$T w o p i p e s r u n n i n g t o g e t h e r c a n f i l l a c i s t e r n i n 2 \frac{8}{11} m i n u t e s . I f o n e p i p e t a k e s o n e m i n u t e m o r e t h a n t h e o t h e r t o f i l l t h e c i s t e r n, f i n d t h e s u m o f i n d i v i d u a l t i m e i n w h i c h e a c h p i p e w i l l f i l l t h e c i s t e r n .$

5 min

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11 min

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6 min

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10 min

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Solution

The correct option is B
11 min

$L e t I s t p i p e f i l l s t h e c i s t e r n i n x m i n I I n d p i p e w i l l f i l l t h e c i s t e r n i n = (x + 1) m i n A c c o r d i n g t o t h e q u e s t i o n, \frac{1}{x} + \frac{1}{x + 1} = \frac{11}{30} \Rightarrow \frac{x + 1 + x}{x (x + 1)} = \frac{11}{30} \Rightarrow \frac{2 x + 1}{x^{2} + x} = \frac{11}{30} \Rightarrow 11 x^{2} + 11 x = 60 x + 30 11 x^{2} + 11 x - 60 x - 30 = 0 11 x^{2} - 49 x - 30 = 0 11 x^{2} - 55 x + 6 x - 30 = 0 \Rightarrow 11 x (x - 5) + 6 (x - 5) = 0 (x - 5) (11 x + 6) = 0 \Rightarrow x = 5, - 6 / 11; x = - 6 / 11 (n e g l e c t e d) S o, t i m e t a k e n b y t h e p i p e s t o f i l l t h e c i s t e r n = 5 m i n . (5 + 1) = 6 m i n .$

$S o, t h e s u m o f i n d i v i d u a l t i m e o f e a c h p i p e = (5 + 6) m i n s = 11 m i n s$

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Two pipes running together can fill a cistern in 2811 minutes. If one pipe takes one minutemore than the other to fill the cistern, find the sum of individual time in which each pipe will fill the cistern.