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A rusty old 20,000 L water tank developed a leak after being filled during the wet season. The leak was not noticed for a number of days, and at that point, "No Nonsense Nick," the plumber, tried to fix the leak. Unfortunately, he made the problem worse, and the leak continued until the tank was empty.

The equations that represent the volume \(V\) (in liters) of water in the tank over a period of time \(t\) (in days) for the two "leaking situations" are given below. Assume that no rain fell after the initial leak occurred. Note: Leak 2 will begin when Leak 1 ends.

- Leak 1: \(V = -850t + 20000\)
- Leak 2: \(V = -1490t + 23850\)

a) How many days had passed until the first leak was detected?

Answer :

The first leak was detected approximately 23.53 days after the initial filling of the tank.

To determine how many days had passed until the first leak was detected, we need to find the time t when the volume V of water in the tank reaches the point where Leak 1 ends and Leak 2 begins.

The equation representing the volume of water in the tank during Leak 1 is V = -850t + 20000. The tank started with 20000 liters of water and was empty when Leak 1 ended, so we set V to 0 and solve for t:

Set the equation to zero: 0 = -850t + 20000

Add 850t to both sides: 850t = 20000

Solve for t: t = 20000 / 850

Calculate t: t ≈ 23.53 days

Therefore, approximately 23.53 days after the initial filling of the tank.

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Rewritten by : Barada