MATH PIPES AND CISTERN

                               PIPE  AND CISTERN / PIPE AND TANK

       It  think everyone see tank and pipe  on their roof or public organization.  Basically water or liquid are filled with pipe  in the tank. These pipe may be off/ on  while the filling the tank.

              Fill the  tank      ----  > + sign

               Empty the tank  -- >  -  sign

                                                               Method are use as  used in the TIME AND WORK chapter

    HOW TO SOLVE THE QUESTION :: choose any method for solving the question

Method 1::    by one day concept:

                         Example:  IF A  pipe can fill  a tank in 10 hours and pipe B can fill the tank in  15 hours .when both pipes are opened simultaneously ,then how much time required to fill the tank?.

                         Description: in this method  days are   converted into 1 hour work  i.e.

                          A pipe one hour work =1/10 (because normally in  1hour  it will  only 1/10 part of total work)

                          similarly     B  one hour  work=1/15

                        then both A   and  B   1 hour work = 1/10 +1/15

                                                                                    =3+2/30 

                                                                                    =5/30

                                                                                   =1/6

                                      Thus both A  and B  pipe opened simultaneously then tank  can fill  1/6 part of work in one hour .Now just reverse it to find total no of hour to finish the work   i.e.  6 hour.

                        

Method 2:: by LCM concept:

                           Example:  IF A  pipe can fill  a tank in 10 hours and pipe B can fill the tank in  15 hours .when both pipes are opened simultaneously ,then how much time required to fill the tank?.

                             Description: in this method LCM will take for hours  then based upon total work each person per work hour is calculated  i.e

                             Now   total work done by A and B  =3w/d+ 2w/d

                                                           =5w/d

                                                            Pipe A and B  will do 5 work per hours of total work 30 to fill the tank.

                                                      5   work can do work in --------> 1 hour

                                                    30  work can do work in ------- >   30/5 hour

                                                                                              -------- > 6 hour

                                                                                        QUESTION

         

 QUESTION 1 : PIPE A  and B can fill a tank in 10 hours and 12 hours respectively ,but  pipe C can empty the same 

                        tank in 15 hours. When all pipes are opened how much time to fill/ empty the tank ?.

Solution:

                   A pipe work in 1 hour   to fill the tank= +(1/10)

                  B pipe work in 1 hour to fill the tank =+(1/12)

                  C pipe work in 1 hour to empty the tank=  -(1/15)

                            Now add all above equation

             (A+B+C )  work   in one hour to fill/empty the tank=+(1/10)+(1/12)-(1/15)

                                                                                                        =(6+5-4)/60

                                                                                                        =7/60

       NOW

                  Reverse it   i.e 60/7 hours.

QUESTION 2 ::  A pipe can fill a tank in 12 hours and other pipe can fill in 15 hours but third pipe can empty the tank in 6 hours .First two pipe are kept open for 5 hours in the beginning then third pipe is also opened .Find the time when tank will be empty ?.

Solution:

              (A + B +C)    work in one hours =(1/12 ) +(1/15)   -  (1/6)

                                                                    =  - 1/60

                              Therefore (A + B +C) simultaneously  open pipe will empty the tank  .i.e it will empty tank in  one part of its 60 part.   

       Now   (A + B)    work in one hours= (1/12 ) +(1/15)= 9/60

                   (A + B)    work in 5  hours=9*5/60

                                                         =45/60

                           (A + B ) will fill tank in 45 part of its 60 part.   

            Thus  remaining work that is done by    (A + B +C)  after 5  hours= 1-(45/60)

                                                                                                                              =15/60

                          But  no use of  reaming work here  since (A + B +C)  are empty the tank

                Now    45 part is full in tank out of 60. Will be empty in 45 hours since  (A + B +C) empty -1 liter every hours