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I came up with this problem: $150$ workers were employed to do a particular work. On first day, $150$ workers worked. On second day, $146$.. and each subsequent day, workers kept on decreasing by 4. Now, it took 8 more days to complete the work (days required now = number of days $150$ workers would have taken if all $150$ worked each day $+ 8$). So, we have to find the number of days it took to complete the work.

Here's my solution:

Let number of days taken by 1 worker to complete the work be $n$.
So, 1 worker's 1 day work = $1/n$

And, 150 workers' 1 day's work = $150/n$
So, time required by 150 workers to complete the work = $n/150$ (Considering work to be $1$, I can't represent work mathematically other than 1, or a whole)

Now, work done in first day = $150/n$
In second day = $146/n$ .. and so on

So, total work done = $150/n + 146/n + 142/n..... = (150 + 146 + 142...)/n = 1$

Now, in the numerator, we can apply the sum to $n$ terms of an arithmetic progression formula, taking $a = 150 , d = -4$

And here, since number of days are 8 more, so number of days = $n/150 + 8$

So, we have this: $(n/150 + 8)\{300 + (n/150 + 7)(-4)\}/n = 1$
And so, $(n/150 + 8)\{300 + (n/150 + 7)(-4)\} = n$

This solution seems to be correct to me, but when I solve the equation, I am getting irrational roots, but I am expected to get an integral answer.

So I just wanted to ask, is there any problem in this or is it correct?

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It is so much easier to use the concept of man-days.

Let the time taken with reducing manpower = $n$ days

Equating man-days taken, $150(n-8) = 150 + 146 + 142 + .... +[150- (n-1)4]$

Using Gauss' formula for sum of an A.P.

$150(n-8) = (n/2)(304-4n)$

$n = 25$

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$150 + 146 + 142 + ... + 54 = 2550 = (25 - 8) \times 150$.

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