One pump fills a tank two times as fast as another pump. If the pumps work together they fill the tank in 18 minutes. How long does it take each pump working alone to fill the tank?

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Solution

Dear student,

$L e t t h e s l o w e r p i p e a l o n e f i l l t h e \tan k i n x m i n u t e s . T h e n, f a s t e r p i p e w i l l f i l l i t i n \frac{x}{2} m i n u t e s . S o, \frac{1}{x} + \frac{2}{x} = \frac{1}{18} \Rightarrow \frac{1 + 2}{x} = \frac{1}{18} \Rightarrow x = 18 \times 3 \Rightarrow x = 54 m i n u t e s T h u s, t h e s l o w e r p i p e a l o n e c a n f i l l t h e \tan k i n 54 m i n u t e s . a n d f a s t e r p i p e w i l l f i l l i t i n \frac{54}{2} = 27 m i n u t e s .$

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One pump fills a tank two times as fast as another pump. If the pumps work together they fill the tank in 18 minutes. How long does it take each pump working alone to fill the tank?

Dear student, Let the slower pipe alone fill the tank in x minutes. Then, faster pipe will fill it in x2 minutes. So,1x+2x=118⇒1+2x=118⇒x=18×3⇒x=54 minutesThus, the slower pipe alone can fill the tank in 54 minutes.and faster pipe will fill it in 542 =27 minutes. Regards