Pipe A, B and C are kept open and together fill a tank in t minutes. Pipe A is kept open throughout, pipe B is kept open for the first 10 minutes and then closed. Two minutes after pipe B is closed , pipe C is opened and is kept open till the tank is full. Each pipe fills an equal share of the tank. Furthermore, it is known that if pipe A and Bare kept open continuously, the tank would be filled completely in t minutes. How long will it take C alone to fill the tank ?

Let the total portion of tank to be filled be '3x' units.
Let the amount of units filled per minute by A,B and C be a,b and c respectively.
We are given that, at+10b+(t-12)c = 3x
Also, each pipe fills equal share of tank i.e. x units
at=x, 10b=x, (t-12)c=x. 
$$\therefore\ $$ b=$$\ \frac{\ x}{10}$$
Also, at+bt=3x
bt=2x, substitute b=$$\ \frac{\ x}{10}$$, t=20 min.
(t-12)c=x
c=$$\ \frac{\ x}{8}$$
Let k be time taken by c to fill the tank alone,
ck=3x
k=$$\ \ \frac{\ 3x}{c}$$=24 minutes.
Option B is the answer.