The dimensions of a rectangular box are in the ratio of 2 : 3 : 4

Solution:

Suppose that the dimensions be $\mathrm{x}$ multiple of each other.

The dimensions are in the ratio $2: 3: 4$.

Hence, length $=2 \mathrm{x} \mathrm{m}$

Breadth $=3 \mathrm{x} \mathrm{m}$

Height $=4 \mathrm{x} \mathrm{m}$

So, total surface area of the rectangular box $=2 \times($ length $\times$ breadth $+$ breadth $\times$ height $+$ length $\times$ height $)$

$=2 \times(2 \mathrm{x} \times 3 \mathrm{x}+3 \mathrm{x} \times 4 \mathrm{x}+2 \mathrm{x} \times 4 \mathrm{x})$

$=2 \times\left(6 \mathrm{x}^{2}+12 \mathrm{x}^{2}+8 \mathrm{x}^{2}\right)$

$=2 \times\left(26 \mathrm{x}^{2}\right)$

$=52 \mathrm{x}^{2} \mathrm{~m}^{2}$

Also, the cost of covering the box with paper at the rate Rs $8 / \mathrm{m}^{2}$ and Rs $9.50 / \mathrm{m}^{2}$ is Rs 1248 .

Here, the total cost of covering the box at a rate of Rs $8 / \mathrm{m}^{2}=8 \times 52 \mathrm{x}^{2}=$ Rs $416 \mathrm{x}^{2}$

And the total cost of covering the box at a rate of Rs $9.50 / \mathrm{m}^{2}=9.50 \times 52 \mathrm{x}^{2}=\mathrm{Rs} 494 \mathrm{x}^{2}$

Now, total cost of covering the box at the rate Rs $9.50 / \mathrm{m}^{2}-$ total cost of covering the box at the rate Rs $8 / \mathrm{m}^{2}=1248$

$\Rightarrow 494 \mathrm{x}^{2}-416 \mathrm{x}^{2}=1248$

$\Rightarrow 78 \mathrm{x}^{2}=1248$

$\Rightarrow \mathrm{x}^{2}=\frac{1248}{78}=16$

$\Rightarrow \mathrm{x}=\sqrt{16}=4$

Hence, length of the rectangular box $=2 \times \mathrm{x}=2 \times 4=8 \mathrm{~m}$

Breadth $=3 \times \mathrm{x}=3 \times 4=12 \mathrm{~m}$

Height $=4 \times \mathrm{x}=4 \times 4=16 \mathrm{~m}$