$23 \cdot 29$ For a cube, each vertex and the midpoints of the three edges meeting at that vertex form a tetrahedron. The remaining part of the cube after cutting off these eight tetrahedra is called a defective cube. The ratio of the volume of this defective cube to the original cube is closest to
(A) $75 \%$.
(B) $78 \%$.
(C) $81 \%$.
(D) $84 \%$.
(E) $87 \%$.
(43rd American High School Mathematics Examination, 1992)
［Solution］Let the edge length of the cube be 2. Then the volume of each small regular tetrahedron is
$$
V_{\text {tetrahedron }}=\frac{1}{3} S h=\frac{1}{3} \cdot\left(\frac{1}{2} \cdot 1 \cdot 1\right) \cdot 1=\frac{1}{6} \text {. }
$$

Also, $V_{\text {cube }}=2^{3}=8$, then
$V_{\text {remaining cube }}: V_{\text {cube }}=\left(8-\frac{8}{6}\right): 8=5: 6 \approx 84 \%$. Therefore, the answer is $(D)$.