Amount EMPTIED by pipe R in one hour = 1

As pipe P fills the entire cistern in 2 hours,

PART of cistern FILLED by pipe P in one hour = ½

As pipe Q fills the entire cistern in 2.5 hours,

Amount filled by pipe Q in one hour = 1/2.5

For the first half an hour only pipe P is open.

∴ Part of cistern filled by pipe P in starting ½ hour = ½ × ½ = ¼

For the next half an hour, pipe P and pipe Q both are open.

∴ Part of cistern filled by pipe P and pipe Q in next half an hour $(= \frac{1}{2}\left( {\frac{1}{2} + \frac{2}{5}} \right) = \frac{1}{4} + \frac{1}{5} = \frac{9}{{20}})$

∴ Total pat of cistern filled after one hour $(= \frac{1}{4} + \frac{9}{{20}} = \frac{7}{{10}})$

After 1 hour all three PIPES are open.

Let x be the number of hours for which all 3 pipes are open

∴ At the end of x hours, the cistern will be completely empty

∴ $(\frac{7}{{10}} + x\left( {\frac{1}{2} + \frac{1}{{2.5}} - \frac{1}{1}} \right) = 0)$

$(\Rightarrow \frac{7}{{10}} - \frac{x}{{10}} = 0)$

⇒ x = 7

∴ All three tanks are open for 7 hours.

⇒ Tank will get empty 7 hours after 7:00 am.