Question

Two taps together can fill a tank in 75/8 hours the with larger diameter takes 10hours less than the smaller one to fill tank separately find the time in which each of them can separately fill the tank solve it in two variables

Answer 1

hope it helps big answer Sol : Let the tap of the larger diameter fills the tank alone in (x – 10) hours. In 1 hour, the tap of the smaller diameter can fill 1/x part of the tank. In 1 hour, the tap of the larger diameter can fill 1/(x – 10) part of the tank. Two water taps together can fii a tank in 75 / 8 hours. But in 1 hour the taps fill 8/75 part of the tank. 1 / x + 1 / (x – 10) = 8 / 75. ( x – 10 + x ) / x ( x – 10) = 8 / 75. 2( x – 5) / ( x2 – 10 x) = 8 / 75. 4x2 – 40x = 75x – 375. 4x2 – 115x + 375 = 0 4x2 – 100x – 15x + 375 = 0 4x ( x – 25) – 15( x – 25) = 0 ( 4x -15)( x – 25) = 0. x = 25, 15/ 4. But x = 15 / 4 then x – 10 = -25 /4 which is not possible since time But x = 25 then x – 10 = 15. Larger diameter of the tap can the tank 15 hours and smaller diameter of the tank can fill the tank in 25 hours.

Answer 2

Answer:

Here is your answer

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Step-by-step explanation:

Sol : Let the tap of the larger diameter fills the tank alone in (x – 10) hours. In 1 hour, the tap of the smaller diameter can fill 1/x part of the tank. In 1 hour, the tap of the larger diameter can fill 1/(x – 10) part of the tank. Two water taps together can fii a tank in 75 / 8 hours. But in 1 hour the taps fill 8/75 part of the tank. 1 / x + 1 / (x – 10) = 8 / 75. ( x – 10 + x ) / x ( x – 10) = 8 / 75. 2( x – 5) / ( x2 – 10 x) = 8 / 75. 4x2 – 40x = 75x – 375. 4x2 – 115x + 375 = 0 4x2 – 100x – 15x + 375 = 0 4x ( x – 25) – 15( x – 25) = 0 ( 4x -15)( x – 25) = 0. x = 25, 15/ 4. But x = 15 / 4 then x – 10 = -25 /4 which is not possible since time But x = 25 then x – 10 = 15. Larger diameter of the tap can the tank 15 hours and smaller diameter of the tank can fill the tank in 25 hours.

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