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Three pipes A, B and C can fill a tank in 12 hours. All the pipes started working together and after 3 hours, C is closed. If A and B can fill the remaining part in 10 hours, then the number of hours taken by C alone to fill the tank is .
Let efficiencies of A,B,C be a,b,c respectively.
Given a+b+c =Â $$\frac{W}{12}$$ => Total work = 12(a+b+c)
They worked together for 3 hours. So work done in 3 hours = 3(a+b+c)
Remaining= 9(a+b+c) which is filled by a and b only in 10 hours
So $$a+b\ =\frac{\left(9\left(a+b+c\right)\right)}{10}$$
On solving a+b=9cÂ
So W= 12(9c+c)=120c
Time taken by C alone =Â $$\frac{120c}{c}$$=120 hours
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