A tank is in the form of a cuboid . It holds a maximum of `540 m^(3)`water . If the tank is 8 m long and 15 m wide, then how many meters deep must the water be when the tank is `(2)/(3)` full

Options

A. 2
B. `4.5`
C. `2.5`
D. 3

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Determine the volume of water when the tank is two-thirds full.
The maximum volume of the tank is given as \(540 \, m^3\).

To find the volume when the tank is two-thirds full, we calculate:
\[ \text{Volume when two-thirds full} = \frac{2}{3} \times 540 \, m^3 \]
Calculating this gives:
\[ \text{Volume when two-thirds full} = \frac{2 \times 540}{3} = \frac{1080}{3} = 360 \, m^3 \]

### Step 2: Use the formula for the volume of a cuboid.
The volume \(V\) of a cuboid is given by the formula:
\[ V = L \times B \times H \]
Where:
- \(L\) = length of the tank = \(8 \, m\)
- \(B\) = breadth of the tank = \(15 \, m\)
- \(H\) = height (or depth) of the water, which we need to find.

### Step 3: Set up the equation with the known values.
We know the volume of water when the tank is two-thirds full is \(360 \, m^3\). Therefore, we can set up the equation:
\[ 360 = 8 \times 15 \times H \]

### Step 4: Calculate \(H\).
First, calculate \(8 \times 15\):
\[ 8 \times 15 = 120 \]
Now, substitute this back into the equation:
\[ 360 = 120 \times H \]
To find \(H\), divide both sides by \(120\):
\[ H = \frac{360}{120} = 3 \, m \]

### Conclusion:
The depth of the water when the tank is two-thirds full is \(3 \, m\).

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