Cone volume change — height tripled and radius halved: A right circular cone’s height is increased by 200% (so it becomes three times), while its base radius is reduced by 50%. What is the resulting change in the cone’s volume?

Introduction / Context:
The cone volume V = (1/3)π r^2 h scales with r^2 and h. When r and h change by given factors, the new volume is obtained by multiplying the old volume by the product of those factors, respecting the square on r.

Given Data / Assumptions:

• h becomes 3h (200% increase means h → h + 2h = 3h).
• r becomes 0.5r (50% reduction).
• All else unchanged.

Concept / Approach:
New/old factor f = (r′/r)^2 * (h′/h) = (0.5)^2 * 3 = 0.25 * 3 = 0.75. Therefore, V decreases by 25%.

Step-by-Step Solution:
V′ = (1/3)π (0.5r)^2 (3h) = (1/3)π * 0.25 r^2 * 3h = 0.75 * VPercent change = (0.75 − 1) * 100% = −25%

Verification / Alternative check:
Example: r = 2, h = 2 ⇒ V = (1/3)π*4*2 = (8/3)π; new r = 1, new h = 6 ⇒ V′ = (1/3)π*1*6 = 2π; ratio 2π : (8/3)π = 0.75.

Why Other Options Are Wrong:
They ignore the squared dependence on r or misinterpret “200% increase.”

Common Pitfalls:
Adding percentage changes rather than multiplying the correct factors.

Final Answer:
decreases by 25%