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

First find the actual dimensions of the cuboid using the sum-of-edges condition, then compare surface areas.
Updated On: Jul 21, 2026
  • 1 : 6
  • 4 : 11
  • 1 : 8
  • 2 : 9
Show Solution

The Correct Option is C

Solution and Explanation

This can be solved faster with a direct formula instead of counting the number of small cubes.
Step 1: Find the dimensions as before. From the sum of edges, \(l+b+h=52\), and with ratio 3:4:6, the scale factor is 4, giving \(l=12\), \(b=16\), \(h=24\) cm, and volume \(V = 4608\) cm\(^3\).
Step 2: Use a shortcut formula for the total surface area of the small cubes. If a cube of volume V is cut into small cubes of side a, the number of small cubes is \(V/a^3\), and each has surface area \(6a^2\). So the total surface area of all the small cubes is \((V/a^3)\times 6a^2 = 6V/a\). Here \(a = 2\), so the total small-cube surface area is \(6 \times 4608 / 2 = 13824\) cm\(^2\), without ever needing to compute the actual count of 576 cubes.
Step 3: Find the surface area of the cuboid directly. \(2(lb+bh+hl) = 2(192+384+288) = 1728\) cm\(^2\).
Step 4: Take the ratio. \(1728 : 13824 = 1 : 8\), the same result as before.\[\boxed{1:8}\]
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