A large tank is filled by a large drainage pipe. when the tank is filled all the outer pipes of the tank that is 10 are opened and tank gets empty in 2.5 hours. When only 5 outer pipes of the tank were opened then the tank gets empty in 5.5 hours.Then what will the time required to empty the tank when only 3 outer pipes were opened ?

• Let the volume of tank be : x lit
  Let the tank be filled by drainage pipe at the rate of : y lit/hr
  
  So, when the tank is fully filled and then all the 10 outer pipes of tank are opened (where as the drainage is still active and it is being filled with a rate of y lit/hr) , the tank gets empty in 2.5 hrs.
  So, the total volume of water drain out by 10 pipes in 2.5 hr is : x + 2.5y
  So the rate of drain out per pipe is : (x + 2.5y)/10 x 2.5 .....(i)
  
  When 5 pipes are used to drain out the water , the tank gets empty in 5.5 hrs,
  So, the total volume of water drain out by 5 pipes in 5.5 hr is : x + 5.5y
  So the rate of drain out per pipe is : (x + 5.5y)/5 x 5.5 .....(ii)
  
  Equating equation (i) & (ii),
  (x + 2.5y)/25 = (x + 5.5y)/27.5
  or, 11(x + 2.5y) = 10(x + 5.5y)
  or, 11x + 27.5y = 10x + 55y
  or, x = 27.5y
  
  Therefore when 3 pipes are used to empty the tank,
  let the time taken be : t hrs
  So, the total volume of water drain out by 3 pipes in t hr is : x + ty
  So the rate of drain out per pipe is : (x + ty)/3t
  
  Therefore; (x + ty)/3t = (x + 2.5y)/25
  or, (27.5y + ty)/3t = (27.5y + 2.5y)/25
  or, (27.5 + t)/3t = 30/25 = 1.2
  or, 27.5 = t = 3.6t
  or, 2.6t = 27.5
  or, t = 27.5/2.6 = 10.57hrs
• 9 years agoHelpfull: Yes(4) No(0)
• one pipe can empty tank in 2.5*10=25h
  so 5 pipes can empty tank in 25/5=5h since total 10 pipes so remaining 5 pipes fill the tank
  so some additional work should be done by 5 pipes used to empty tank
  given total required if 5 outer pipes are opened is 5.5h=5+0.5=(5+1/2)h
  additional time=10/5=1/2........
  if 3 outer pipes are opened then time to empty tank is (25/3)+(10/3)= 11.66h
• 9 years agoHelpfull: Yes(2) No(10)
• 1)part of water drained in an hour by a pipe=2.5/10=0.25;2)part of water drained by a pipe=5.5/5=1.1;then the series as AP with d as 0.17 is 0.25,0.42,0.59,0.76,0.93,1.1,1.27,1.44====>x/3=1.44=>x=4.32
  10 9 8 7 6 5 4 3
• 9 years agoHelpfull: Yes(0) No(2)
• 10 pipes work-hour=1/2.5
  5 pipes work-hour=1/5.5
  Therefore,(10-5)pipes work-hour=(1/2.5)-(1/5.5)
  = 60/275
  Then, 3 pipes work-hour=(60/275)*(3/5) ------------(Unitary method)
  = 36/275
  So,time required to empty the tank by 3 pipes is= 275/36
  =7.63 Hours
• 9 years agoHelpfull: Yes(0) No(1)
• Let the volume of tank be : x lit
  Let the tank be filled by drainage pipe at the rate of : y lit/hr
  
  So, when the tank is fully filled and then all the 10 outer pipes of tank are opened (where as the drainage is still active and it is being filled with a rate of y lit/hr) , the tank gets empty in 2.5 hrs.
  So, the total volume of water drain out by 10 pipes in 2.5 hr is : x + 2.5y
  So the rate of drain out per pipe is : (x + 2.5y)/10 x 2.5 .....(i)
  
  When 5 pipes are used to drain out the water , the tank gets empty in 5.5 hrs,
  So, the total volume of water drain out by 5 pipes in 5.5 hr is : x + 5.5y
  So the rate of drain out per pipe is : (x + 5.5y)/5 x 5.5 .....(ii)
  
  Equating equation (i) & (ii),
  (x + 2.5y)/25 = (x + 5.5y)/27.5
  or, 11(x + 2.5y) = 10(x + 5.5y)
  or, 11x + 27.5y = 10x + 55y
  or, x = 27.5y
  
  Therefore when 3 pipes are used to empty the tank,
  let the time taken be : t hrs
  So, the total volume of water drain out by 3 pipes in t hr is : x + ty
  So the rate of drain out per pipe is : (x + ty)/3t
  
  Therefore; (x + ty)/3t = (x + 2.5y)/25
  or, (27.5y + ty)/3t = (27.5y + 2.5y)/25
  or, (27.5 + t)/3t = 30/25 = 1.2
  or, 27.5 = t = 3.6t
  or, 2.6t = 27.5
  or, t = 27.5/2.6 = 10.57hrs
• 9 years agoHelpfull: Yes(0) No(1)