Two pipes A and B can fill a tank in 20 hours and 30 hours respectively. Both the pipes are opened to fill the tank but when the tank is $\frac{1}{3}$ full, a leak develops in the tank through which $\frac{1}{3}$ water supplied by both the pipes goes out. How much time will the tank take to be full?

dsuc.created: 2 years ago | dsuc.updated: 1 year ago

dsuc.updated: 1 year ago

$I n 1 h o u r, A a n d B t o g e t h e r c a n f i l l = \frac{1}{20} + \frac{1}{30} = \frac{3 + 2}{60} = \frac{5}{60} = \frac{1}{12} p a r t o f t h e t a n k ∴ B e f o r e l a e k a g e, b o t h p i p e w o u l d t a k e t o f i l l \frac{1}{3} p a r t o f t h e \tan k = \frac{\frac{1}{3}}{\frac{1}{12}} = \frac{1}{3} \times \frac{12}{1} = 4 h o u r s N o w, i n 1 h o u r, b o t h p i p e s c a n f i l l u p t h e \tan k e f f e c t i v e l y = \frac{1}{12} \times \frac{2}{3} = \frac{1}{18} p a r t R e m a i n t i n g (1 - \frac{1}{3}) = \frac{2}{3} p a r t o f t h e \tan k w i l l t a k e = \frac{\frac{2}{3}}{\frac{1}{18}} = \frac{2}{3} \times \frac{18}{1} = 12 h o u r s ∴ T o t a l t i m e r e q u i r e d = 4 + 12 = 16 h o u r s (a n s .)$