Time and Work, Pipes and Cisterns: Methods and Worked Examples

Time and work questions ask how long a job takes when one or more people, machines or pipes work at fixed speeds. They appear in almost every Lok Sewa maths and mental ability paper. Only two simple methods are needed, the one-day work method and the LCM method, and this note shows both with clear worked examples, including wages and pipes.

1. The basic idea: one day’s work

The whole job is taken as 1 complete work.

• If a person can finish the work in n days, then in one day he does 1/n of the work.
• If a person does 1/n of the work in one day, he finishes the whole work in n days.
• Work done in several days = one day’s work × number of days.

Worked examples:

• Asha can finish a job in 12 days. In one day she does 1/12 of it. In 3 days she does 3/12 = 1/4 of the job.
• Bikash does 1/9 of a job each day. He needs 9 days to finish it.

2. Working together

When people work together, add their one-day works. The time taken together is 1 divided by this sum.

Two people: if A takes x days and B takes y days, together they take x × y/(x + y) days.

Worked examples:

• A takes 12 days and B takes 18 days. One day’s work together = 1/12 + 1/18 = 3/36 + 2/36 = 5/36. Time = 36/5 = 7.2 days. By the formula: 12 × 18/30 = 216/30 = 7.2 days.
• A, B and C take 10, 15 and 30 days. Together: 1/10 + 1/15 + 1/30 = 3/30 + 2/30 + 1/30 = 6/30 = 1/5. They finish in 5 days.

Finding one person’s time: A and B together take 8 days and A alone takes 12 days. B’s one day’s work = 1/8 − 1/12 = 3/24 − 2/24 = 1/24, so B alone takes 24 days.

Pairs given: A and B take 10 days, B and C take 12 days, C and A take 15 days. Adding: 2 × (A + B + C) = 1/10 + 1/12 + 1/15 = 6/60 + 5/60 + 4/60 = 15/60 = 1/4. So A + B + C = 1/8, and all three together take 8 days.

Half the work: if A does half the work in 5 days, he needs 10 days for the full work.

3. The LCM (total work) method

Fractions can be avoided. Take the total work = LCM of the given days, measured in units.

1. Total work = LCM of the days.
2. Each person’s units per day = total work ÷ his days.
3. Add the units per day for people working together.
4. Time = units of work ÷ units per day.

Worked example: A takes 12 days, B takes 18 days. LCM = 36 units. A does 3 units a day and B does 2 units a day. Together 5 units a day, so time = 36 ÷ 5 = 7.2 days (same answer as before).

Someone leaves: A takes 20 days and B takes 30 days. They work together for 6 days, then A leaves. Total = 60 units; A does 3 and B does 2 units a day. In 6 days they do 6 × 5 = 30 units. Remaining 30 units are done by B in 30 ÷ 2 = 15 days. Total time = 6 + 15 = 21 days.

One starts, other finishes: A can do a job in 20 days. He works for 5 days and does 5/20 = 1/4. B finishes the remaining 3/4 in 9 days, so B alone would take 9 × 4/3 = 12 days.

Efficiency: efficiency is the work done per day. More efficient means less time; time is inversely proportional to efficiency.

• A is twice as efficient as B, and B takes 30 days. Then A takes 15 days. Together: 1/15 + 1/30 = 3/30 = 1/10, so 10 days.
• If B is 50% more efficient than A, the efficiency ratio A : B = 2 : 3, so B takes 2/3 of A’s time. If A takes 15 days, B takes 10 days.

4. Work and wages

When people work together on one job for the same number of days, the total wage is shared in the ratio of the work each person does, which equals the ratio of their efficiencies (units per day).

Worked examples:

• A takes 10 days and B takes 15 days. They work together and receive Rs 7,500. LCM = 30 units; A does 3 and B does 2 units a day. Ratio 3 : 2. A gets 7,500 × 3/5 = Rs 4,500 and B gets Rs 3,000.
• A takes 6 days and B takes 12 days. With the help of C they finish in 3 days and receive Rs 2,400. Total = 12 units; A does 2 and B does 1 unit a day. Together all three do 12 ÷ 3 = 4 units a day, so C does 4 − 3 = 1 unit a day. Work in 3 days: A 6, B 3, C 3 units, ratio 2 : 1 : 1. Shares: A Rs 1,200, B Rs 600, C Rs 600.

Note: wages depend on work done, not on how many days a person takes alone. The faster worker earns more.

5. Men-days and man-hours

If a group of workers of equal ability does a fixed job, the total work stays the same however many workers there are. So:

M1 × D1 = M2 × D2 (M = men, D = days)

With hours per day (H) and different amounts of work (W):

M1 × D1 × H1 / W1 = M2 × D2 × H2 / W2

Worked examples:

• 12 men build a house in 20 days. 16 men will take 12 × 20/16 = 15 days.
• 10 men working 8 hours a day finish a job in 12 days, which is 960 man-hours. 16 men working 6 hours a day need 960 ÷ (16 × 6) = 10 days.
• 15 men dig a 30 m trench in 8 days. Men needed to dig a 50 m trench in 10 days = 15 × 8 × 50/(30 × 10) = 6,000/300 = 20 men.

More men needed: 20 men can do a job in 12 days. To finish it in 8 days, men needed = 20 × 12/8 = 30, so 10 more men must join.

6. Pipes and cisterns

A tank filled by pipes is a time and work question in another form.

• An inlet pipe fills the tank; its work is added (positive).
• An outlet pipe or a leak empties the tank; its work is subtracted (negative).
• If a pipe fills a tank in x hours, it fills 1/x of the tank in one hour.

Worked examples:

• Pipe A fills a tank in 6 hours and pipe B in 12 hours. Together: 1/6 + 1/12 = 3/12 = 1/4, so 4 hours.
• An inlet fills a tank in 10 hours and an outlet empties it in 15 hours. Both open: 1/10 − 1/15 = 3/30 − 2/30 = 1/30, so the tank fills in 30 hours.
• Pipes A and B fill a tank in 12 and 15 hours; pipe C empties it in 20 hours. LCM = 60 units. Per hour: 5 + 4 − 3 = 6 units. All open: 60 ÷ 6 = 10 hours.
• Leak: a pipe normally fills a tank in 8 hours, but because of a leak it takes 10 hours. Leak’s work per hour = 1/8 − 1/10 = 5/40 − 4/40 = 1/40. The leak alone empties the full tank in 40 hours.

The LCM method works here too: take the tank as LCM units and give an outlet a negative number of units per hour.

7. Step-by-step method

1. Write each person’s or pipe’s time to finish the whole job.
2. Turn each into one day’s (or one hour’s) work, either as 1/n or as units using the LCM.
3. Add for workers or inlets; subtract for outlets and leaks.
4. Divide the work left by the work per day to get the time.
5. For wages, share in the ratio of work done. For groups of men, use M1 × D1 × H1 / W1 = M2 × D2 × H2 / W2.