125 identical cubes are cut from a big cube and all the smaller cubes are arranged in a row to form a long cuboid. What is the percentage increase in the total surface area of the cuboid over the total surface area of the cube?
$234\dfrac{1}{3}\%$
$234\dfrac{2}{3}\%$
$117\%$
None of these
Solution
Let side of small cubes be $a$, and side of larger cube be $b$.
$\therefore$ Volume of larger cube $ = 125 \times$ volume of smaller cube
$\Rightarrow b^3 = 125a^3$
$\Rightarrow b = 5a ~~~~~~~~~~~~~~~~~~~~~$......(i)
Total surface area of larger cube $= 6b^2$
$= 6 (5a)^2$
$= 150 a^2$
When these 125 cubes are arranged to form a cuboid
$\Rightarrow$ Breadth and height of cuboid $ =a$
Length of cuboid $= 125a$
Now, total surface of cuboid $= 2 (lb + bh \times lh)$
$ = 2(125a^2 + a^2 + 125a^2)$
$= 502 a^2$
Increase in total surface area = $ \dfrac{502a^2 - 150a^2}{150a^2} \times 100 = 234\dfrac{2}{3}\%$