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The length, breadth and height of a room are in the ratio \(3:2:1\). If the breadth and height are halved while the length is doubled, then the total area of the four walls of the room will


remains the same.


decrease by \(13.64\%\)


decrease by \(15\%\)


decrease by \(30\%\)

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Option(D) is correct

Let the original length, breadth and height of the room be $3x$, $2x$ and $x$ respectively.

Therefore, the new length, breadth and height are $6x$, $x$ and $x/2$ respectively.

Area of four walls = (2 × length × height) + (2 × breadth × height)

Original area of four walls,

\(=(2 \times 3x \times x)+(2 \times 2x \times x)\)

\(= 6x^2 + 4x^2 = 10x^2\)

New area of four walls,

\(=(2 \times 6x \times \frac{x}{2})+(2 \times x \times \frac{x}{2})\)

 \( =6x^2 + x^2 = 7x^2\)

Therefore, Area of wall decreases by

\( =\left[\dfrac{10x^2 - 7x^2}{10x^2} \right] \times 100\)

\(= 30\%\)

Edit: A typo in the final step has been corrected based on the input from KARTIK.

(4) Comment(s)


in this question u have not mentioned which four walls are to be considered .


A room has only four walls. Other two surfaces are called 'roof' and 'floor'. So, I believe the question is okay and does not need any correction.


last step : $10x^2$ instead of $10^2$ :)


Thank you Kartik, corrected the mistake.