two pipes a and b can fill up a half full tank in 1.2hrs.the tank was initially empty.pipe b was kept open for half the time required by pipe a to fill the tank by itself.then, pipe A was kept open for as much time as was required by pipe B to fill up 1/3 of the tank by itself.it was then found that the tank was 5/6 full.the least time in which any of the pipes can fill the tank fully is?

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let a takes x hrs and b takes y hrs, first case b was kept open for x/2 hrs and then a was kept open for y/3 hrs, that means b has done (x/2y) amount of work, and a has done (y/3x) amount of work right?
because pipe a needs x hrs to fill the tank
x hrsby a.....> full work
then in y/3 hrs the amount of work done will be? (y/3x)?
and now, it is given that these two works done by a and b adds up to 5/6 th of the whole work

so we have two equations now

(1/x)+(1/y)=(1/1.2) hrs
and (x/2y)+(y/3x)=5/6 solve and u ll get x and y in hrs

Excellent explanation Satyatej.

approach pls?

4 hours

4.8 hours.