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?
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.
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.
Total work units = \( \text{LCM}(30, 40, 50) \)
Let's find the prime factors:
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.
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.
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.
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.
After 36 days, 564 units of work are done, and 36 units remain. The work starts with A again for the 13th cycle.
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 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 |
| 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 | \( T = \frac{W}{R} \) |
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:
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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