Solving the Cistern Filling Problem in 19 Hours

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Homework Statement

The problem is as follows
" 2 pipes fill cistern with water in 4 and 6 hours respectively.If the first pipe in 4 and 6 hours
respectively . if the first pipe be opened first and the pipes be opened alternatively one at a time for 1 hour each, in how many hours will the cistern be filled up"
My attempt:
I first took that in 1 hour first pipe fills 1/4th of tank and second pipe 1/6 thof tank
.: in 2 hours they fill 1/4+1/6= 5/12th of tank
in 1hour they fill 5/24 th of tank
.: they fill the tank in 24/5 hours
But the problem is my book's answer is 19 hours.I don't know how that is possible because if they alone fill it in 4 and 6 hours then if they are working together the time should be less.
Please help me!

 

[Master7731]

1/4 + 1/6 is = to 5/12 of the tank filled which is if both of them are filling the tank in one hour so you have to get a further 1/4 and 1/6 of these repectively so 1/4*5/12 and 1/6*5/12 and add them to get the time, maybe you'll get the rite answer...

Hope this helps, confusing problem have to admit.

 

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I am really sorry, but could you please be more precise?