Answer :

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

⇒

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⇒ 12x – 54 = x^{2} – 9x

⇒ x^{2} – 21x + 54 = 0

⇒ x^{2} – 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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