The dimensions of a rectangular box are in the ratio of $2 : 3 : 4$ and the difference between the cost of covering it with sheet of paper at the rates of $Rs.\ 8$ and $Rs.\ 9.50$ per $m^2$ is $Rs.\ 1248$. Find the dimensions of the box.

Given:

The dimensions of a rectangular box are in the ratio of $2 : 3 : 4$ and the difference between the cost of covering it with sheet of paper at the rates of $Rs.\ 8$ and $Rs.\ 9.50$ per $m^2$ is $Rs.\ 1248$.

The dimensions of each brick is $22.5\ cm \times 10\ cm \times 7.5\ cm$.

To do:

We have to find the dimensions of the box.

Solution:

Ratio of the dimensions of the cuboidal box $= 2 : 3 : 4$

Let the length of the box be $(l) = 4x$, breadth be $(b) = 3x$ and height be $(h) = 2x$

Therefore,

Total surface area $= 2 (lb + bh + lh)$

$=2(4x \times 3 x+3 x \times 2 x+2 x \times 4 x)$

$=2(12 x^{2}+6 x^{2}+8 x^{2})$

$=2 \times 26 x^{2}$

$=52 x^{2}$

In the first case,

Rate of covering it with paper $=Rs.\ 8$ per $\mathrm{m}^{2}$

In the second case,

Rate of covering with paper $=Rs.\ 9.50$ per $\mathrm{m}^{2}$
Therefore,

Difference in the cost of both ways $=Rs.\ (52 x^{2} \times 9.50-52 x^{2} \times 8)$

$=Rs.\ (494 x^{2}-416 x^{2})$

$=Rs.\ 78 x^{2}$

This implies,

$78 x^{2}=1248$

$x^{2}=\frac{1248}{78}$

$x^2=16$

$x=\sqrt{16}$

$x=4$
Dimensions of the box are $2 \times 4=8 \mathrm{~m}, 3 \times 4=12 \mathrm{~m}$ and $4 \times 4=16 \mathrm{~m}$.

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Updated on: 10-Oct-2022

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