Such questions don't need any sort of calculation. What it need is just thinking. Here's how:
Minimum s can be 10. And the ratio of t to u is always less than 1. Thus the result will be between 10 and 11 i.e less than 11.
On other hand, s can be greater than 11, and thus Quantity A will become greater in that case because the ratio of t to u cannot be negative.

So, information is not sufficient to answer the question.
Choice D correct.

I got this one wrong because I didn't look closely enough and see that it was either greater or equals, and only noticed the greater bit

For me, I calculated this by making s, t, & u equal by applying their Maximum and Minimum values of 20 and 10 respectively. Is this reasoning appropriate?

Answer: D
We assess the biggest and lowest possible amounts for A to be able to compare it with 11.

If we want to maximize A: u should be the lowest and t and s should be the biggest possible. But there might be a tradeoff between u and (s and t).
Minimum value for u can be 12 while t is 11 and s is 10. Then A=S+T/U= 10+11/12 which is less than B, but this equation can have a bigger value, for example when s=11 and t=12 and u is whatever between 13 to 20, the value of A is bigger than 11.
(The maximum value for A is when s is maximum, because u/t is always something less than 2 and doesn’t have so much effect. And it is when s has the minimum value. For u=20, t=19 and s=18 we have the maximum value of A which is 18+19/20.)


If we want to minimize A: u should be maximum value and t&s should be minimum. But again there might be a tradeoff between u and (s and t)
Maximum value that u can have is 20 (as it can be equal to the upper bound(20) and the minimum value for s can be 10 (as it can be equal to the lower bound(10) and t can’t be equal to 10, the Lowes possible value for t can be 11. Thus we have:
A = S+T/U = 10 + 11/20 that is less than 11 (B)

Answer is D. As A is between 10+11/20 and 18+19/20

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