe C can empty the full tank in 20 hours. If all three are opened together, how long will it take to fill the empty tank?
(a) 3 3/4 hours (b) 6 8/13 hours (c) 4 8/13 hours (d) 3 3/19 hours
Q8. Pipes A and B can fill a tank in 20 and 6 hours. Pipe C can empty the full tank in 12 hours. If all three are opened together, how long will it take to fill the empty tank?
(a) 3 1/3 hours (b) 7 1/2 hours (c) 4 8/13 hours (d) 9 1/2 hours
Q9. A can finish a piece of work in 30 days and B in 36 days. In how many days can they finish it working together?
(a) 66 days (b) 17 4/11 days (c) 16 4/11 days (d) 33 days
Q10. Pipes A and B can fill a tank in 15 and 10 hours. Pipe C can empty the full tank in 24 hours. If all three are opened together, how long will it take to fill the empty tank?
(a) 8 hours (b) 10 hours (c) 6 hours (d) 4 4/5 hours
Q11. A and B can do a job in 20 and 15 days respectively. They start together, but A leaves after 5 days. In how many days in total is the job finished?
(a) 6 1/4 days (b) 12 1/4 days (c) 8 4/7 days (d) 11 1/4 days
Q12. A and B can do a job in 40 and 24 days respectively. They start together, but A leaves after 3 days. In how many days in total is the job finished?
(a) 22 1/5 days (b) 15 days (c) 19 1/5 days (d) 23 1/5 days
Q1 – (b) 4 4/29 hours. Tank = LCM(6, 8, 20) = 120 units. A fills 20, B fills 15, C empties 6 units/hour. Net = 29 units/hour → time = 120 ÷ 29 = 4 4/29 hours.
Q2 – (c) 9 1/2 days. Total work = 40 units; A = 1/day, B = 4/day. In 2 days together: 2 × 5 = 10 units. Remaining 30 units by B alone: 30 ÷ 4 = 7 1/2 days. Total = 2 + 7 1/2 = 9 1/2 days.
Q3 – (d) 9 days. A takes 36 ÷ 3 = 12 days. If B does 1 unit/day, A does 3; together 4 units/day. Work = 36 units → 36 ÷ 4 = 9 days.
Q4 – (d) 12 6/7 days. Take total work = LCM(20, 36) = 180 units. A does 9 units/day, B does 5 units/day; together 14. Time = 180 ÷ 14 = 12 6/7 days.
Q5 – (b) 20 hours. Tank = LCM(20, 12, 12) = 60 units. A fills 3, B fills 5, C empties 5 units/hour. Net = 3 units/hour → time = 60 ÷ 3 = 20 hours.
Q6 – (a) 6 2/3 hours. Tank = LCM(20, 8, 40) = 40 units. A fills 2, B fills 5, C empties 1 units/hour. Net = 6 units/hour → time = 40 ÷ 6 = 6 2/3 hours.
Q7 – (c) 4 8/13 hours. Tank = LCM(6, 10, 20) = 60 units. A fills 10, B fills 6, C empties 3 units/hour. Net = 13 units/hour → time = 60 ÷ 13 = 4 8/13 hours.
Q8 – (b) 7 1/2 hours. Tank = LCM(20, 6, 12) = 60 units. A fills 3, B fills 10, C empties 5 units/hour. Net = 8 units/hour → time = 60 ÷ 8 = 7 1/2 hours.
Q9 – (c) 16 4/11 days. Take total work = LCM(30, 36) = 180 units. A does 6 units/day, B does 5 units/day; together 11. Time = 180 ÷ 11 = 16 4/11 days.
Q10 – (a) 8 hours. Tank = LCM(15, 10, 24) = 120 units. A fills 8, B fills 12, C empties 5 units/hour. Net = 15 units/hour → time = 120 ÷ 15 = 8 hours.
Q11 – (d) 11 1/4 days. Total work = 60 units; A = 3/day, B = 4/day. In 5 days together: 5 × 7 = 35 units. Remaining 25 units by B alone: 25 ÷ 4 = 6 1/4 days. Total = 5 + 6 1/4 = 11 1/4 days.
Q12 – (a) 22 1/5 days. Total work = 120 units; A = 3/day, B = 5/day. In 3 days together: 3 × 8 = 24 units. Remaining 96 units by B alone: 96 ÷ 5 = 19 1/5 days. Total = 3 + 19 1/5 = 22 1/5 days.
- Work = rate × time
- LCM units method
- Leak = negative worker
- Wages ∝ work done
