3 taps x y and z can fill a tank in 18,24,36 hrs resp. X is opened first , Y is opened after t hour's and z is opened after 2t hours from the time x was opened. If time taken to fill the tank is 9/2 hours less than double the time it would had taken when x y and z opened simultaneously. Then find t?

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

3 taps x, y, and z can fill a tank in 18, 24, and 36 hours respectively. X is opened first, Y is opened after t hours, and Z is opened after 2t hours from the time X was opened. If the time taken to fill the tank is 9/2 hours less than double the time it would have taken when x, y, and z are opened simultaneously, then find t.

Learn how to solve a work and time problem where three taps can fill a tank in 18, 24, and 36 hours respectively. This problem involves calculating the time t at which each tap is opened and uses concepts of algebra and work rates. Suitable for students in Grades 9-11.

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