Verified Solution • 2025–26 Session moderate

QJEEM24Ma076c: JEE Main 2024 Mathematics - Class 11 JEEM24Ma076c Solution: A software company sets up m number of computer systems to…

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\mathrm{m} is equal to:

Multiple Choice Options

  • A 125
  • B 160
  • C 150
  • D 180

Step-by-Step Verified Solution

To determine the value of m\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 m\mathrm{m} computers, and it is estimated that with these m\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 m\mathrm{m} computers for 17 days:

W=17mW = 17m

When computers crash, the number of working computers each day forms an arithmetic sequence. On the first day, there are m\mathrm{m} computers. On the second day, there are m−4\mathrm{m} - 4 computers, on the third day, there are m−8\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)+…+[m−4×(n−1)]W = m + (m - 4) + (m - 8) + \ldots + \left[m - 4 \times (n - 1)\right]

where nn is the number of days. Here, n=25n = 25 .

Notice that we form an arithmetic series where the first term (a) is m\mathrm{m} and the common difference (d) is -4. The sum of the first n terms of an arithmetic series is:

Sn=n2[2a+(n−1)d]S_n = \frac{n}{2} \left[ 2a + (n - 1)d \right]

Plugging in the values:

S25=252[2m+(25−1)(−4)]S_{25} = \frac{25}{2} \left[ 2m + (25 - 1)(-4) \right]

S25=252[2m−96]S_{25} = \frac{25}{2} \left[ 2m - 96 \right]

S25=252[2m−96]=25(m−48)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)17m = 25(m - 48)

Solving for m\mathrm{m} :

17m=25m−120017m = 25m - 1200

8m=12008m = 1200

m=150m = 150

Thus, the value of m\mathrm{m} is equal to:

Option C: 150

Related Learning Resources & Exam Preparation