Solving the College Students Project Dropout Problem

The problem describes a scenario where a group of college students undertaking a project experiences daily dropouts, affecting the project's completion time. We need to determine the initial number of students based on the expected versus actual project duration and the dropout rate.

Let's define the variables:

• Let N be the initial number of students at the beginning of the project (on Day 1).
• Let's assume the total amount of work required to complete the project is constant.
• Let's assume each student does an equal amount of work per day.

Calculating Total Work

If there were no dropouts and the project was completed in the planned 10 days with the initial N students, the total work required would be proportional to the product of the number of students and the planned duration.

Planned Work = Number of Students $\times$ Planned Days

Planned Work = $\text{N} \times \text{10}$

Planned Work = $\text{10N}$ units (assuming 1 unit of work per student per day)

Calculating Actual Work with Dropouts

In reality, 2 students dropped out every day starting from the end of Day 1. The project took 15 days to complete. The number of students working on each day changed as follows:

• Day 1: N students
• Day 2: N - 2 students
• Day 3: N - 4 students
• Day 4: N - 6 students
• ... and so on.

On any given day, say Day 'k', the number of students would be $\text{N} - \text{2(k-1)}$.

Let's find the number of students on the last day, Day 15:

Students on Day 15 = $\text{N} - \text{2(15-1)}$

Students on Day 15 = $\text{N} - \text{2(14)}$

Students on Day 15 = $\text{N} - \text{28}$

The total actual work done is the sum of the work done by the students each day for 15 days. This forms an arithmetic series:

Actual Work = (Students on Day 1) + (Students on Day 2) + ... + (Students on Day 15)

Actual Work = $\text{N} + (\text{N}-2) + (\text{N}-4) + \dots + (\text{N}-28)$

This is an arithmetic series with:

• Number of terms (days) = 15
• First term = N
• Last term = N - 28

The sum of an arithmetic series is given by the formula:

Sum = $\frac{\text{Number of terms}}{2} \times (\text{First term} + \text{Last term})$

Actual Work = $\frac{15}{2} \times (\text{N} + (\text{N}-28))$

Actual Work = $\frac{15}{2} \times (2\text{N} - 28)$

Actual Work = $15 \times \frac{(2\text{N} - 28)}{2}$

Actual Work = $15 \times (\text{N} - 14)$

Actual Work = $\text{15N} - \text{210}$ units

Equating Planned Work and Actual Work

The total amount of work required to complete the project remains the same, regardless of how many students work on it daily or how long it takes. Therefore, we can equate the Planned Work and the Actual Work:

Planned Work = Actual Work

$\text{10N} = \text{15N} - \text{210}$

Solving for N

Now, we solve the equation for N to find the initial number of students:

$\text{10N} = \text{15N} - \text{210}$

Subtract 10N from both sides:

$0 = \text{15N} - \text{10N} - \text{210}$

$0 = \text{5N} - \text{210}$

Add 210 to both sides:

$\text{210} = \text{5N}$

Divide both sides by 5:

$\text{N} = \frac{210}{5}$

$\text{N} = 42$

So, the number of students at the beginning of the project was 42.

Verification

Let's verify if this number makes sense:

• Initial students: 42
• Planned Work (42 students for 10 days): $42 \times 10 = 420$ units
• Students per day for 15 days: 42, 40, 38, 36, 34, 32, 30, 28, 26, 24, 22, 20, 18, 16, 14.
• Actual Work (Sum of students for 15 days): This is an arithmetic series with 15 terms, first term 42, last term 14.
• Actual Work = $\frac{15}{2} \times (42 + 14) = \frac{15}{2} \times 56 = 15 \times 28 = 420$ units.

The Planned Work (420 units) matches the Actual Work (420 units), confirming our solution.

Revision Table: College Project Dropout Problem

Additional Information: Work and Time Problems

Work and time problems often involve scenarios where a certain amount of work needs to be completed by a group of people or machines. Key concepts include:

• Total Work: Often considered constant for a specific task. It can be represented as the product of the number of workers and the time taken, assuming a constant work rate per worker.
• Work Rate: The amount of work done by one worker (or machine) in a unit of time (e.g., per day, per hour).
• Variable Workers: Problems like this one involve the number of workers changing over time due to people joining, leaving, or dropping out. In such cases, the total work is the sum of the work done by the varying number of workers over their respective periods.
• Arithmetic Series: When the number of workers changes by a constant amount each day (like dropping out 2 students daily), the daily work contribution forms an arithmetic progression, and the total work over several days is the sum of this series. The sum of an arithmetic series is $\text{S}_n = \frac{n}{2}(\text{a}_1 + \text{a}_n)$, where $\text{n}$ is the number of terms, $\text{a}_1$ is the first term, and $\text{a}_n$ is the last term.

Solving these problems typically involves setting up an equation based on the principle that the total work required is fixed. By calculating the total work under different conditions (planned vs. actual) and equating them, we can solve for unknown variables like the number of workers or the time taken.

In this specific college project problem, the daily reduction in the number of students leads to an arithmetic series of daily work contributions, making the sum formula particularly useful.