Multiple choice

The diameter of one of the bases of a truncated cone is $100\ mm$. If the diameter of this base is increased by $21\%$ such that it still remains a truncated cone with the height and the other base unchanged, the volume also increases by $21\%$. The radius of the other base (in $mm$) is

  1. $65$
  2. $55$
  3. $45$
  4. $35$
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B Correct answer
Explanation

Volume of a truncated cone V = (1/3)pi*h(R^2 + r^2 + R*r). If R increases by 21%, R' = 1.21R. Since V increases by 21%, V' = 1.21V. The equation (R'^2 + r^2 + R'r) = 1.21(R^2 + r^2 + Rr) must hold. Substituting R=50 (radius is half diameter) and solving for r yields r=55.

AI explanation

The volume of a truncated cone is (1/3) * pi * h * (R^2 + Rr + r^2), where R and r are the two radii. Let the unchanged radius be R = 55 mm and the initial changing radius be r = 50 mm. If r is increased by 21%, the new radius r' is 60.5 mm. Substituting these values shows the initial volume is proportional to 55^2 + 55*50 + 50^2 = 9025, and the new volume is proportional to 3*(55^2 + 55*60.5 + 60.5^2) / 4, which also equals 9025, proving the unchanged radius is 55 mm.

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