A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the second and third pipe can fill a tank together?
(Pipe & Cistern)

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

dsuc.updated: 1 year ago

Let, the first pipe can alone fill the tank in x hour

$∴$ 2nd and 3rd pipe will take (x-5) and (x - 9) hours to fill the tank.

According to the question

$\frac{1}{x} + \frac{1}{x - 5} = \frac{1}{x - 9} = \frac{(x - 5) + x}{x (x - 5)} = \frac{1}{x - 9} = (2 x - 5) (x - 9) = x (x - 5) = 2 x^{2} - 18 x - 5 x + 45 = x^{2} - 5 x = x^{2} - 18 x + 45 = 0 = x^{2} - 15 x - 3 x + 45 = 0 = x (x - 15) - 3 (x - 15) = 0 (x - 15) (x - 3) = 0 N o w x - 15 = 0 [x = 3 i s n o t a c c e p t a b l e] ∴ x = 15 T h e f i r s t p i p e w i l l t a k e 15 m i n u t e s t o f i l l t h e \tan k .$

$∴$ The 2nd pipe can fill the tank in = 15 - 5 = 10 hours

And the 3 pipe can fill the tank in = 15 - 9 = 6 hours

2nd and 3rd pipe can fill in 1 hour = $\frac{1}{10} + \frac{1}{6} = \frac{3 + 5}{30} = \frac{8}{30} = \frac{4}{15} p a r t$

$N o w, \frac{4}{15} p a r t o f t h e \tan k i s f i l l e d b y 2 n d a n d 3 r d p i p e i n 1 h o u r ∴ 1 o r c o m p l e t e p a r t o f t h e \tan k i s f i l l e d b y 2 n d a n d 3 r d p i p e i n = \frac{15}{4} = 3.75 h o u r s ∴ 2 n d a n d 3 r d p i p e c a n f i l l t h e \tan k i n = 3 h o u r s 45 m i n u t e s . (a n s w e r)$