A,B and C can do a piece of work in 30 days, 40 days and 50 days, respectively. Beginning with A, if A, B and C do the work alternatively then in how many days will the work be finished?
Understanding the Work and Time Problem
This problem involves calculating the total time taken to complete a piece of work when three individuals, A, B, and C, work on it alternatively. Each person takes a different amount of time to finish the entire work individually. They work in a specific order: A first, then B, then C, and this sequence repeats until the work is done.
Calculating Total Work Units (LCM Method)
To solve problems like this, we first need a common unit for the total work. This is usually done by finding the Least Common Multiple (LCM) of the individual times taken by each person. The LCM represents the total number of work units that need to be completed.
- A completes the work in 30 days.
- B completes the work in 40 days.
- C completes the work in 50 days.
Total work units = \( \text{LCM}(30, 40, 50) \)
Let's find the prime factors:
- \(30 = 2 \times 3 \times 5\)
- \(40 = 2^3 \times 5\)
- \(50 = 2 \times 5^2\)
LCM is found by taking the highest power of all prime factors involved:
\(\text{LCM}(30, 40, 50) = 2^3 \times 3 \times 5^2 = 8 \times 3 \times 25 = 24 \times 25 = 600\)
So, let the total work be 600 units.
Determining Individual Work Rates (Efficiency)
Now that we have the total work units, we can find out how many units each person completes per day. This is their work rate or efficiency.
- A's work rate = Total work / Days A takes = \( \frac{600}{30} = 20 \) units/day
- B's work rate = Total work / Days B takes = \( \frac{600}{40} = 15 \) units/day
- C's work rate = Total work / Days C takes = \( \frac{600}{50} = 12 \) units/day
Calculating Work Done in One Alternative Cycle
The individuals A, B, and C work alternatively, starting with A. One full cycle consists of A working on day 1, B working on day 2, and C working on day 3. After 3 days, the cycle repeats.
Work done in one cycle (3 days) = Work by A in 1 day + Work by B in 1 day + Work by C in 1 day
Work done in one cycle = \( 20 + 15 + 12 = 47 \) units.
Calculating Full Cycles and Remaining Work
We need to find out how many full cycles of A, B, C are completed before the work is almost finished. We divide the total work by the work done in one cycle.
Number of full cycles = \( \lfloor \frac{\text{Total Work}}{\text{Work done in one cycle}} \rfloor = \lfloor \frac{600}{47} \rfloor \)
\(600 \div 47\)
| Operation | Result | Remainder |
|---|---|---|
| \(47 \times 10\) | 470 | \(600 - 470 = 130\) |
| \(47 \times 2\) | 94 | \(130 - 94 = 36\) |
| Total (10+2) | \(47 \times 12 = 564\) | 36 |
\(600 = 47 \times 12 + 36\)
This means 12 full cycles of A, B, C working are completed.
Days taken for 12 full cycles = \(12 \text{ cycles} \times 3 \text{ days/cycle} = 36 \text{ days}\).
Work done in 12 full cycles = \(12 \times 47 = 564\) units.
Remaining work = Total work - Work done in full cycles = \(600 - 564 = 36\) units.
Completing the Remaining Work
After 36 days, 564 units of work are done, and 36 units remain. The work starts with A again for the 13th cycle.
- Day 37: A works. A does 20 units. Remaining work = \(36 - 20 = 16\) units.
- Day 38: B works. B does 15 units. Remaining work = \(16 - 15 = 1\) unit.
- Day 39: C is scheduled to work. C's work rate is 12 units/day. The remaining work is 1 unit.
Time taken by C to finish the remaining 1 unit of work = \( \frac{\text{Remaining work}}{\text{C's work rate}} = \frac{1}{12} \) days.
Total Time Taken to Finish the Work
Total days = Days for 12 full cycles + Days A worked on remaining + Days B worked on remaining + Days C worked on remaining
Total days = \( 36 \text{ days} + 1 \text{ day (by A)} + 1 \text{ day (by B)} + \frac{1}{12} \text{ days (by C)} \)
Total days = \( 36 + 1 + 1 + \frac{1}{12} = 38 + \frac{1}{12} = 38\frac{1}{12} \) days.
So, the work will be finished in \(38\frac{1}{12}\) days.
| Worker | Days to Finish Alone | Work Rate (units/day) |
|---|---|---|
| A | 30 | 20 |
| B | 40 | 15 |
| C | 50 | 12 |
| Stage | Days Taken | Work Done | Cumulative Work | Remaining Work |
|---|---|---|---|---|
| 1 Cycle (A, B, C) | 3 | 47 | 47 | \(600 - 47 = 553\) |
| 12 Cycles (A, B, C) | \(12 \times 3 = 36\) | \(12 \times 47 = 564\) | 564 | \(600 - 564 = 36\) |
| Day 37 (A) | 1 | 20 | \(564 + 20 = 584\) | \(36 - 20 = 16\) |
| Day 38 (B) | 1 | 15 | \(584 + 15 = 599\) | \(16 - 15 = 1\) |
| Day 39 (C) | \(1/12\) | 1 | \(599 + 1 = 600\) | \(1 - 1 = 0\) |
| Total | \(36 + 1 + 1 + 1/12 = 38\frac{1}{12}\) | 600 | 600 | 0 |
Work and Time Revision Table
| Concept | Description | Formula/Method |
|---|---|---|
| Total Work | Represented as a quantity, often LCM of individual times. | \( \text{LCM}(\text{Time}_1, \text{Time}_2, ...) \) |
| Work Rate (Efficiency) | Amount of work done by a person in one unit of time (e.g., 1 day). | \( \text{Work Rate} = \frac{\text{Total Work}}{\text{Time Taken}} \) |
| Alternative Work | Individuals work in sequence, not simultaneously. | Calculate work done per cycle of workers. |
| Work Done | Work Rate × Time | \( W = R \times T \) |
| Time Taken | Total Work / Work Rate | \( T = \frac{W}{R} \) |
Additional Information on Work and Time Problems
Work and Time problems are a common topic in quantitative aptitude. Understanding the relationship between work, time, and efficiency (work rate) is key. Here are some points to remember:
- Work is generally considered constant in a specific problem unless stated otherwise.
- Efficiency and Time are inversely proportional. If a person is more efficient, they take less time to complete the same amount of work.
- When people work together simultaneously, their work rates are added up.
- When people work alternatively, you calculate the work done in one cycle (which spans over the number of days equal to the number of workers in the cycle) and then find out how many cycles are needed.
- Always handle the remaining work carefully after full cycles, as the next person in the sequence will start the remainder.
These principles help in solving various types of work and time problems, whether involving individuals, groups, or machines working together or alternatively.
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