Firstly, let me just remark that problems by McGraw-Hill are generally terrible. So don't take them as a very accurate gauge of how the GRE works.
This one is not too bad though. When we find the area of circles, we don't really care about diameters; we care about radii. We could quickly eliminate 3 answer choices by thinking along these lines: If the diameter is an integer, then the radius must also be an integer, or end in .5. For example, if the diameter was 10, the radius must be 5, but if the diameter was 9, the radius would be 4.5. So the area, which is πr^2, should either be a perfect square (if the radius was an integer) or end in .25 (if the radius ended in .5). C, D, and E fulfill neither of those categories. So it's either A or B.
We've also been told that 6 ≤ d ≤ 10. We still don't care about
d, so I'd replace it with
2r and we'd get 6 ≤ 2r ≤ 10. Dividing all three parts gets us 3 ≤ r ≤ 5. So the radius must be 3, 4, or 5, and the areas could be 9π, 16π, or 25π. So it's B.
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