Mensuration Practice Test 8

Let length, breadth and height of the room be 3x, 2x and x respectively.
Area of the room = 2
[(l × h)+ (b × h)] = 2 [(3x × x)+ (2x × x)]
= $2\left(3{\mathit{x}}^2+2{\mathit{x}}^2\right)$
= $10{\mathit{x}}^2$
New dimensions of room, i.e., length, breadth and height = 6x, x and $\frac{\mathit{x}}{2}$ respectively.
New area of the room = 2 [(l × h) + (b × h)]
= $2\left[\left(6\mathit{x}\times \frac{\mathit{x}}{2}\right)+\left(\mathit{x}\times \frac{\mathit{x}}{2}\right)\right]$
= $2\left[3{\mathit{x}}^2+\frac{{\mathit{x}}^2}{2}\right]$
= $2\left(\frac{7{\mathit{x}}^2}{2}\right)$ = $7{\mathit{x}}^2$
Percentage decrease in area = $\frac{3{\mathit{x}}^2}{10{\mathit{x}}^2}\times 100$ = 30 %