If length and breadth of a cuboid are increased by 20%, then by how much percent the height should be reduced to keep the volume same ?

Options

A. 40
B. 44
C. `30(5)/(9)`
D. `33(4)/(7)`

Answer

To solve the problem step by step, we need to determine how much the height of a cuboid should be reduced to maintain the same volume after increasing the length and breadth by 20%.

### Step 1: Define the Original Dimensions
Let:
- Length of the cuboid = \( L \)
- Breadth of the cuboid = \( B \)
- Height of the cuboid = \( H \)

### Step 2: Calculate the Original Volume
The original volume \( V \) of the cuboid is given by the formula:
\[ V = L \times B \times H \]

### Step 3: Calculate the New Dimensions After Increase
If the length and breadth are increased by 20%, the new dimensions will be:
- New Length = \( 1.2L \) (which is \( L + 0.2L \))
- New Breadth = \( 1.2B \) (which is \( B + 0.2B \))

### Step 4: Set Up the Equation for the New Volume
Let the new height be \( xH \) (where \( x \) is the factor by which the height is reduced). The new volume \( V' \) is given by:
\[ V' = (1.2L) \times (1.2B) \times (xH) \]

### Step 5: Equate the Original Volume and New Volume
Since we want the volumes to remain the same, we set the original volume equal to the new volume:
\[ L \times B \times H = (1.2L) \times (1.2B) \times (xH) \]

### Step 6: Simplify the Equation
Cancelling \( L \), \( B \), and \( H \) from both sides (assuming they are not zero), we get:
\[ 1 = 1.2 \times 1.2 \times x \]
\[ 1 = 1.44x \]

### Step 7: Solve for \( x \)
To find \( x \), rearrange the equation:
\[ x = \frac{1}{1.44} = \frac{100}{144} = \frac{25}{36} \]

### Step 8: Determine the Reduction in Height
The original height is \( H \) and the new height is \( \frac{25}{36}H \). The reduction in height is:
\[ \text{Reduction} = H - \frac{25}{36}H = \left(1 - \frac{25}{36}\right)H = \frac{11}{36}H \]

### Step 9: Calculate the Percentage Reduction
To find the percentage reduction in height:
\[ \text{Percentage Reduction} = \left(\frac{\text{Reduction}}{H}\right) \times 100 = \left(\frac{\frac{11}{36}H}{H}\right) \times 100 = \frac{11}{36} \times 100 \]
\[ = \frac{1100}{36} \approx 30.56\% \]

### Conclusion
Thus, the height should be reduced by approximately \( 30.56\% \) to keep the volume the same.