Time and Work: Complete Master Guide for UPSC/MPSC CSAT (English)

Time and Work: Comprehensive Guide for CSAT

The ‘Time and Work’ topic is a staple in UPSC and MPSC CSAT quantitative aptitude sections. The core principle of Time and Work revolves around the efficiency of individuals or groups and the time they take to complete a specific task. Mastering this topic requires understanding the inverse relationship between time and efficiency, and learning to apply methods like the ‘LCM method’ (Least Common Multiple) which makes calculations remarkably faster than traditional fractional methods.

1. Core Concepts and Formulas

Basic Principles:

• Work = Efficiency × Time
• Efficiency is the amount of work done in one unit of time (e.g., one day, one hour).
• Efficiency is inversely proportional to the time taken. If person A takes twice as long as person B to do a work, A’s efficiency is half of B’s.
• If A can do a piece of work in n days, then A’s 1 day’s work = 1/n.

The LCM Method (Pro Technique)

Instead of dealing with fractions like 1/x + 1/y, we assume the total work to be a convenient number—specifically, the Least Common Multiple (LCM) of the given time periods. This converts all work into ‘units’ and efficiency into ‘units per day’.

• Step 1: Find the LCM of the given days. Let this LCM be the ‘Total Work Units’.
• Step 2: Calculate 1 day’s work (Efficiency) for each person by dividing the Total Work by their respective days.
• Step 3: Add/Subtract efficiencies as per the question to find the combined efficiency.
• Step 4: Total Time = Total Work / Combined Efficiency.

MDH Rule (Chain Rule)

When comparing groups of people working for certain hours and days to complete some work, we use the formula:

(M_1 × D_1 × H_1) / W_1 = (M_2 × D_2 × H_2) / W_2

Where:
M = Number of Men/Workers
D = Number of Days
H = Working Hours per day
W = Amount of Work done (if constant, W_1 = W_2 and it cancels out)

2. Solved Examples with Step-by-Step Explanations

Example 1 (Basic LCM Method): A can do a work in 10 days and B can do the same work in 15 days. In how many days can they complete the work if they work together?

Step-by-step Solution:

1. Days taken by A = 10, B = 15.
2. Find the Total Work: LCM of 10 and 15 is 30. Let Total Work = 30 units.
3. Calculate Efficiency (Units per day):
   • A’s efficiency = 30 / 10 = 3 units/day.
   • B’s efficiency = 30 / 15 = 2 units/day.
4. Combined Efficiency of (A + B) = 3 + 2 = 5 units/day.
5. Total Time taken = Total Work / Combined Efficiency = 30 / 5 = 6 days.

Example 2 (Leaving Work Midway): A and B can do a piece of work in 12 days and 18 days respectively. They begin together, but A leaves after 3 days. In how many days will B complete the remaining work?

Step-by-step Solution:

1. Total Work = LCM(12, 18) = 36 units.
2. A’s efficiency = 36 / 12 = 3 units/day.
3. B’s efficiency = 36 / 18 = 2 units/day.
4. Combined efficiency (A + B) = 3 + 2 = 5 units/day.
5. Work done in first 3 days = 5 × 3 = 15 units.
6. Remaining Work = 36 – 15 = 21 units.
7. B completes remaining work alone. Time = Remaining Work / B’s efficiency = 21 / 2 = 10.5 days.

Example 3 (MDH Rule): If 12 men can build a wall in 24 days working 8 hours a day, in how many days can 16 men build the same wall working 6 hours a day?

Step-by-step Solution:

1. Use MDH formula: M_1 × D_1 × H_1 = M_2 × D_2 × H_2 (Since Work is same, W_1 = W_2)
2. Here, M_1 = 12, D_1 = 24, H_1 = 8.
3. And, M_2 = 16, D_2 = ?, H_2 = 6.
4. Substitute values: 12 × 24 × 8 = 16 × D_2 × 6.
5. 2304 = 96 × D_2
6. D_2 = 2304 / 96 = 24 days.

3. Pro-Tips to Avoid Common Mistakes

• Ditch Fractions: Whenever possible, avoid the fractional method (1/x). It leads to complex calculations and high chances of silly mistakes. Stick to the LCM method.
• Alternative Work Schedules: If people work on alternate days, be careful. Calculate the work done in a ‘cycle’ of 2 days, and then multiply to get close to the total work without exceeding it.
• Negative Work: In problems involving pipes and cisterns (a variation of time and work), remember that a leak or emptying pipe does negative work. Subtract its efficiency.
• MDH Formula Application: Ensure that the ‘Work’ (W) parameter is correctly placed in the denominator. If the second group has to do ‘double the work’, set W_2 = 2 × W_1.

4. Practice Questions

1. P can do a work in 20 days and Q can do the same work in 30 days. How long will they take to complete it together? (Answer: 12 days)
2. A is twice as good a workman as B and together they finish a piece of work in 14 days. In how many days can A alone finish the work? (Answer: 21 days)
3. 15 men take 20 days to complete a job working 8 hours a day. How many hours a day should 20 men work to complete the job in 12 days? (Answer: 10 hours)
4. A, B, and C can complete a piece of work in 24, 6 and 12 days respectively. Working together, they will complete the same work in how many days? (Answer: 3.42 or 24/7 days)