There are three taps A,B and C.they can fill the tank in 10,20,and 25 hrs respectively.At first all of them are opened simultaneously.Then after 2 hrs
tap C is closed and A and B are kept running.After the 4rth hour tap B is closed.The remaining work is done by tap A alone.Find the percentage of the work done by Tap A by itself?

• Ans is 72%
  percentage of work done by c in 2 days is 2/25*100=8%
  percentage of work done by b in four days is 4/20*100=20%
  so percentage of work done by A alone is 100-8-20=72%
• 10 years agoHelpfull: Yes(11) No(4)
• Ans is 72%
  percentage of work done by c in 2 days is 2/25*100=8%
  percentage of work done by b in four days is (4+2)/20*100=30%
  so percentage of work done by A alone is 100-8-30=62%
• 10 years agoHelpfull: Yes(5) No(2)
• a percentage work = 100/10 = 10%
  b =100/20 = 5% and c= 100/25=4%
  all of them opened for 2hrs,and then a and b for another 2hrs (given after 4th hrs) then a done alone
  so we can write
  => (10+5+4)*2 + (10+5)*2 + work percentage by a alone = 100
  =>38+30+work by a alone = 100
  => work done by a alone = 100-68 = 32
  
  so 32 is the answer.
• 10 years agoHelpfull: Yes(3) No(13)
• Ans is 72%
  We have to calculate entire work done by A
  Ans given by bhaskar calculate work of A after 4 hr(i.e 32%).. it do not consider before 4hr (which will be 40%)
  Hence final ans will be 72%
• 10 years agoHelpfull: Yes(3) No(2)
• ans=32
  work done by a,b and c in 2 hours=19/50
  work done by a and b in another 2 hr=3/10
  19/50+3/10+x=1
  x=16/50*100%
  32
• 10 years agoHelpfull: Yes(1) No(6)
• 2% only
  B'coz supposed 100 liter water present in tank
  2 hours worked by all three =38
  4 hours work by a and b= 60
  And rest by only a means(100-98)=2
  Then works has by itself a = 2%
• 6 years agoHelpfull: Yes(1) No(1)
• right answer is 32%
  
• 10 years agoHelpfull: Yes(0) No(7)
• Work done by A is 62%
  1st method:
  For 2 hours three taps can fill
  = 2(1/10+1/20+1/25)
  = 38/100
  
  For 4 hours A and B worked
  4(1/10+1/20)
  = 60/100(or 3/5)
  
  Remaining cap of tank is
  1-(38/100)-(60/100)
  = 2/100
  
  2/100 part must be filled by A itself.
  
  Finally A percentage of work is
  = 2(1/10)+4(1/10)+2/100
  = 62/100(62%)
  
  Second method:
  Lonely
  Tap A fill tank in 10h
  Tap B fill tank in 20h
  Tap C fill tank in 25h
  
  Total cap of tank is 100 units( LCM of 10, 20, 25)
  For 1 hour
  A can fill 10 units(100/10),
  B can do 5 units(100/20),
  C can do 4 units(100/25).
  
  Firstly 3 taps runs for 2 hours
  Completed units are 2(10+5+4) = 38 units
  
  We have 62 units remaining
  For the second time A and B work for 4 hours
  Completed units are 4(10+5)
  60 units of work is completed .
  
  Finally A alone work for 2 units(100-38-60)
  1 hour-----10 u
  0.2 hours--2 units
  
  Total work by A is (10(2+4)+2)/100
  62/100(62%)
• 10 years agoHelpfull: Yes(0) No(7)
• actually mr tirupathi u have considered for the second equation as they have worked for 4 hrs but actually a and b combinely worked for 2 hrs only so your equation is wrong
• 8 years agoHelpfull: Yes(0) No(0)
• Ans will be 2%
  Supposed take are is 100%
  Fill by a,b,c in 2 hour=38%
  Fill by a,b in 4 hours=60%
  Fill by only itself by a= 100% - 98%
  = 2% ans
• 6 years agoHelpfull: Yes(0) No(1)
• A's one day work = 1/10
  B's one day work = 1/20
  C's one day work = 1/25
  (A+B+C)'s one day work = (1/10+1/20+1/25) =19/100
  (A+B)'s one day work = (1/10 +1/20) = 3/20
  
  work is done by A+B+C in 2 hours is 19/50
  work is done by A+B in 4 hours is 3/5
  the remaining work is (1-(19/50+3/5)) = 1/50
  
  in 6 hours, A do ((1/10)*6) works i.e 3/5 works.
  total work is done by A by itself = (1/50)+(3/5)=31/50
  %age of the work done by tap A by itself is ((31/50)/1)*100 i.e 62%.
• 2 years agoHelpfull: Yes(0) No(0)