Given the following data for three projects with the number of workers and days required:

| Project | Case 1: Number of workers | Case 1: Days required | Case 2: Number of workers | Case 2: Days required |
|---------|---------------------------|----------------------|---------------------------|----------------------|
| A | X | 88 | X + 8 | 66 |
| B | Y | Y - 1 | Y + 6 | Y - 6 |
| C | Z² | 75 | M² | 108 |

If initially 0.5Y workers work on Project B for 10 days, and then an additional 0.5Y workers join, how many total days will it take to complete Project B?

Choose the correct option.

Question

Given the following data for three projects with the number of workers and days required:

| Project | Case 1: Number of workers | Case 1: Days required | Case 2: Number of workers | Case 2: Days required |
|---------|---------------------------|----------------------|---------------------------|----------------------|
| A       | X                         | 88                   | X + 8                     | 66                   |
| B       | Y                         | Y - 1                | Y + 6                     | Y - 6                |
| C       | Z²                        | 75                   | M²                        | 108                  |

If initially 0.5Y workers work on Project B for 10 days, and then an additional 0.5Y workers join, how many total days will it take to complete Project B?

Choose the correct option.

Accepted Answer

**Solution:**

1. Let the total number of workers for Project B be *Y*.

2. Initially, half of the workers, i.e., *0.5Y*, work for 10 days.

3. Work done in these 10 days = Number of workers × Days = $0.5Y 	imes 10 = 5Y$ units of work.

4. After 10 days, another 0.5Y workers join, so total workers become $0.5Y + 0.5Y = Y$.

5. Let the remaining days required to complete the project be *A*.

6. Total work for Project B can be calculated using the formula: Number of workers × Days required = Total work.

From Case 1, days required = $Y - 1$, so total work = $Y 	imes (Y - 1)$ units.

7. Work done so far = $5Y$ units, so remaining work = $Y(Y - 1) - 5Y = Y^2 - Y - 5Y = Y^2 - 6Y$ units.

8. The remaining work is done by *Y* workers in *A* days:

$$Y 	imes A = Y^2 - 6Y$$

$$A = Y - 6$$

9. Total days to complete the project = initial 10 days + remaining days *A* = $10 + (Y - 6) = Y + 4$.

10. Given from Case 2, days required = $Y - 6$, and from the problem, the total days required is 40.

So, $Y + 4 = 40 \Rightarrow Y = 36$.

11. Therefore, total days = 40.

**Answer:** 40 days

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