# 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 in which each tap can separately fill the tank.

Let the time taken by tap of smaller diameter to fill the tank be x hours

The time taken by tap of larger diameter to fill the tank = x – 9 hours

Let the volume of the tank be V

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

⇒ Volume of tank filled by tap of smaller diameter in 1 hour = V/x

⇒ Volume of tank filled by tap of smaller diameter in 6 hours =

Similarly, Volume of tank filled by tap of larger diameter in 6 hours =

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

⇒

⇒

⇒

⇒ 12x – 54 = x2 – 9x

⇒ x2 – 21x + 54 = 0

⇒ x2 – 18x – 3x + 54 = 0

⇒ x(x – 18) – 3(x – 18) = 0

⇒ (x – 18)(x – 3) = 0
⇒ (x – 18) = 0 and  (x – 3) = 0

⇒ x = 18 or x = 3

For x = 3,  time taken by tap of larger diameter is negative which is not possible

Thus,  x = 18

Hence the time taken by tap of smaller diameter to fill the tank be 18 hours

The time taken by tap of larger diameter to fill the tank = 18 – 9 = 9hours

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