A cone has height h and radius r. The cone is melted and recast into a smaller cone whose height is h⁄2 and radius is r⁄3 What fraction of the original volume is unused?

Correct — A.
Cone volume = (1⁄3)πr²h, so it scales with r² × h
Radius factor, squared: (1⁄3)² = 1⁄9
Height factor: 1⁄2

New cone's share: 1⁄9 × 1⁄2 = 1⁄18 of the original metal
Unused: 1 − 1⁄18 = 17⁄18 → option (a)

Why the others are wrong

• (b)
  15⁄16 — 15⁄16 unused means the new cone holds 1⁄16 of the metal. Its real share is (1⁄3)² × 1⁄2 = 1⁄18, a smaller piece, so more than 15⁄16 is left over.
• (c)
  12⁄17 — 12⁄17 cannot come from these scale factors. Products of 1⁄3 and 1⁄2 give denominators made of 2s and 3s, such as 18, never 17. The used share is 1⁄18.
• (d)
  16⁄19 — 16⁄19 would put 3⁄19 of the metal, about 16%, into the new cone. The real share is 1⁄18, about 5.6%, because the radius factor is squared before the height halves it.

Volume scales by the product of the scale factors, with the radius factor squared. A cone's volume is (1⁄3)πr²h, and the (1⁄3)π cancels in any comparison.

Shrinking the radius to r⁄3 cuts the volume to 1⁄9. Halving the height halves that again, to 1⁄18.

The small cone uses 1⁄18 of the melted metal, so the rest, 17⁄18, is left over.

Key facts

• Volume of a cone = (1⁄3)πr²h
• Scaling a cone's radius by k multiplies its volume by k², and scaling its height by m multiplies it by m
• If radius and height both scale by k, the new cone is similar and the volume scales by k³

Common traps

• Forgetting to square the radius factor: 1⁄3 × 1⁄2 = 1⁄6 used, giving 5⁄6 unused
• Answering the used share, 1⁄18, when the question asks what is unused

No directly related past PYQ was found.