This is actually one of my favorite problems in all of the OGs! There's a lot to learn about good PS process here.
I would *NOT* recommend calculating here. I think it's unrealistic to think that someone without a solid quant background would execute that flawlessly in 2 minutes. You are definitely NOT expected to know the square root of 576 off the top of your head, without a calculator!
Brent's solution of factoring out a 16 or David's solution of doing prime factorization are the best approaches here.
But... there is something you can and should do before you start solving, though. With PS questions, you should always scan the answer choices first, and see if you can eliminate any that are nonsensical or obvious traps.
The problem gives us a single root sign with a complicated sum underneath. We know that we're never allowed to split the root of a sum into the sum of roots:
So, we can automatically eliminate D. There's no way to get a sum of 2 roots from the root of one sum.
We can also eliminate A, since it's unsimplified. sqrt(20) can be simplified to 2(sqrt(5)). The GMAT will never have an unsimplified root in a correct answer. (You may also have noticed that A would just represent the first half of the sqrt in the problem - another reason to eliminate it).
Now that we're left with B, C, and D, we know that the right answer must be an integer. This implies that what's underneath the root sign must be a perfect square.
The square of 25 would have to be an odd integer, ending in 5. If all of the terms under the root sign are even, there's no way we'll end up with an odd result. Eliminate C.
Now, looking at B and E, we can get the right answer with very little effort. 32 is a number in the problem, making it less likely to be right. But also, when we square 32, we'll get some number with a units digit of 4. When we square 24, we'll get some number with a units digit of 6.
See how quickly you can find the units digit of what's under the root sign: 6*0 = 0, 8*2=6, 0 + 6 = 6. We need a units digit of 6.
The answer must be
B.
