Question:

A software company sets up m number of computer systems to finish an assignment in 17 days. If 4 computer systems crashed on the start of the second day, 4 more computer systems crashed on the start of the third day and so on, then it took 8 more days to finish the assignment. The value of m is equal to :

Updated On: Nov 3, 2025

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The Correct Option is B

Approach Solution - 1

To solve this problem, we need to understand the impact of computer system crashes on the completion time of the assignment.

Initially, the company plans to use \(m\) computer systems to complete the assignment in 17 days. With computer crashes occurring each day after the first, we know that:

On the first day, no computer system crashes.
On the second day, 4 systems crash.
On the third day, 4 more systems crash, and this pattern continues daily.

The assignment actually took 8 more days longer than planned. Hence, the assignment was completed in \(17 + 8 = 25\) days.

We need to set up an equation to determine the initial number of systems \(m\).

Work Done Analysis

Let’s calculate the total work in terms of "computer-days" required to complete the assignment without crashes:

The total work required is \(17m\) computer-days.

Accounting for Crashes

Now consider the reduction in computer-days due to crashes:

Day 1: Work done by \(m\) systems.
Day 2: Work done by \(m - 4\) systems.
Day 3: Work done by \(m - 8\) systems.
... and so forth until the end of the 25 days, following the pattern.

Therefore, the total work done over 25 days is:

\(m + (m - 4) + (m - 8) + \ldots + (m - 4 \times (d-1))\) where \(d\) is the day number, totaling 25.

Summation

The series \(m + (m - 4) + (m - 8) + \cdots\) can be simplified with an arithmetic series where:

\(a_1 = m\)
Common difference \(d = -4\)
We are given \(25\) terms.

The last term, \(l = m - 4 \times (25 - 1)\).

The sum of the first \(n\) terms of an arithmetic series is given by:

\(S_n = \frac{n}{2} \times (a + l)\)

Substitute in the values to set up the equation:

\(25m - 4 \times (0 + 1 + 2 + \cdots + 24) = 17m\)

The term \((0 + 1 + 2 + \cdots + 24)\) is a sum of the first 24 natural numbers:

\(\frac{24 \times (24 + 1)}{2} = 300\)

So the equation simplifies to:

\(25m - 4 \times 300 = 17m\)

Solving for \(m\) gives:

\(25m - 17m = 1200\)

\(8m = 1200\)

\(m = 150\)

Thus, the initial number of computer systems, \(m\), is 150.

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Approach Solution -2

\[ 17m = m + (m - 4) + (m - 4 \times 2) + \dots + (m - 4 \times 24) \]

\[ 17m = 25m - 4(1 + 2 + \dots + 24) \]

\[ 8m = 4 \times \frac{24 \times 25}{2} = 150 \]