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Quantitative Aptitude

Time & Work

Simplification & Surds5 / 51Linear Equations
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Concepts (3)

Wages are directly proportional to work done. Calculate individual work shares or efficiencies (often using LCM method) to distribute total wages accurately and quickly.

Core Formula

The fundamental principle of Work & Wages is that wages are directly proportional to the work done. If individuals work for the same duration, wages are proportional to their efficiencies. This is crucial for SSC CGL.

  • Wages ∝ Work Done
  • If total work is done by multiple people, their share of wages will be in the ratio of the work they complete.
  • Ratio of Wages = Ratio of Work Done
  • If they work for the same number of days, Ratio of Wages = Ratio of Efficiencies.
  • If they work for different number of days, Ratio of Wages = Ratio of (Efficiency × Days Worked).

Worked Example 1

A, B, and C can complete a piece of work in 10, 12, and 15 days respectively. They start working together, and the total wages for the work are ₹6000. How much will A get if they complete the work together?

  • Step 1: Find individual efficiencies (LCM Method). Assume total work = LCM(10, 12, 15) = 60 units. A's efficiency = 60/10 = 6 units/day B's efficiency = 60/12 = 5 units/day C's efficiency = 60/15 = 4 units/day

  • Step 2: Find the ratio of their efficiencies. Since they work together for the same duration, the ratio of wages is the ratio of their efficiencies. Ratio of efficiencies (A:B:C) = 6:5:4

  • Step 3: Distribute total wages in this ratio. Total parts = 6 + 5 + 4 = 15 parts A's share = (6/15) * ₹6000 = ₹2400

Worked Example 2

P and Q undertake to do a piece of work for ₹4500. P alone can do it in 8 days and Q alone in 12 days. With the assistance of R, they complete the work in 4 days. What is R's share of the money?

  • Step 1: Find individual efficiencies. Assume total work = LCM(8, 12, 4) = 24 units. P's efficiency = 24/8 = 3 units/day Q's efficiency = 24/12 = 2 units/day (P+Q+R)'s combined efficiency = 24/4 = 6 units/day

  • Step 2: Find R's efficiency. R's efficiency = (P+Q+R)'s efficiency - (P's efficiency + Q's efficiency) R's efficiency = 6 - (3 + 2) = 6 - 5 = 1 unit/day

  • Step 3: Find the ratio of their efficiencies. Ratio of efficiencies (P:Q:R) = 3:2:1

  • Step 4: Distribute total wages in this ratio. Total parts = 3 + 2 + 1 = 6 parts R's share = (1/6) * ₹4500 = ₹750

Shortcuts & Tricks

  1. LCM Method for Efficiency: Always use the LCM of days to determine total work and individual efficiencies. This avoids fractions and simplifies calculations, boosting speed.
  2. Ratio of Work Done: If people work for different durations or complete different parts of the work, calculate the exact work done by each person and distribute wages in that ratio directly. This is often the fastest approach.
  3. Per Day Work/Wages: Sometimes, it's quicker to calculate the per-day wage for a unit of work (e.g., ₹X per unit) and then multiply by the total units of work done by each individual.
  4. Direct Proportionality: Remember Wages ∝ Work Done. If someone does twice the work, they get twice the wages. This direct relationship saves time on complex calculations.

Common Mistakes

  1. Distributing Wages Equally: Assuming everyone gets an equal share, regardless of their efficiency or actual work done. This is a common trap.
  2. Ignoring Time Factor: If individuals work for different numbers of days, simply using efficiency ratio is incorrect. The ratio should be (Efficiency × Days Worked). Always account for the time spent.
  3. Incorrectly Calculating Combined Efficiency: Especially when a third person joins or leaves, ensure the combined efficiency is correctly calculated for the relevant period. A small error here can lead to a completely wrong answer.
  4. Confusing Work Done with Time Taken: A person taking less time is more efficient and does more work in a given period, thus earning more. Don't mix up inverse and direct proportionality.

Derivation (brief)

The principle of Work & Wages is rooted in fair compensation for labor. If an individual performs a certain amount of work, they are compensated for that effort. When multiple individuals contribute to a single task, their compensation should be proportional to their individual contribution. Let W be the total work, T be the total time taken, and E be the efficiency. We know W = E × T. If a total wage ₹X is paid for W units of work, then the wage per unit of work is ₹X/W. If person A does W_A units of work, their wage Wage_A = (W_A/W) × ₹X. Since W_A = E_A × T_A, then Wage_A ∝ E_A × T_A. This means wages are directly proportional to the product of efficiency and time worked. If T_A is constant for all, then Wage_A ∝ E_A.

Advanced Examples

A contractor undertakes to complete a road in 100 days and employs 110 men. After 40 days, he finds that only 1/4th of the work is completed. How many additional men must he employ to finish the work on time? If the total contract value is ₹2,00,000, and the initial 110 men are paid for 40 days, what would be the total wages for the additional men for the remaining period if all men are paid equally per day?

  • Step 1: Analyze initial work and remaining requirements. M1*D1/W1 = M2*D2/W2 110 men work for 40 days to complete 1/4th of the work. Remaining work = 1 - 1/4 = 3/4. Remaining days = 100 - 40 = 60 days.

  • Step 2: Calculate total men required for remaining work. (110 * 40) / (1/4) = (M_total * 60) / (3/4) 110 * 40 * 4 = M_total * 60 * (4/3) 110 * 40 = M_total * 20 M_total = (110 * 40) / 20 = 220 men Additional men needed = 220 - 110 = 110 men.

  • Step 3: Calculate wages for additional men. Total work done by initial 110 men = 1/4. Total work done in remaining 60 days by 220 men = 3/4. Since all men are paid equally per day, wages are proportional to (Men × Days). Man-days for initial work = 110 men × 40 days = 4400 man-days (for 1/4 work). Man-days for remaining work = 220 men × 60 days = 13200 man-days (for 3/4 work). Total contract value = ₹2,00,000. Wage for 1 man-day = ₹2,00,000 / (4400 + 13200) = ₹2,00,000 / 17600 = ₹11.36 (approx). Wages for additional 110 men for 60 days = 110 men × 60 days × ₹11.36/man-day = 6600 man-days × ₹11.36/man-day = ₹75,000 (approx). Alternatively, using ratios: Work done by additional men in remaining period = (110 men * 60 days) / (Total man-days for entire project) = 6600 / 17600 = 3/8 of total work. So, wages = (3/8) * ₹2,00,000 = ₹75,000.

Variation Types

  1. Individual Work & Combined Wages: The most common type, where individuals work together, and wages are distributed based on their efficiency or work done. (Example 1).
  2. Partial Work & Wages: Scenarios where someone leaves, joins, or only completes a fraction of the work. Wages are distributed based on the actual contribution.
  3. Contract-based Problems: A total sum is agreed upon for a project, and wages are calculated based on the proportion of work completed by each party or individual. (Advanced Example).
  4. Men, Women, Children: Problems involving different efficiency levels for different groups (e.g., 2 men = 3 women). Wages are distributed based on their effective work units.

Time-Saving Methods

  1. Unitary Method: Calculate the wage for one unit of work or one man-day. Then multiply by the total units/man-days contributed by each person. This is highly effective for complex scenarios.
  2. Ratio Method: Directly establish the ratio of work done or (Efficiency × Days) and distribute the total wages in that ratio. This is often the fastest and most accurate method for competitive exams.
  3. Avoid Calculating Total Days: In many problems, you don't need to find the total days to complete the work if everyone works together. Just the ratio of efficiencies is enough if they work for the same duration, saving precious time.
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Pipes & Cisterns applies Time & Work to tanks. Inlet pipes do positive work, outlet pipes negative. Use combined rates or LCM method for quick, accurate solutions.

Core Formula

In Pipes & Cisterns problems, the fundamental concept is work rate. If a pipe can fill or empty a tank in 'T' hours, its work rate (the fraction of the tank filled or emptied in one hour) is 1/T. Inlet pipes have a positive work rate, while outlet pipes (or leaks) have a negative work rate.

  • Individual Pipe Rate: If a pipe fills a tank in T hours, its rate is 1/T (tank/hour).
  • Individual Outlet Rate: If a pipe empties a tank in T hours, its rate is -1/T (tank/hour).
  • Combined Rate: For multiple pipes working simultaneously, sum their individual rates. If the combined rate is positive, the tank fills; if negative, it empties. Combined Rate = Rate₁ + Rate₂ + ... + Rateₙ
  • Time Taken: The total time to fill or empty the tank is 1 / Combined Rate.

Worked Example 1

Pipe A fills a tank in 12 hours, and Pipe B fills it in 18 hours. If both pipes are opened simultaneously, in how many hours will the tank be filled?

Step-by-step solution:

  1. Rate of Pipe A: 1/12 tank/hour (filling).
  2. Rate of Pipe B: 1/18 tank/hour (filling).
  3. Combined Rate: (1/12) + (1/18) = (3/36) + (2/36) = 5/36 tank/hour.
  4. Time to Fill: 1 / (5/36) = 36/5 = 7.2 hours.

Worked Example 2

Pipe P can fill a tank in 10 hours, and Pipe Q can empty the same tank in 15 hours. If both pipes are opened at the same time, how long will it take to fill the empty tank?

Step-by-step solution:

  1. Rate of Pipe P: 1/10 tank/hour (filling).
  2. Rate of Pipe Q: -1/15 tank/hour (emptying).
  3. Combined Rate: (1/10) + (-1/15) = (3/30) - (2/30) = 1/30 tank/hour.
  4. Time to Fill: 1 / (1/30) = 30 hours.

Shortcuts & Tricks

LCM Method (Unitary Method): This is the most efficient method for competitive exams as it avoids fractions.

  1. Assume the total capacity of the tank as the LCM (Least Common Multiple) of the individual times taken by each pipe.
  2. Calculate the 'efficiency' (units of work per hour) of each pipe by dividing the total capacity by its respective time.
  3. For outlet pipes, their efficiency is negative.
  4. Sum the efficiencies to get the combined efficiency.
  5. Time = Total Capacity / Combined Efficiency.

Applying LCM Method to Example 2:

  • Pipe P fills in 10 hours, Pipe Q empties in 15 hours.
  • LCM(10, 15) = 30 units. (Assume tank capacity is 30 units).
  • Efficiency of P: 30 units / 10 hours = +3 units/hour (filling).
  • Efficiency of Q: 30 units / 15 hours = -2 units/hour (emptying).
  • Combined Efficiency: +3 - 2 = +1 unit/hour.
  • Time to Fill: 30 units / 1 unit/hour = 30 hours.

Common Mistakes

  1. Incorrect Sign for Outlet Pipes: Forgetting to assign a negative work rate or efficiency to pipes that empty the tank.
  2. Adding Times Instead of Rates: Students sometimes incorrectly add or subtract the total times taken by pipes instead of their work rates.
  3. Overlooking Initial State: Not considering if the tank is initially empty, full, or partially filled, which can affect the total work required.
  4. Alternating Pipes Error: For alternating pipe problems, failing to adjust the target work for the last cycle to prevent overfilling or over-emptying the tank.

Derivation (brief)

The fundamental principle of 'Pipes & Cisterns' is an application of 'Time & Work'. If a task (filling a tank) is considered '1 unit of work', and it takes 'T' time to complete, then the 'rate' at which work is done is Work / Time = 1/T. This rate represents the fraction of the tank filled or emptied per unit of time (e.g., per hour).

Advanced Examples

1. Alternating Pipes (Emptying a Full Tank): A tank when full can be emptied by an outlet pipe A in 5.6 hours, while an inlet pipe B can fill the same empty tank in 7 hours. If pipes A and B are turned on alternatively for 1 hour each starting with pipe A when the tank is full, how long will it take to empty the tank?

Step-by-step solution:

  1. Convert to fractions: Pipe A empties in 5.6 hours = 28/5 hours. Pipe B fills in 7 hours.
  2. LCM Method: Let the tank capacity be LCM(28, 7) = 28 units.
  3. Efficiencies:
    • Pipe A (emptying): 28 units / (28/5) hours = -5 units/hour.
    • Pipe B (filling): 28 units / 7 hours = +4 units/hour.
  4. Cycle Analysis: Pipes work alternatively for 1 hour each, starting with A.
    • Hour 1 (A): -5 units (Tank has 28 - 5 = 23 units).
    • Hour 2 (B): +4 units (Tank has 23 + 4 = 27 units).
    • Net change in 2 hours (1 cycle) = -5 + 4 = -1 unit.
  5. Target Adjustment: We need to empty 28 units. However, A will empty the tank completely in its turn. So, we calculate cycles to empty (Total Capacity - A's 1-hour work).
    • Target for cycles = 28 - 5 = 23 units.
  6. Cycles Calculation: To empty 23 units at -1 unit/2 hours:
    • Number of cycles = 23 units / 1 unit/cycle = 23 cycles.
    • Time for 23 cycles = 23 cycles * 2 hours/cycle = 46 hours.
  7. Remaining Work: After 46 hours, 23 units have been emptied. The tank now holds 28 - 23 = 5 units. It's A's turn.
  8. Final Step: In the 47th hour, Pipe A empties the remaining 5 units in 1 hour.
  9. Total Time: 46 hours + 1 hour = 47 hours.

2. Leakage Problem: Two pipes A and B can fill a tank in 15 hours and 20 hours respectively. A third pipe C can empty it in 30 hours. If all three pipes are opened simultaneously, but pipe C is closed after 5 hours, in what time will the tank be filled?

Step-by-step solution:

  1. LCM Method: LCM(15, 20, 30) = 60 units (Tank capacity).
  2. Efficiencies:
    • Pipe A: 60/15 = +4 units/hour.
    • Pipe B: 60/20 = +3 units/hour.
    • Pipe C: 60/30 = -2 units/hour.
  3. Combined Efficiency (A+B+C): 4 + 3 - 2 = +5 units/hour.
  4. Work in first 5 hours: All three pipes work for 5 hours.
    • Work done = 5 units/hour * 5 hours = 25 units.
  5. Remaining Capacity: 60 - 25 = 35 units.
  6. Pipe C closed: Now only A and B are working.
  7. Combined Efficiency (A+B): 4 + 3 = +7 units/hour.
  8. Time to fill remaining: 35 units / 7 units/hour = 5 hours.
  9. Total Time: 5 hours (initial) + 5 hours (remaining) = 10 hours.

Variation Types

  • Multiple Pipes: More than two inlet/outlet pipes.
  • Partially Filled/Empty Tanks: Problems starting with a tank that is already partially filled or needs to be partially emptied.
  • Pipes Opened/Closed: Pipes are opened or closed after a certain duration, requiring calculation of work done in stages.
  • Efficiency Ratios: Pipes whose filling/emptying rates are given in terms of ratios (e.g., A is twice as fast as B).
  • Leaks: An additional outlet that reduces the effective filling rate.

Time-Saving Methods

  • Always use the LCM Method: It's significantly faster than fraction-based calculations, especially with multiple pipes.
  • Relative Speed for Alternating Problems: For alternating pipes, calculate the net work done in one full cycle (e.g., 2 hours for A and B). Then, adjust the total work by subtracting the last pipe's individual contribution to avoid overshooting.
  • Unitary Approach: Think in terms of 'units of work' rather than abstract fractions. This makes calculations more intuitive and less prone to errors.
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Time & Work fundamentally links work, efficiency (rate), and time. Use the LCM method to find total work and calculate individual efficiencies for quick problem-solving.

Core Formula

At the heart of Time & Work problems is the relationship:

Work = Efficiency × Time

Where:

  • Work is the total task to be completed.
  • Efficiency (or Rate) is the amount of work done per unit of time (e.g., units/day, units/hour).
  • Time is the duration taken to complete the work.

From this, we can derive: Efficiency = Work / Time Time = Work / Efficiency

For competitive exams, we often assume 'Total Work' as a convenient number, typically the Least Common Multiple (LCM) of the individual times given. This converts fractional work rates into whole numbers, simplifying calculations significantly.

Worked Example 1

A can complete a piece of work in 10 days, and B can complete the same work in 15 days. In how many days will they complete the work together?

Solution:

  1. Assume Total Work: Take the LCM of 10 and 15. LCM(10, 15) = 30 units. (This is our 'Total Work').
  2. Calculate Individual Efficiencies:
    • A's efficiency = Total Work / Time taken by A = 30 units / 10 days = 3 units/day.
    • B's efficiency = Total Work / Time taken by B = 30 units / 15 days = 2 units/day.
  3. Calculate Combined Efficiency:
    • Combined efficiency (A+B) = A's efficiency + B's efficiency = 3 + 2 = 5 units/day.
  4. Calculate Time Together:
    • Time = Total Work / Combined Efficiency = 30 units / 5 units/day = 6 days.

Worked Example 2

A and B together can complete a work in 12 days. A alone can complete the same work in 20 days. In how many days can B alone complete the work?

Solution:

  1. Assume Total Work: Take the LCM of 12 and 20. LCM(12, 20) = 60 units.
  2. Calculate Efficiencies:
    • (A+B)'s combined efficiency = Total Work / Time taken by (A+B) = 60 units / 12 days = 5 units/day.
    • A's efficiency = Total Work / Time taken by A = 60 units / 20 days = 3 units/day.
  3. Calculate B's Efficiency:
    • B's efficiency = (A+B)'s efficiency - A's efficiency = 5 units/day - 3 units/day = 2 units/day.
  4. Calculate Time for B Alone:
    • Time for B alone = Total Work / B's efficiency = 60 units / 2 units/day = 30 days.

Shortcuts & Tricks

  • LCM Method is King: Always use the LCM of the given 'days' to represent the 'Total Work'. This avoids fractions and makes calculations much faster and less error-prone. This is the single most important shortcut.
  • Combined Time Formula (for 2 people): If A takes 'x' days and B takes 'y' days, together they take (x * y) / (x + y) days. While useful, the LCM method is more versatile for 3 or more people or when efficiencies are given.
  • Efficiency is inversely proportional to Time: If a person is twice as efficient, they take half the time to do the same work. This relationship is crucial for understanding efficiency ratios.

Common Mistakes

  1. Adding Days Directly: A common blunder is to add the individual times (e.g., A takes 10 days, B takes 15 days, so together 10+15=25 days). This is incorrect. You must work with rates/efficiencies.
  2. Confusing Work Done with Rate: Students sometimes mix up the total work done with the rate of work. Remember, rate is work per unit time, while total work is the entire task.
  3. Calculation Errors with LCM: While the LCM method simplifies, errors can still occur if the LCM is calculated incorrectly or if individual efficiencies are miscalculated.
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