If the side of a cube is tripled, what is the ratio of the new total surface area to the original, and the ratio of the new volume to the original volume?

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1. Let the original side be $s$. The new side is $3s$.

2. Original total surface area = $6s^2$. New surface area = $6(3s)^2 = 6\cdot9s^2 = 54s^2$. So the ratio is **$54s^2:6s^2 = 9:1$.**

3. Original volume = $s^3$. New volume = $(3s)^3 = 27s^3$. So the ratio is **$27s^3:s^3 = 27:1$.**

FINAL_ANSWER: a

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