Question bank / Inferences

Reading and Writing · Information and IdeasHard

Planetary scientists often estimate the age of a rocky surface by counting its impact craters. Because craters build up over time, an older surface generally has more of them. Geologic processes such as lava flows, however, can erase craters by burying them. On Jupiter's moon Io, which has hundreds of active volcanoes, spacecraft images have revealed no impact craters at all. This observation suggests that ______

Which choice most logically completes the text?

A

no asteroids or comets have ever struck Io.

B

Io formed much more recently than Jupiter's other moons.

C

volcanic activity renews Io's surface faster than craters can build up on it.

D

counting craters is not a useful way to estimate the age of any surface.

Show answer and explanation

Answer: C

Io is covered with active volcanoes, and lava can bury craters. Having no craters at all means Io's surface is being repaved faster than impacts can mark it, so the surface is young. That is choice C.

Choice B confuses the age of Io's surface with the age of the moon itself; a young surface can sit on an old body.

Why the other choices are wrong

  • A. This choice is too extreme. The text states craters are erased by lava, not that impacts never occur. A lack of visible craters does not prove no asteroids or comets have ever struck the moon.
  • B. This confuses the age of the surface with the age of the moon itself. A young surface can exist on an old body if volcanic activity constantly renews the ground, which is what the text describes.
  • D. This overgeneralizes the finding. The text only shows that crater counting is unreliable for Io specifically because of its volcanoes, not that the method is useless for any surface.
Ask Owl about this questionNext question

Got it wrong? On SAT Owl, Owl looks at the answer you chose and explains the exact mistake, then gives you similar questions until it sticks. Get your predicted SAT score free, or take a free full-length practice test.