Question

There are two taps opening into a tank. If both are opened, the tank would be full in 12 minutes. The time taken for it to fill with only the small tap open, is 10 minutes more than the time to fill with only the large tap open. What is the time taken to fill the tank with only the small tap open?

Answer 1

Let the time taken by the large tap to fill the tank be x minutes.

∴ Tank filled by the large tap in 1 minute =

Time taken by the small tap to fill the tank = (x + 10) minutes

∴ Tank filled by the small tap in 1 minute =

Time taken by both the taps to fill the tank = 12 minutes

∴ Tank filled by both the taps in 1 min =

Adding 49 to both the sides to make L.H.S a perfect square:

⇒ x² + 14x + 49 = 120 + 49

⇒ x² + 2 × 7 × x + (7)² = 169

⇒ (x + 7)² = 169 {Using a² + b² + 2ab = (a + b)²}

⇒ x + 7 =

⇒ x + 7 =

⇒ x = 7

⇒ x = 7 + 13 or 7 − 13

⇒ x = 20 or x = − 6

As time cannot be in negative, x = 20.

Therefore, the time taken by the small tap to fill the tank is (20 + 10) minutes = 30 minutes