Question

(x-2) men can do a job in x days and (x+7) men can do a job in ${\frac{3}{4}}^{th}$ of the same work in (x-10) days. The no. of days in which (x+4) men can finish the job is___

Answer 1

Total work is constant.
$\therefore$ Total work = (x-2)(x) = $\frac{4}{3}$*(x+7)(x-10)
$\Rightarrow x^2-6x-280=0$
Solving for x; we get x = 20.
Now 18 men takes 20 days to complete work
$\therefore$ 24 men will take $\frac{20\times 18}{24}$ days = 15 days

Alternatively:
$(x-2)$ $\to x$ days
$(x+7)$ $\to (x-10)$ days for ${\frac{3}{4}}^{th}$ of work
$\Rightarrow (x-2)$ will take $\frac{3}{4}x$ days for ${\frac{3}{4}}^{th}$ of work
So $\frac{x-2}{x+7}=\frac{x-10}{3x/4}$
$\Rightarrow x^2-6x-280=0$
So, x = 20
Now 18 men takes 20 days to complete work
$\therefore$ 24 men will take $\frac{20\times 18}{24}$ days = 15 days