There is a water-cask with three different water-taps.

With the smallest tap, the water-cask can be filled in 30 minutes.
With the middle tap, the water-cask can be filled in 20 minutes.
With the largest tap, the water-cask can be emptied in 15 minutes.

The water- ask is one-fourth filled. Now each of the 3 taps is opened turn by turn for a minute each in the sequence, smallest, largest, middle, smallest, largest, middle, smallest, largest, middle.

In how much time would the water-cask be first filled?

• Simply lets suppose tank is of 60 liter.
  1/4th = 15 liter is already filled so remaining 45 liter has to be filled.
  cycle is of 2 Liter - 4 Liter + 3 Liter.
  So average 1 liter filled in 3 min.
  After 43*3 min = 129 min it will be 15+43 = 58 liter
  and then next 1 min the 2 liter will fill the tank.
  So total time 129+1 = 130
• 10 years agoHelpfull: Yes(26) No(19)
• If x be the capacity of water cask then 3x/4 capacity is to be filled.
  let k be the no. of required turns need to fill.
  then
  k(x/30+x/15+x/20) = 3x/4
  => k= 5
  since one turn contains 3 min ( one min for each)
  total time= 5*3= 15 minutes
• 10 years agoHelpfull: Yes(19) No(7)
• 135 minute is the right answer.....
  in 3 minute all the three pipe fill 1/60 part..
  in 6 minute...............2/60 part
  in 9 minute .............3/60 part
  in 135 minute..............45/60 part that is 3/4 part...which is remaining in the cask...
  thats why 135 is the right answer....
• 10 years agoHelpfull: Yes(15) No(4)
• 132 min 30 seconds
  
  In 1 min
  smallest tap fills--> 1/30
  middle tap fills--> 1/20
  largest tap empties--> 1/15
  
  since 1/4 of water-cask is filled,remaining = 3/4
  
  In 3 minutes the 3 taps fills
  1/30 - 1/15 + 1/20 =(2-4+3)/60 = 1/60
  
  Let each tap be opened for x minutes to fill the water-cask
  
  1/60*x = 3/4
  x = 45
  
  Since the taps are opened turn by turn for a minute
  therefore in 44*3 =132min the fraction of water-cask filled = 1/60*44
  
  Remaining fraction to be filled = 3/4-44/60
  =1/60
  133rd min the smallest tap is opened and the remaining 1/60 is filled in 30 seconds.
• 10 years agoHelpfull: Yes(14) No(10)
• If x be the capacity of water cask then 3x/4 capacity is to be filled.
  let k be the no. of required turns need to fill.
  then
  k(x/30+x/15+x/20) = 3x/4
  => k= 5
  since one turn contains 3 min ( one min for each)
  total time= 5*3= 15 minutes
• 10 years agoHelpfull: Yes(7) No(4)
• A=1/30 ,b=1/20 ,c=-1/15
  In 3 minutes all can fill=1/30+1/20-1/15=1/60
  so,in 1 minute all can fill 1/3*60=1/180
  Now to fill 3/4 part
  Hence,let pipe fill for for first time in x minute,then
  x*(1/180)=3/4 which gives 135 minutes
  
• 10 years agoHelpfull: Yes(4) No(0)
• Excuse guyss, smallest tap itself can filled in 30 minutes then how can result greater than 30 min??
  I think answer is 16.25
  1/4th already filled
  3/4 need to be filled
  PIPE 1 can be done 3/4 in ==30min*(3/4)==22.5min
  PIPE 2 in 15min
  PIPE 3 in 11.25
  now average of 3 pipes
  (22.5+15+11.25)/3==16.25
  
• 10 years agoHelpfull: Yes(3) No(2)
• 15 minutes
• 10 years agoHelpfull: Yes(3) No(2)
• it shuld be 45 minutes
• 10 years agoHelpfull: Yes(2) No(3)
• Simply lets suppose tank is of 60 liter.
  1/4th = 15 liter is already filled so remaining 45 liter has to be filled.
  cycle is of 2 Liter - 4 Liter + 3 Liter.
  So average 1 liter filled in 3 min.
  After 43*3 min = 129 min it will be 15+43 = 58 liter
  and then next 1 min the 2 liter will fill the tank.
  So total time 129+1 = 130
• 10 years agoHelpfull: Yes(1) No(0)
• Let 60 L capacity of tank
  1/4 of 60 = 15L already filled
  so, remaining 45 L to be filled
  In 1 min smallest tap can fill 2 L
  In 1 min Middle tap can fill 3L
  In 1 min Largest tap can fill 4 L
  so in 3 min they can fill (2,4,3)L = 9L
  similarly, next 3 min they fill (2,4,3)=9 L
  next 3 min (2,4,3)=9L
  next 3 min (2,4,3)=9L
  next 3 min (2,4,3)=9L
  so total time= 15 min
• 10 years agoHelpfull: Yes(1) No(1)
• ANN THERESSA IAM CONFUSED TO UNDERSAND THIS 1/30 - 1/15 + 1/20 =(2-4+3)/60 = 1/60 HOW YOU THINK SUBTRACT LARGEST ONE?
  
  
  Read more: http://www.m4maths.com/placement-puzzles.php?ISSOLVED=Y&SOURCE=&TOPIC=&SUB_TOPIC=#ixzz2wxH86i9U
  Follow us: m4maths on Facebook
• 10 years agoHelpfull: Yes(0) No(1)
• If a pipe can fill a tank in x hours and another pipe can empty the full tank in y hours (where y > x), then on opening both the pipes, then
  
  the net part filled in 1 hour = 1/x - 1/y
  
  Here the largest tap empties the water-cask in 15 minutes. So I subtracted 1/15.Since the water-cask is already 1/4th filled, there will be enough water in the water-cask for the largest tap to empty at the begining.
• 10 years agoHelpfull: Yes(0) No(0)
• ANN THERASA I GO WITH UR SOLUTION
  except that ans is 135 minutes .actually u have put 1/60 part is completed by pipe a in 30sec.but in 3minutes 3tapes fills 1/60 part.so in ur answer after u found x=45 multiply by 3 u got answer
  
• 10 years agoHelpfull: Yes(0) No(0)
• y shuld i multiply it with 3?
• 10 years agoHelpfull: Yes(0) No(0)
• Given that 1/4 is filled so remaining part is 3/4.
  Now we find how much part of total quantity is filled by every tap in 1 minute is:-
  smallest tape= 1/30;middle tape:-1/20 nd largest tape:-1/15...
  given that it filed in order of S,L nd M tape....so now total water filled in three(3) minute is:-
  (1/30+1/20+1/15)=9/60=3/20.
  so it's filled in 3 mintuts..so in one (1) minute=(3/20)*3=1/20.
  It means that 1/20 part filles in one minute so we have to filed 3/4 part so calculation is ....
  (3/4)/(1/20)=15 minutes .
  I think it is definitely corect answer 15 minutes.
• 10 years agoHelpfull: Yes(0) No(3)
• I think it will take 8 minutes to fill..
  Bcz if the tap works altogether then these 3 tap can fill the 3/20 part of the tank in 1 minute.. and now adding the individual tap effort simultaneously sequence giving in the question.. it will take just 8 mins.. and it is the absolute answer..
• 10 years agoHelpfull: Yes(0) No(0)
• let The total volume is V.
  Since (1/4)V is already filled.
  left volume=(3/4)V=(45/60)V
  small tap can fill (1/30)V in one minute.
  In the same way middle tap and largest tap can fill (1/20)V and (1/15)V respectively in a minute.
  Now we can see that they fill [(1/30)+(1/20)+(1/15)]V=(9/60)V
  Thus if they continued for 15 minutes.. which means these 3 are repeated 5 times. then volume filled is (9*5)/60=(45/60)V which is the required volume to be filled..hence ans is 15 min.
• 9 years agoHelpfull: Yes(0) No(0)
• Let the capacity of tank be 60 units (i.e. the lcm of time taken by each tap)
  so, largest tap would fill in 1 min = 60/15 = 4 units
  middle tap in 1 minute = 60 / 20 = 3 units
  smallest tap in 1 minute = 60 / 30 = 2 units of tank
  as tank is 1/4th filled , therefore empty capacity = 60-(1/4*60) = 45 units
  in 3 minutes tank filled = (2 + 4 + 3 = )9 units
  so whole tank will be filled in 45 / 9 = 5 cycles of 3 min each
  therefore time taken in filling the tank = 3*5 = 15 minutes
• 9 years agoHelpfull: Yes(0) No(0)
• You are not reading the question properly. The largest tank is emptying the water cask. Its not filling it. So its contribution has to be subtracted not added. Also the taps are opened turn by turn and The answer has to be 135 minutes
• 9 years agoHelpfull: Yes(0) No(0)
• You are not reading the question properly. The largest tank is emptying the water cask. Its not filling it. So its contribution has to be subtracted not added. Also the taps are opened turn by turn and The answer has to be 135 minutes
• 9 years agoHelpfull: Yes(0) No(0)
• 15 is answer
• 9 years agoHelpfull: Yes(0) No(0)
• Most of the people here have not read the question properly.Largest tank is emptying the water so we need to subtract it (1/30 -1/15 +1/20)=1/60
  
• 9 years agoHelpfull: Yes(0) No(0)
• can we think in percentage bassis
  
• 9 years agoHelpfull: Yes(0) No(0)
• 135 is my answer
• 9 years agoHelpfull: Yes(0) No(0)
• for first 3 min (1/30+1/20-1/15) = 1/60
  for 1 min work = 1/180
  remining work = 3/4
  3/4*180 = 135 min
• 9 years agoHelpfull: Yes(0) No(0)