Three cubes have edge lengths $3 \mathrm{~cm}, 12 \mathrm{~cm}$, and $x \mathrm{~cm}$. The average volume of the three cubes is $700 \mathrm{~cm}^{3}$. The value of $x$, rounded to the nearest integer, is
(A) 6
(B) 10
(C) 8
(D) 9
(E) 7
Since the average volume of three cubes is $700 \mathrm{~cm}^{3}$, their total volume is $3 \times 700 \mathrm{~cm}^{3}$ or $2100 \mathrm{~cm}^{3}$.

The volume of a cube with edge length $s \mathrm{~cm}$ is $s^{3} \mathrm{~cm}^{3}$.

Therefore, $3^{3}+12^{3}+x^{3}=2100$ and so $27+1728+x^{3}=2100$ or $x^{3}=345$.

Since $x^{3}=345$, then $x \approx 7.01$, which is closest to 7 .

ANSWER: (E)