Question

If the length, breadth and height of the room are in the ratio 3:2:1. The breadth and height of the room are halved and length of the room is doubled. Then area of the four walls of the room will,

• A. decrease by 13.64%
• B. decrease by 15%
• C. decrease by 18.75%
• D. decrease by 30%

Answer 1

The Correct Option is D

Let's analyze the problem step by step. The original dimensions of the room are in the ratio 3:2:1. Let's assume the length, breadth, and height of the room are \(3x\), \(2x\), and \(x\) respectively.

The formula for the area of the four walls of a cuboid is given by \(2h(l+b)\).

Initially, the area of the four walls is:

\(2 \cdot x \cdot ((3x) + (2x)) = 2x \cdot 5x = 10x^2\).

After the changes:

• The length \(L\) is doubled: \(2 \cdot 3x = 6x\).
• The breadth \(B\) and height \(H\) are halved: \(\frac{2x}{2} = x\) and \(\frac{x}{2} = \frac{x}{2}\).

The new area of the four walls is:

\(2 \cdot \frac{x}{2} \cdot (6x + x) = x \cdot 7x = 7x^2\).

Now, we find the percentage decrease in the area:

\(\frac{\text{initial area} - \text{new area}}{\text{initial area}} \times 100 = \frac{10x^2 - 7x^2}{10x^2} \times 100 = \frac{3x^2}{10x^2} \times 100 = 30\%\).

Therefore, the area of the four walls of the room decreases by 30%.