Let's say r = # of packages of rice and c = # of packages of corn.

S1::

17c + 13r = 282

Suppose we ONLY had packages of corn. Then we would have more than 16 packages, since 17*16 = 272.

Suppose we ONLY had packages of rice. Then we would have less than 22 packages, since 13*22 = 286.

So we (c + r) must be between 17 and 21, inclusive.

Now let's write 17c + 13r as 13c + 13r + 4c. This is really 13(c + r) + 4c, so we have

13(c + r) + 4c = 282

Notice that 4c is even. Since 282 is also even, we must be adding TWO EVEN TERMS. Hence 13(c + r) is even, which means (c + r) is even.

Since 17 ≤ c + r ≤ 21, this gives (c + r) = 18 or (c + r) = 20. (c + r) = 18 works, and gives c = 12. (c + r) = 20 does NOT work, as it gives c = 5.5, which is not an integer. So there is only ONE solution, c = 12 and r = 6; SUFFICIENT.

S2:: Obviously not sufficient alone.

Key takeaway here: if C is too easy in DS, be very suspicious!! You can certainly solve the system of two equations 17c + 13r = 282 and c = 2r, so that is TOO EASY. Since S2 doesn't work, the answer is very likely A, even if you can't say why on test day.