A man undertakes to do a work in 150 days. He employs 200 men. He finds that only a quarter of the work is done in 50 days. How many additional men should he employ so that the whole work is finished in time?
100
This problem is about how the amount of work done relates to the number of men working and the time they spend. We are given a situation where a task needs to be completed within a specific timeframe, and we need to figure out how to adjust the workforce to meet the deadline after a certain amount of work has been done.
The core idea behind these types of problems is that the total "work effort" required to complete a task is constant. Work effort can be thought of as the product of the number of workers, the time they work, and their efficiency. Assuming efficiency remains constant, the total work effort is proportional to (Number of Men) \(\times\) (Number of Days).
If the work done is also considered, the relationship can be expressed as:
\[ \frac{M_1 \times D_1}{W_1} = \frac{M_2 \times D_2}{W_2} \] Where:
Let's break down the information given in the problem:
Now, let's determine the details for the second phase:
Using the formula \( \frac{M_1 \times D_1}{W_1} = \frac{M_2 \times D_2}{W_2} \), we can plug in the values:
\[ \frac{200 \times 50}{\frac{1}{4}} = \frac{M_2 \times 100}{\frac{3}{4}} \]
Let's solve for \(M_2\):
\[ \frac{200 \times 50}{\frac{1}{4}} = 200 \times 50 \times 4 = 40000 \]
\[ \frac{M_2 \times 100}{\frac{3}{4}} = M_2 \times 100 \times \frac{4}{3} = M_2 \times \frac{400}{3} \]
So, the equation becomes:
\[ 40000 = M_2 \times \frac{400}{3} \]
To find \(M_2\), we rearrange the equation:
\[ M_2 = 40000 \times \frac{3}{400} \]
\[ M_2 = \frac{40000 \times 3}{400} = \frac{100 \times 3 \times 400}{400} = 100 \times 3 \]
\[ M_2 = 300 \]
This means a total of 300 men are needed for the remaining 100 days to finish the work on time.
The question asks for the number of additional men required. The initial number of men was 200. The total number of men needed for the rest of the project is 300.
Additional men = Total men needed - Initial men
Additional men = 300 - 200 = 100 men.
Therefore, 100 additional men should be employed to ensure the whole work is finished within the original 150-day deadline.
| Phase | Men (M) | Days (D) | Work (W) |
|---|---|---|---|
| Phase 1 (Given) | 200 | 50 | \(\frac{1}{4}\) |
| Phase 2 (Required) | \(M_2\) (to find) | 100 | \(\frac{3}{4}\) |
| Concept | Explanation | Formula/Relationship |
|---|---|---|
| Work Rate | Amount of work done per unit of time by one person/unit. | Work = Rate \(\times\) Time |
| Man-Days | A unit representing the amount of work one man can do in one day. Total work is often measured in man-days. | Total Work = Number of Men \(\times\) Number of Days (assuming constant rate) |
| Inverse Proportion | If the number of men increases, the time taken to complete the same amount of work decreases (assuming constant rate). | \( M_1 \times D_1 = M_2 \times D_2 \) (for same work W) |
| Direct Proportion | If the amount of work increases, the time taken or the number of men required increases (assuming other factors are constant). | \( \frac{W_1}{D_1} = \frac{W_2}{D_2} \) (for same men M) or \( \frac{W_1}{M_1} = \frac{W_2}{M_2} \) (for same days D) |
Work and time problems can come in various forms, including:
The key to solving these problems is often to first determine the total amount of work (sometimes in "units" or "man-days") or the individual rates, and then calculate how much work is done or needs to be done under different conditions. The formula \( \frac{M_1 \times D_1}{W_1} = \frac{M_2 \times D_2}{W_2} \) is a powerful tool for problems involving changes in the number of workers and time for a given amount of work.
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