Time and Work, Pipes and Cisterns

This section deals with problems involving the time taken to complete tasks by individuals or groups, and by pipes filling or emptying tanks. While they appear different, the core concept is the rate at which work is done.

Time and Work

Basic Concepts

The fundamental principle here is that the amount of work done is proportional to the rate of work and the time spent. If 'W' is the work, 'R' is the rate, and 'T' is the time, then:

W = R × T

In most problems, the total work is considered as 1 unit. Therefore, the rate of work is the reciprocal of the time taken to complete the work.

Rate (R) = Work (W) / Time (T)

If a person can complete a work in 'x' days, their rate of work is 1/x per day.

Key Principles and Formulas

1. Individual Rate: If A can complete a work in 'a' days, then A's rate of work is 1/a per day. Similarly, if B can complete the same work in 'b' days, B's rate is 1/b per day.

2. Combined Rate (Working Together): If A and B work together, their combined rate is the sum of their individual rates: (1/a + 1/b) per day.

3. Time Taken Together: The time taken by A and B to complete the work together is the reciprocal of their combined rate: 1 / (1/a + 1/b) = ab / (a + b) days.

4. Working Alternately: If they work on alternate days (e.g., A on day 1, B on day 2, A on day 3, etc.), the calculation becomes more complex. We calculate the work done in one cycle (e.g., 2 days) and then determine how many cycles are needed.

5. Efficiency: Efficiency is directly proportional to the rate of work. If A is twice as efficient as B, A completes twice the work in the same amount of time, or completes the same work in half the time. If A's efficiency is 'k' times B's efficiency, then A's rate is 'k' times B's rate.

Example Scenario

Let's say Person A can complete a piece of work in 10 days, and Person B can complete the same work in 15 days.

A's rate = 1/10 work per day.

B's rate = 1/15 work per day.

If they work together, their combined rate = 1/10 + 1/15.

To add these fractions, find a common denominator, which is 30:

Combined rate = 3/30 + 2/30 = 5/30 = 1/6 work per day.

The time taken by A and B to complete the work together is the reciprocal of their combined rate:

Time together = 1 / (1/6) = 6 days.

Working with Multiple People

If A, B, and C can complete a work in 'a', 'b', and 'c' days respectively, their combined rate is (1/a + 1/b + 1/c) per day.

Time taken together = 1 / (1/a + 1/b + 1/c).

Working on Alternate Days

Consider A takes 10 days and B takes 15 days, and they work on alternate days, starting with A.

A's rate = 1/10

B's rate = 1/15

Work done in 2 days (A on day 1, B on day 2) = 1/10 + 1/15 = 3/30 + 2/30 = 5/30 = 1/6 of the work.

To complete the entire work (1 unit), we need to find how many such 2-day cycles are required.

Number of cycles = Total Work / Work per cycle = 1 / (1/6) = 6 cycles.

Since each cycle is 2 days, the total time = 6 cycles × 2 days/cycle = 12 days.

Important Note: If the work is not perfectly divisible, the last person might finish the work on their turn, and the time calculation needs adjustment. For example, if work done in 2 days was 1/7, and we need to complete 1 unit, we'd see how many full 2-day cycles fit into 1 unit. If it's 5 cycles (10 days), work done is 5/7. The remaining work is 2/7. The next person to work is A. A's rate is 1/10. Time for A to do 2/7 work = (2/7) / (1/10) = 20/7 days. Total time = 10 + 20/7 days.

Leaving the Work

If a person leaves the work after some days, we calculate the work done by that person and then find the time taken by the remaining people to complete the rest of the work.

Example: A takes 10 days, B takes 15 days. They work together for 3 days. Then A leaves. How long will B take to finish the remaining work?

Combined rate of A and B = 1/10 + 1/15 = 1/6 work per day.

Work done in 3 days = (1/6) × 3 = 3/6 = 1/2 of the work.

Remaining work = 1 - 1/2 = 1/2.

B's rate = 1/15 work per day.

Time for B to finish the remaining work = Remaining Work / B's Rate = (1/2) / (1/15) = (1/2) × 15 = 7.5 days.

Efficiency Based Problems

If A is 'x' times as efficient as B, it means A's rate is 'x' times B's rate. If B takes 't' days, A will take t/x days.

Example: A is twice as efficient as B. If B can complete a work in 12 days, how long will A take?

A's efficiency = 2 × B's efficiency.

A's rate = 2 × B's rate.

B's rate = 1/12.

A's rate = 2 × (1/12) = 1/6.

Time for A = 1 / (1/6) = 6 days.

Work Done by Men, Women, and Children

These problems often involve different rates for men, women, and children. The key is to establish a relationship between their working capacities.

Example: 3 men can do a work in 10 days, and 5 women can do the same work in 12 days. How long will 2 men and 3 women take to do the same work?

Work done by 3 men in 10 days = 1 unit.

Work done by 3 men in 1 day = 1/10.

Work done by 1 man in 1 day = (1/10) / 3 = 1/30.

Work done by 5 women in 12 days = 1 unit.

Work done by 5 women in 1 day = 1/12.

Work done by 1 woman in 1 day = (1/12) / 5 = 1/60.

Now, we need the work done by 2 men and 3 women in 1 day:

Work by 2 men in 1 day = 2 × (1/30) = 2/30 = 1/15.

Work by 3 women in 1 day = 3 × (1/60) = 3/60 = 1/20.

Combined work by 2 men and 3 women in 1 day = 1/15 + 1/20.

Common denominator is 60: 4/60 + 3/60 = 7/60.

Time taken by 2 men and 3 women = 1 / (7/60) = 60/7 days.

Work and Wages

Wages are usually distributed in proportion to the amount of work done. If individuals work for different durations or at different rates, their share of wages will reflect their contribution.

Example: A and B can complete a work in 10 and 15 days respectively. They complete the work for a sum of ₹4500. What is A's share?

A's rate = 1/10

B's rate = 1/15

Ratio of their rates = (1/10) : (1/15) = 3 : 2 (multiplying by LCM of 10 and 15, which is 30).

This ratio (3:2) represents the ratio of work done by A and B in the same amount of time. Since they complete the entire work together, their shares will be in this ratio.

Total parts = 3 + 2 = 5.

A's share = (3/5) × ₹4500 = ₹2700.

B's share = (2/5) × ₹4500 = ₹1800.

Pipes and Cisterns

Basic Concepts

This topic is a direct analogy to Time and Work. Instead of people doing work, we have pipes filling or emptying a cistern (tank).

Inlet Pipes: These pipes fill the cistern. If a pipe can fill a cistern in 'x' hours, its rate of filling is +1/x of the cistern per hour.

Outlet Pipes (or Leaks): These pipes empty the cistern. If a pipe can empty a cistern in 'y' hours, its rate of emptying is -1/y of the cistern per hour.

The total capacity of the cistern is considered as 1 unit.

Key Principles and Formulas

1. Filling Rate: If pipe A can fill a cistern in 'a' hours, its rate is +1/a per hour.

2. Emptying Rate: If pipe B can empty a cistern in 'b' hours, its rate is -1/b per hour.

3. Combined Rate (Filling/Emptying): If multiple pipes are working, their rates are added algebraically. For example, if pipe A fills and pipe B empties, the net rate is (1/a - 1/b) per hour.

4. Time Taken: The time taken to fill or empty the cistern is the reciprocal of the net rate. If the net rate is positive, it means the cistern is being filled. If negative, it's being emptied.

Example Scenario

Pipe A can fill a cistern in 4 hours. Pipe B can empty the same cistern in 6 hours.

A's rate = +1/4 (filling)

B's rate = -1/6 (emptying)

If both pipes are opened simultaneously, the net rate is:

Net Rate = 1/4 - 1/6

Common denominator is 12:

Net Rate = 3/12 - 2/12 = 1/12 of the cistern per hour.

Since the net rate is positive, the cistern is being filled.

Time taken to fill the cistern = 1 / (1/12) = 12 hours.

Working with Multiple Pipes

If pipes A, B, C fill in 'a', 'b', 'c' hours respectively, and pipe D empties in 'd' hours:

Net Rate = (1/a + 1/b + 1/c) - 1/d.

Time taken to fill = 1 / Net Rate.

Cistern Partially Filled or Empty

If a cistern is already partially filled, we calculate the remaining capacity to be filled and use the net filling rate.

Example: A cistern has a capacity of 1000 liters. Pipe A can fill it in 5 hours. Pipe B can empty it in 10 hours. If the cistern is half-filled, and both pipes are opened, how long will it take to fill the remaining half?

A's rate = +1/5

B's rate = -1/10

Net rate = 1/5 - 1/10 = 2/10 - 1/10 = 1/10 of the cistern per hour.

Remaining capacity to fill = 1/2 of the cistern.

Time = Remaining Capacity / Net Rate = (1/2) / (1/10) = (1/2) × 10 = 5 hours.

Leaks in the Cistern

Leaks are treated as outlet pipes.

Example: A tap can fill a cistern in 8 hours. Due to a leak, it takes 10 hours to fill the cistern. How long would the leak alone take to empty the full cistern?

Rate of tap filling = +1/8.

Combined rate of tap and leak = +1/10 (since it takes longer to fill, the leak is slowing down the filling process).

Let the leak's rate be -1/L (where L is the time taken by the leak to empty).

So, (1/8) + (-1/L) = 1/10

1/8 - 1/L = 1/10

1/L = 1/8 - 1/10

Common denominator is 40:

1/L = 5/40 - 4/40 = 1/40.

L = 40 hours.

The leak alone can empty the cistern in 40 hours.

Cisterns with Fixed Capacity

Sometimes, problems specify the capacity (e.g., in liters) and the rate of flow (e.g., liters per minute/hour). In such cases, we can work with capacities directly.

Example: A pipe of flow rate 10 liters/minute fills a tank of capacity 12000 liters. There is a leak in the tank which empties at 4 liters/minute. How long will it take to fill the tank?

Effective filling rate = Filling rate - Emptying rate

Effective filling rate = 10 liters/minute - 4 liters/minute = 6 liters/minute.

Time to fill = Total Capacity / Effective Filling Rate

Time = 12000 liters / 6 liters/minute = 2000 minutes.

To convert to hours: 2000 minutes / 60 minutes/hour = 200/6 = 100/3 hours = 33 hours and 20 minutes.

Important Considerations for Both Topics

1. Units: Ensure consistency in units of time (days, hours, minutes). Convert if necessary.

2. LCM Method: For problems involving whole numbers of days/hours, using the Least Common Multiple (LCM) of the times taken can simplify calculations. The LCM represents a convenient 'total work' or 'total capacity' unit.

Example (Time & Work): A takes 10 days, B takes 15 days. LCM(10, 15) = 30. Assume total work = 30 units.

A's rate = 30 units / 10 days = 3 units/day.

B's rate = 30 units / 15 days = 2 units/day.

Combined rate = 3 + 2 = 5 units/day.

Time taken together = Total Work / Combined Rate = 30 units / 5 units/day = 6 days.

Example (Pipes & Cisterns): Pipe A fills in 4 hours, B empties in 6 hours. LCM(4, 6) = 12. Assume cistern capacity = 12 units.

A's filling rate = 12 units / 4 hours = +3 units/hour.

B's emptying rate = 12 units / 6 hours = -2 units/hour.

Net filling rate = 3 - 2 = +1 unit/hour.

Time to fill = Total Capacity / Net Filling Rate = 12 units / 1 unit/hour = 12 hours.

3. Positive vs. Negative Rates: Always be mindful of whether a pipe/person is adding to the work (filling/helping) or removing from it (emptying/hindering). Use positive signs for filling/helping and negative signs for emptying/hindering.

4. Problem Interpretation: Read the question carefully to understand who is working, for how long, and what is the exact question being asked (time to complete, work done, share of wages, etc.).