Problem 8 - Algebra for JEE ADVANCED

Question

150 workers were engaged to finish a piece of work in a certain number of days. Four workers stopped working on the second day, four more workers stopped their work on the third day and so on. It took 8 more days to finish the work. Then the number of days in which the work was completed is a. 29 days b. 24 days c. 25 days d. none of these

Step-by-Step Solution

Answer

The work was completed in 33 days.

Step 1: Define Variables

Let \( n \) be the number of days in which the work was originally supposed to be completed, and \( N \) be the total amount of work to be done. Each worker completes \( \frac{N}{150n} \) work per day.

Step 2: Determine Work Completion Initially

Originally, 150 workers were meant to complete the work in \( n \) days, implying \( 150 \times n \times \frac{N}{150n} = N \), meaning the whole work was to be completed.

Step 3: Calculate Work Done With Reduced Workforce

From the second day onward, workers decrease by 4 per day. Hence, on the second day, there are 146 workers, on the third day, 142 workers, and so forth.

Step 4: Setup Equation with Extended Days

The new scenario took \( n+8 \) days, and during these \( n+8 \) days, the total number of workers decreased by 4 every day after the first day. The sum of workers over the days forms an arithmetic series. Hence, we calculate the total worker-days as: \( 150 + 146 + 142 + \, ... \, \) until the last few days when workers are constantly decreasing by 4.

Step 5: Calculate Total Worker-Days

The last term of this arithmetic series is the result after \( n+8-1 \) days, which is \( 150 - 4(n+7-1) \). Set up the equation for the total worker-days equalling \( n \times 150 \).

Step 6: Solve for n

Calculate the arithmetic sum formula equating it to the initial work completion scenario. Solve for \( n \) using the formula for the sum of an arithmetic progression \( S = \frac{k}{2}(2a + (k-1)\cdot d) \).

Step 7: Solve Equation

Simplify the equation and solve for \( n \). Substitute \( n = 25 \) which satisfies this equation, hence the total number of days required with decreasing workforce ending results in \( n+8 = 33 \, \) days.

Key Concepts

Arithmetic Series

In mathematics, an arithmetic series is the sum of the terms of an arithmetic sequence. An arithmetic sequence is a series of numbers where each term after the first is derived by adding a constant, called the "common difference," to the previous term.
For example, the sequence 3, 5, 7, 9 is an arithmetic sequence where each term increases by 2. To find the sum of this arithmetic series, you can use the formula:

• Sum, \( S_n = \frac{n}{2} (2a + (n-1) \cdot d) \)
• where \( n \) is the number of terms, \( a \) is the first term, and \( d \) is the common difference.

In our problem, the number of workers decreases each day, forming an arithmetic sequence, and the solution involves calculating an arithmetic series to determine worker-days over the extended completion period.

Work and Time Problems

Work and time problems are common applications in algebra, requiring you to find how long it will take to complete a task or how much work can be completed in a given time.
They often involve rates, which describe how quickly work is done:

• This could mean workers per day, tasks per hour, etc.
• The key is understanding the relationship between the number of workers and the time needed to complete a job.

In the given problem, as workers leave, there are fewer hands to share the load, extending the work period. By equating worker-days (the product of the number of workers and days worked) with the total work required, you can solve for the unknown time variable.

Equation Solving

Solving equations is fundamental in algebra and requires finding the values for variables that make the equation true. It often involves manipulating equations to isolate the variable.
In our example:

• The initial number of days, \( n \), was set up, equating the work done to the total work required.
• A second equation was constructed with decreasing worker numbers modeled by an arithmetic series.
• By relating these two expressions, we formed a solvable equation using the sum formula for arithmetic series.

The key is an understanding of critical algebraic techniques, such as distribution, factoring, and simplifying expressions to extract values for variables.

Progressions in Mathematics

Progressions, particularly arithmetic progressions, are sequences of numbers with a specific, consistent pattern.
Understanding progressions is essential, as shown in problems like this one, where the number of workers fit into an arithmetic sequence.

• An arithmetic progression has a constant difference between consecutive terms.
• This systematic decrease helped us model and solve the problem by calculating a total of worker-days over time.

Grasping the concept of progression is crucial as it extends beyond mere mathematical exercises, linking directly to real-world scenarios involving repetitive change over time, such as gradual increases or decreases in workforce, savings, or consumption.