1. Two taps can fill a tank in 18 and 24 min respectively. When both the taps are opened find when the first tap be turned off so that the tank may be filled in 12 min.

* a. 6 min
* b. 10 min
* c. 9 min
* d. 12 min

• Lets assume first tap is opened for t min.
  we know that the second tap will be opened for whole 12 min and from question we know that first and second tap can fill 1/18 and 1/24 part separately in 1 min respectively.
  so that in t min 1st tap will fill up=1/18*t=t/18.
  in 12 min 2nd tap will fill up=1/24*12=1/2.
  so, if we assume total tank is 1 part as whole.
  then,t/18+1/2=1
  =>(t+9)/18=1
  =>t=9
  Answer:(c)9 min.
• 12 years agoHelpfull: Yes(7) No(2)
• Second tap is open for 12 mins.
  It will fill half the tank.
  Remaining half of tank is filled by first tap .
  It fills full tank in 18 min .
  hence it will fill half tank in 9 mins.
  
  So after 9 mins, first tap is turned off.
• 12 years agoHelpfull: Yes(7) No(4)
• let tank capacity be 72 lit. (lcm of 18,24).
  first tap fill tank in 18 min so it fill 4lit/min
  second tap fill in 24 min, so it fill as 3lit/min
  they combinedly fill tank at rate of 7lit/min.
  let x be time of closing first tap.
  upto x min both taps are opened and from x min to 12 min i.e (12-x)min second tap is opened.
  so, 7x+3(12-x)=72 => x= 9 min.
• 12 years agoHelpfull: Yes(4) No(1)