Part of the tank filled when all the three pipes are opened for 10 hours
= 10 x \(\big(\frac{1}{15}+\frac{1}{20}-\frac{1}{25}\big)\) = 10 x \(\big(\frac{200+150-120}{300}\big)\) = \(\frac{230}{300}\) = \(\frac{23}{30}\)
Remaining empty part = 1 - \(\frac{23}{30}\) = \(\frac{7}{30}\)
Part of the tank filled in 1 hour by pipes (A and B) = \(\frac{1}{15}\) + \(\frac{1}{20}\) = \(\frac{4+3}{60}\) = \(\frac{7}{60}\)
\(\because\) \(\frac{7}{60}\) part is filled in 1 hour
\(\frac{7}{30}\) part is filled in 2 hours
\(\therefore\) Total time to fill the tank = 10 + 2 = 12 hours