An algebra problem by Sparsh Ghimire

If x = number of men employed and y = portion of work to be done by 1 man in 1 day then:

60xy = $\frac{3}{5}$

And since the remaining number of days (90-60) is 30 , and there is now x+15 men working to finish the remainder of the work(1 - $\frac{3}{5}$ ), which is $\frac{2}{5}$ , then:

30(x+15)y = $\frac{2}{5}$

(30x+450)y = $\frac{2}{5}$

30xy+450y = $\frac{2}{5}$

60xy + 900y = $\frac{4}{5}$

And we can now subtract our other equation (60xy = $\frac{3}{5}$ ) from our new equation (60xy + 900y = $\frac{4}{5}$ ) and to get rid of the 60xy and get:

900y = $\frac{1}{5}$

y = $\frac{1}{4500}$

Now we can plug in y into our equation 60xy = $\frac{3}{5}$ and get :

60x( $\frac{1}{4500}$ ) = $\frac{3}{5}$

$\frac{60x}{4500}$ = $\frac{3}{5}$

$\frac{x}{75}$ = $\frac{3}{5}$

$\boxed{x = 45}$