Let N is the number of basket, X is the number of oranges in each basket.

We will have X * N = 20, remember both X & N are integer

so we can guess X & N will be one of the following pairs (interchangeable)

1 - 20, 2 - 10, 4 - 5

I. We will choose 1 - 20, 1 is X and 20 is N. If we halve the number of basket (then now 10) we will have X is 2. So it's sufficient

II. Insufficient.

So answer is A. It's a little bit tricky to solve the problem

There are 20 oranges, and B baskets. Since each basket has at least 1 orange, and the number of oranges per basket is the same, there are only a limited number of possibilities; B must be a factor of 20.

So B can be 1, 2, 4, 5, 10 or 20.

(1) if B was halved, there would be twice as many oranges per basket. This is true for B=2, 4, 10, 20

(2) if B were doubled, there would not be enough oranges to satisfy the constraint. This is only true for B=20 (one orange per basket). If you doubled B=4 to eight baskets, you could not split the oranges evenly, but you could put one orange in each.

Statement (2) is sufficient, and Statement (1) is not sufficient.

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praxis is right, I don't pay attention to 2-10 and 4-5 case. and even maximum 20.

Thanks praxis