A solid cylinder (radius 6 cm) has five cylindrical holes (radius 1.5 cm each) drilled throughout its length. Approximately what percent of volume is removed?

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Accepted Answer

1. The heights cancel, so fraction removed = 5 × (area of small circle) / (area of big circle).

2. Compute with radii $r=1.5$, $R=6$:  $5\cdot\dfrac{1.5^2}{6^2}=5\cdot\dfrac{2.25}{36}=5\cdot\dfrac{1}{16}=\dfrac{5}{16}$.

3. As a percent: $\dfrac{5}{16}	imes100\%=31.25\%$. So choice **(d)** is correct.

FINAL_ANSWER: d

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