We can easily handle it. Like, we know that sum of A and C would have to be >= 10 so that we can carry 1 for Position E as it's not 0. But if we take 5 and 6 as A and C then E and F become same which is invalid ( all have to be different). But we if take 6 and 4 as A and C then we get F =0 and E = 1. We are left with 2, 3 and 5. We can do it simply, talking B, D as 2, 3 and adding them would give 5=G.

If we add two 2-digit numbers and the sum is a 3-digit number, then the 3-digit number must start with a 1.
So, E = 1

In order for the sum to be a 3-digit number, A+C must be greater than 9
So, we have two options:
EITHER A and C are 5 and 6, OR A and C are 4 and 6
If A and C are 5 and 6, then F = 1, but we already know that E = 1
So, it MUST be the case that A and C are 4 and 6, which means F = 0

The three digits unaccounted for are 2, 3 and 5
Since we can see that B + D = G, it must be the case that G = 5

Answer: D

Cheers,
Brent

Can we expect something like this in the actual test?

I saw a similar question in the BB book. However, based on power prep plus e GRe mentor course, I did not spot any question like this

very low if not zero probability to see it