Question:medium

A solid metallic cuboid with sides in the ratio 3 : 4 : 6 is melted to form smaller cubes with sides 2 cm. If the sum of the length of the edges of the cuboid is 208 cm, then what is the ratio of the surface area of the original cuboid to the total surface area of the smaller cubes?

Show Hint

Use the edge-sum to find the cuboid's actual dimensions, then compare its total surface area to the combined surface area of all the small cubes it is melted into.
Updated On: Jul 20, 2026
  • 1 : 6
  • 4 : 11
  • 1 : 8
  • 2 : 9
  • 3 : 11
Show Solution

The Correct Option is C

Solution and Explanation

There is a shortcut that avoids computing the number of cubes and both surface areas separately, by using the general relation between a solid's volume, its surface area, and the size of the pieces it is cut into.

As before, from the edge sum $4(l+w+h)=208$ with ratio $3:4:6$, we get $l=12$, $w=16$, $h=24$ (all in cm).

When a solid of volume $V$ is cut into $N$ identical cubes of side $s$, we have $N = V/s^3$, and the combined surface area of the cubes is $N \times 6s^2 = \frac{V}{s^3}\times 6s^2 = \frac{6V}{s}$.

So the ratio we want is:
$$\frac{\text{Surface area of cuboid}}{\text{Total surface area of small cubes}} = \frac{2(lw+wh+hl)}{6V/s} = \frac{s\big(2(lw+wh+hl)\big)}{6V}$$

Volume $V = 12\times16\times24 = 4608$ cm$^3$, and $2(lw+wh+hl) = 2(192+384+288) = 1728$ cm$^2$, with $s=2$ cm.

Ratio $= \dfrac{2\times1728}{6\times4608} = \dfrac{3456}{27648}$

Dividing numerator and denominator by 3456: $\dfrac{3456}{27648} = \dfrac{1}{8}$

This confirms the same result via a purely algebraic shortcut, without separately counting the 576 cubes.
\[\boxed{1:8}\]
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