Let y be the rate per hour of the initial 3 taps.
The 2 identical new taps can individually fill the tank in 40 hours.
Let the capacity of tank be x m^3.
Therefore, the rate per hour of new taps = x/40.
Now,
2*x/40 + 3*y = 30
Since all the taps when kept open for 1 hour fill 30 m^3 water, then the 3 initial taps will fill (x-60) m^3 in 2.5 hours.
Therefore,
3*2.5*y = x - 60
On solving the above two linear equations, we get
x = 120 m^3 and y = 8 m^3/hr.