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 \[\mathrm{m}\] is equal to:
Solutions · 1
Official solution
To determine the value of \[\mathrm{m}\], we need to formulate the problem using some basic concepts of arithmetic progression and work. Let's first understand the nature of the problem:
Initially, there are \[\mathrm{m}\] computers, and it is estimated that with these \[\mathrm{m}\] computers, the assignment can be completed in 17 days.
However, due to the crash of 4 computers every day starting from the second day onward, the total time taken extends by 8 days, making it 25 days in total.
To begin with, let's define the total work (W) in terms of the number of computers and days:
The total work (W) is given by:
The amount of work completed each day with \[\mathrm{m}\] computers for 17 days:
W=17m
When computers crash, the number of working computers each day forms an arithmetic sequence. On the first day, there are \[\mathrm{m}\] computers. On the second day, there are \[\mathrm{m} - 4\] computers, on the third day, there are \[\mathrm{m} - 8\] computers, and so on. We need to sum this series until 25 days are completed.
This can be formulated as:
Total work done over 25 days with decrement in the number of computers:
\[W = m + (m - 4) + (m - 8) + \ldots + \left[m - 4 \times (n - 1)\right]\]
where n is the number of days. Here, n=25.
Notice that we form an arithmetic series where the first term (a) is \[\mathrm{m}\] and the common difference (d) is -4. The sum of the first n terms of an arithmetic series is:
\[S_n = \frac{n}{2} \left[ 2a + (n - 1)d \right]\]
Plugging in the values:
\[S_{25} = \frac{25}{2} \left[ 2m + (25 - 1)(-4) \right]\]
\[S_{25} = \frac{25}{2} \left[ 2m - 96 \right]\]
\[S_{25} = \frac{25}{2} \left[ 2m - 96 \right] = 25(m - 48)\]
This work should be equivalent to the work calculated earlier, so:
17m=25(m−48)
Solving for \[\mathrm{m}\]:
17m=25m−12008m=1200m=150
Thus, the value of \[\mathrm{m}\] is equal to:
Option C: 150