We are given three taps, $x$, $y$, and $z$, which can fill a tank in 18, 24, and 36 hours, respectively. The taps are opened at different times, and the total time to fill the tank is affected by these delays. Let's analyze this step-by-step.

Step 1: Define the rate of each tap

• The rate of tap $x$ is $\frac{1}{18}$ of the tank per hour.
• The rate of tap $y$ is $\frac{1}{24}$ of the tank per hour.
• The rate of tap $z$ is $\frac{1}{36}$ of the tank per hour.

Step 2: Time when each tap is opened

• Tap $x$ is opened at $t = 0$ (initially).
• Tap $y$ is opened after $t$ hours.
• Tap $z$ is opened after $2t$ hours.

Let the total time to fill the tank be $T$. This is the time from when tap $x$ is opened until the tank is filled. We know:

$T = 2t - \frac{9}{2}$

Step 3: Correct solution is