Time and Work

The 'Time and Work' section is a crucial part of the quantitative aptitude syllabus for competitive exams like RRB NTPC. It tests your ability to understand and solve problems related to the time taken by individuals or groups to complete a task, working at different rates. This topic often involves concepts like efficiency, combined work, and work done by multiple people over a period. Mastering this topic requires a clear understanding of the fundamental principles and the ability to apply them to various scenarios.

Basic Concepts

The core idea behind 'Time and Work' problems is the relationship between the amount of work done, the rate at which it is done, and the time taken.

Work (W): This is the total task that needs to be completed. It can be represented as a unit, for example, '1 unit of work'.

Time (T): This is the duration taken to complete the work.

Rate or Efficiency (R): This is the amount of work done per unit of time. It represents how fast someone or something can complete a task.

The fundamental formula connecting these three is:

Work = Rate × Time

From this, we can derive:

  • Rate = Work / Time
  • Time = Work / Rate

In most 'Time and Work' problems, the 'Work' is considered as a single unit (i.e., W = 1). Therefore, the relationship simplifies to:

1 = Rate × Time

This implies that Rate and Time are inversely proportional. If the rate of work increases, the time taken decreases, and vice versa, assuming the total work remains constant.

Key Takeaway: The faster you work (higher rate), the less time you need to complete the same job.

Calculating Individual Rates

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

Example: If Ram can complete a work in 10 days, then Ram's rate of work per day is 1/10th of the work. This means in one day, Ram completes 1/10th of the total job.

Conversely, if a person's rate of work is 1/y per day, then the total time taken to complete the work is 'y' days.

Example: If Sita's rate of work is 1/15th of the work per day, she will take 15 days to complete the entire work.

Shortcut: If a person takes 'D' days to complete a work, their efficiency (work done per day) is 1/D.

Combined Work - Multiple People Working Together

When two or more people work together, their rates of work add up to form a combined rate.

Suppose Person A can complete a work in 'a' days and Person B can complete the same work in 'b' days.

  • Rate of Person A per day = 1/a
  • Rate of Person B per day = 1/b

When they work together, their combined rate per day is the sum of their individual rates:

Combined Rate = (Rate of A) + (Rate of B) = 1/a + 1/b

To find the total time taken by them to complete the work together, we use the formula:

Time = Work / Combined Rate

Since Work = 1,

Time taken together = 1 / (Combined Rate) = 1 / (1/a + 1/b)

Simplifying the denominator: 1/a + 1/b = (b + a) / ab

So, Time taken together = 1 / [(a + b) / ab] = ab / (a + b) days.

Formula for Two People: If A takes 'a' days and B takes 'b' days, time taken together = (a × b) / (a + b) days.

Example: Ram can complete a work in 10 days, and Sita can complete the same work in 15 days. How long will it take for them to complete the work together?

  • Ram's rate = 1/10 work per day
  • Sita's rate = 1/15 work per day
  • Combined rate = 1/10 + 1/15 = (3 + 2) / 30 = 5/30 = 1/6 work per day
  • Time taken together = 1 / (1/6) = 6 days.

Using the shortcut formula: Time = (10 × 15) / (10 + 15) = 150 / 25 = 6 days.

Work Done by Multiple People (More than Two)

The principle remains the same. If A, B, and C can complete a work in 'a', 'b', and 'c' days respectively, their individual rates are 1/a, 1/b, and 1/c.

Their combined rate is: 1/a + 1/b + 1/c

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

Example: A can do a piece of work in 6 days, B in 8 days, and C in 12 days. In how many days can they together finish the work?

  • A's rate = 1/6
  • B's rate = 1/8
  • C's rate = 1/12
  • Combined rate = 1/6 + 1/8 + 1/12
  • To add these fractions, find the LCM of 6, 8, and 12, which is 24.
  • Combined rate = (4/24) + (3/24) + (2/24) = (4 + 3 + 2) / 24 = 9/24 = 3/8 work per day.
  • Time taken together = 1 / (3/8) = 8/3 days.

So, they can finish the work together in 8/3 days, or 2 and 2/3 days.

Working in Alternating Days

Sometimes, problems involve individuals working on alternate days. For example, A works on day 1, B on day 2, A on day 3, B on day 4, and so on.

To solve these, calculate the total work done by them in one cycle (e.g., 2 days in the A, B alternating case).

Example: A can do a work in 10 days and B can do the same work in 15 days. They work in alternate days, starting with A. In how many days will the work be completed?

  • A's rate = 1/10
  • B's rate = 1/15
  • Work done by A in 1 day = 1/10
  • Work done by B in 1 day = 1/15
  • In 2 days (1 cycle of A then B):
  • Work done = Work by A + Work by B = 1/10 + 1/15 = (3 + 2) / 30 = 5/30 = 1/6 of the work.
  • So, in 2 days, 1/6 of the work is completed.
  • To complete the whole work (1 unit), they need 6 such cycles.
  • Total time = 6 cycles × 2 days/cycle = 12 days.

Important Note: This assumes the work is perfectly divisible. If the remaining work after several full cycles is less than what one person can do in a day, you need to calculate the specific day.

Let's consider a variation: A can do a work in 6 days, B in 5 days. They work on alternate days, starting with A. In how many days will the work be completed?

  • A's rate = 1/6
  • B's rate = 1/5
  • Work in 2 days (A then B) = 1/6 + 1/5 = (5 + 6) / 30 = 11/30.
  • Total work = 1.
  • Let's see how many full cycles of 2 days are needed.
  • If we do 2 cycles (4 days): Work done = 2 * (11/30) = 22/30. Remaining work = 1 - 22/30 = 8/30.
  • On the 5th day, A starts working. A does 1/6 work per day.
  • Remaining work = 8/30. A's rate = 1/6 = 5/30.
  • Since 8/30 > 5/30, A can complete a portion of his work on the 5th day.
  • Time taken by A to do 8/30 work = Work / Rate = (8/30) / (5/30) = 8/5 days.
  • Total time = 4 days (for 2 cycles) + 8/5 days (by A) = 4 + 1.6 = 5.6 days.

This calculation can be tricky. A simpler way for such cases:

  • Work done in 2 days = 11/30.
  • Total work = 1.
  • Number of cycles = Total Work / Work per cycle = 1 / (11/30) = 30/11 ≈ 2.72 cycles.
  • So, we can complete 2 full cycles.
  • Time for 2 cycles = 2 * 2 = 4 days.
  • Work done in 4 days = 2 * (11/30) = 22/30.
  • Remaining work = 1 - 22/30 = 8/30.
  • On the 5th day, A starts. A's rate is 1/6.
  • Time taken by A to finish the remaining 8/30 work = (8/30) / (1/6) = (8/30) * 6 = 48/30 = 8/5 days.
  • Total time = 4 days + 8/5 days = 4 + 1.6 = 5.6 days.
Alternating Days Strategy: Calculate work done in one cycle (e.g., 2 days for A then B). Determine how many full cycles can be completed without exceeding the total work. Calculate remaining work. Determine who works next and calculate the fraction of the day they take to finish the remaining work.

Men, Women, and Children

These problems often involve different types of workers (men, women, children) who have different efficiencies. The key is to establish the relationship between their efficiencies.

Example: 3 men can do a work in 10 days. 5 women can do the same work in 12 days. 10 children can do the same work in 15 days. How many days will it take for 1 man, 1 woman, and 1 child to complete the work together?

  • Step 1: Find the total work units.
  • Total work by 3 men in 10 days = 3 men × 10 days = 30 man-days.
  • Total work by 5 women in 12 days = 5 women × 12 days = 60 woman-days.
  • Total work by 10 children in 15 days = 10 children × 15 days = 150 child-days.
  • Let's assume the total work is the LCM of the man-days, woman-days, and child-days, or a value that makes calculations easy. A common approach is to equate them. Let W be the total work.
  • W = 3 men × 10 days => 1 man's work per day = W / 30
  • W = 5 women × 12 days => 1 woman's work per day = W / 60
  • W = 10 children × 15 days => 1 child's work per day = W / 150
  • Let's take W = 300 units (LCM of 30, 60, 150 is 300).
  • Total work = 300 units.
  • 1 man's rate = 300 / 30 = 10 units/day.
  • 1 woman's rate = 300 / 60 = 5 units/day.
  • 1 child's rate = 300 / 150 = 2 units/day.
  • Step 2: Find the combined rate of 1 man, 1 woman, and 1 child.
  • Combined rate = 10 + 5 + 2 = 17 units/day.
  • Step 3: Calculate the time taken together.
  • Time = Total Work / Combined Rate = 300 units / 17 units/day = 300/17 days.

Alternative using ratios of efficiency:

  • 3M = 10 Days => Total work = 3M × 10 = 30 M-days
  • 5W = 12 Days => Total work = 5W × 12 = 60 W-days
  • 10C = 15 Days => Total work = 10C × 15 = 150 C-days
  • Equating the work: 30 M = 60 W = 150 C
  • Divide by 30: 1 M = 2 W = 5 C
  • This means 1 man's efficiency is equal to 2 women's efficiency, and 1 woman's efficiency is equal to 5 children's efficiency.
  • Let's express everything in terms of the least efficient unit, say Children (C).
  • 1 W = 5 C
  • 1 M = 2 W = 2 * (5 C) = 10 C
  • Now, convert the total work into 'child-days'. Use any of the given conditions. Let's use 10 children working for 15 days.
  • Total Work = 10 C × 15 days = 150 C-days.
  • We need to find the time for 1 Man + 1 Woman + 1 Child.
  • Their combined efficiency in terms of children = 1 M + 1 W + 1 C = 10 C + 5 C + 1 C = 16 C.
  • Time = Total Work / Combined Efficiency = 150 C-days / 16 C = 150/16 days = 75/8 days.

Wait, there's a discrepancy. Let's recheck the first method's LCM. LCM(30, 60, 150) = 300.

Man-rate = 300/30 = 10. Woman-rate = 300/60 = 5. Child-rate = 300/150 = 2. Combined Rate = 10 + 5 + 2 = 17. Time = 300/17 days.

Let's recheck the ratio method. 3M = 60W = 150C. Divide by 30. 1M = 2W = 5C. This implies: 1M = 2W 1W = 5C 1M = 5C (This is wrong, 1M = 2W = 2(5C) = 10C)

Okay, the ratio is correct: 1M : 2W : 5C. Let's express everything in terms of M. 1W = 1/2 M 1C = 1/5 M Total work from 3M in 10 days = 30 M-days. Total work from 5W in 12 days = 5W * 12 = 5 * (1/2 M) * 12 = 30 M-days. Total work from 10C in 15 days = 10C * 15 = 10 * (1/5 M) * 15 = 30 M-days. All consistent. Total work = 30 M-days.

Now, we need the time for 1M + 1W + 1C. Their combined efficiency in terms of M = 1M + 1W + 1C = 1M + (1/2 M) + (1/5 M) = M * (1 + 1/2 + 1/5) = M * (10/10 + 5/10 + 2/10) = M * (17/10). So, their combined rate is 17/10 Man-equivalents per day.

Time = Total Work / Combined Rate = (30 M-days) / (17/10 M/day) = 30 * (10/17) days = 300/17 days.

Both methods now agree. The first method using LCM is often more intuitive if you can find a suitable LCM. The ratio method is more robust.

Men, Women, Children Strategy: 1. Calculate the total work in terms of 'person-days' (e.g., man-days, woman-days) for each group. 2. Equate these total works to find the ratio of efficiencies between different types of workers (e.g., M:W:C). 3. Express the required group (e.g., 1 man, 1 woman, 1 child) in terms of a single unit of efficiency (e.g., all in terms of men or children). 4. Calculate the total work using one of the initial conditions. 5. Divide the total work by the combined efficiency of the required group to find the time taken.

Work Done and Left

In some problems, people work for a certain number of days, and then leave, or the work is stopped. You need to calculate the remaining work and the time taken to complete it by others.

Example: A can do a work in 20 days, and B can do it in 30 days. They work together for 4 days. After that, B leaves. In how many days will A complete the remaining work?

  • A's rate = 1/20
  • B's rate = 1/30
  • Combined rate of A and B = 1/20 + 1/30 = (3 + 2) / 60 = 5/60 = 1/12 work per day.
  • Work done by A and B together in 4 days = Combined rate × Time = (1/12) × 4 = 4/12 = 1/3 of the work.
  • Remaining work = Total Work - Work Done = 1 - 1/3 = 2/3 of the work.
  • Now, A has to complete the remaining work alone.
  • A's rate = 1/20 work per day.
  • Time taken by A to complete the remaining work = Remaining Work / A's Rate
  • Time = (2/3) / (1/20) = (2/3) × 20 = 40/3 days.

So, A will take 40/3 days (or 13 and 1/3 days) to complete the remaining work.

Work Left Strategy: 1. Calculate the work done by the group(s) for the specified period. 2. Subtract the work done from the total work (1) to find the remaining work. 3. Calculate the time needed by the remaining person/group to complete this remaining work using their respective rates.

Pipes and Cisterns (Related Concept)

The 'Pipes and Cisterns' topic is very similar to 'Time and Work'. Here, instead of people working, pipes fill or empty a cistern.

  • Inlet Pipes: These pipes fill the cistern. Their rate is considered positive. If a pipe can fill a cistern in 'x' hours, its filling rate is 1/x cistern per hour.
  • Outlet Pipes (or Leak): These pipes empty the cistern. Their rate is considered negative. If a pipe can empty a cistern in 'y' hours, its emptying rate is -1/y cistern per hour.

Example: Pipe A can fill a cistern in 10 hours, and Pipe B can empty it in 15 hours. If both pipes are opened simultaneously, in how many hours will the cistern be filled?

  • A's filling rate = +1/10 cistern per hour.
  • B's emptying rate = -1/15 cistern per hour.
  • Combined rate = (1/10) + (-1/15) = 1/10 - 1/15
  • Combined rate = (3 - 2) / 30 = 1/30 cistern per hour.
  • Since the combined rate is positive, the cistern will be filled.
  • Time to fill = Total Work / Combined Rate = 1 cistern / (1/30 cistern per hour) = 30 hours.
Pipes & Cisterns Analogy: Inlet pipes are like workers who complete work (positive rate). Outlet pipes/leaks are like workers who undo work (negative rate).

Efficiency-Based Problems

Problems where efficiency is directly compared.

Example: A is twice as efficient as B. Together, they can complete a work in 15 days. In how many days can A alone complete the work?

  • Efficiency of A = 2 × Efficiency of B.
  • Let efficiency of B be 'x' units per day. Then efficiency of A is '2x' units per day.
  • Combined efficiency = Efficiency of A + Efficiency of B = 2x + x = 3x units per day.
  • Total work = Combined Efficiency × Time = (3x units/day) × 15 days = 45x units.
  • Now, find the time for A alone to complete this work.
  • Time for A = Total Work / Efficiency of A = (45x units) / (2x units/day) = 45/2 days.

Using the rate method:

  • A is twice as efficient as B. This means A does twice the work B does in the same time.
  • If B takes 'b' days, A takes 'a' days. Efficiency is inversely proportional to time.
  • Efficiency(A) / Efficiency(B) = Time(B) / Time(A)
  • 2 / 1 = b / a => b = 2a.
  • So, if A takes 'a' days, B takes '2a' days.
  • Time taken together = (a × 2a) / (a + 2a) = 2a² / 3a = 2a / 3 days.
  • We are given that they complete the work together in 15 days.
  • So, 2a / 3 = 15
  • 2a = 45
  • a = 45/2 days.
  • This 'a' is the time A takes alone.
Efficiency Shortcut: If A is 'n' times as efficient as B, then A takes 1/n times the time B takes. Also, if they work together, the ratio of work done is equal to the ratio of their efficiencies.

Important Considerations and Common Pitfalls

  • Units: Ensure consistency in units (days, hours, minutes).
  • Inverse Proportionality: Remember that rate and time are inversely proportional when the work is constant.
  • LCM Calculation: Accurate calculation of LCM is key for adding fractions of work.
  • Fractional Days/Work: Be comfortable working with fractions for time and work done.
  • Misinterpreting "Working Together": Sometimes, it means simultaneous work, other times it could imply sequential work or alternating work. Read carefully.
  • Negative Work: For emptying pipes or tasks that undo previous work, use negative rates.

By understanding these concepts and practicing a variety of problems, you can effectively tackle the 'Time and Work' section in your RRB NTPC exam.

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