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Two water taps together can fill a tank in 6 hours. The tap of larger diameter takes 9 hours less than the smaller one to fill the tank separately. Find the time which each tap can separately fill the tank.

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Let the tap of smaller diameter fill the tank in x hours.

∴ Time taken by the tap of larger diameter to fill the tank = (x - 9)h

Suppose the volume of the tank be V.

Volume of the tank filled by the tap of smaller diameter in x hours = F

∴ Volume of the tank filled by the tap of smaller diameter in 1 hour = F/x

⇒ Volume of the tank filled by the tap of smaller diameter in 6 hour = F/x x 6 

Similarly

Volume of the tank filled by the tap of larger diameter in 6 hours = F/(x - 9) x 6

Now, 

Volume of the tank filled by the tap of smaller diameter in 6 hours + Volume of the tank filled by the tap of larger diameter in 6 hours = V

For x = 3, time taken by the tap of larger diameter to fill the tank is negative which is not possible.

∴ x = 18

Time taken by the tap of smaller diameter to fill the tank = 18 h

Time taken by the tap of larger diameter to fill the tank = (18 - 9) = 9h

Hence, the time taken by the taps of smaller and larger diameter to fill the tank is 18 hours and 9 hours, respectively.

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