Here is another set on another 123 tricky questions covering allmost all topics in PS and DS.

If you have any doubt then please bring it on the forum. Lets analyse together.

How are you supposed to actually learn how to solve these questions? I could spend an hour trying to solve some of the questions and even that I will probably still be stumped....is there a good way to learn the tricks to solve these toughies?

These questions are for practice. You don't have to 'learn' the method for each one. First figure out if you are ready for them i.e. can you solve the easier questions comfortably? Are your basics in place? Let's say, can you solve most OG12 questions in less than a minute? Once you know that you have tamed the easy ones, then go for the difficult ones.
You just have to try and solve them and if you are unable to, then check out the explanations. If you have doubts in a question, post it and people will give their take on it. With every new question, you will learn something new. You will get different ways of figuring out the answer since most people have their favored mechanisms. Finally most questions in GMAT are based on a handful of basics. Once you have seen 20 different applications of the same concept, it doesn't matter how the concept is presented to you. You will know how to deal with it.
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AB + CD = AAA, where AB and CD are two-digit numbers and AAA is a
three digit number; A, B, C, and D are distinct positive integers. In the
addition problem above, what is the value of C?

In this - how can u say that AAA should be 111 - i dint get it.

A B
+C D
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A A A


Notice here that B+D ends with A (there could be a carry over so I cannot say that B+D = A)
Also, A + C ends with A.
When can this happen? e.g. Think 2 + x = .....2
What can you say about x? We can say that it must end with 0. Only then will you have 2 at the end.

Since all letters represent positive integers, C cannot be 0. Then it must be 9 and there must have been a carry over from previous addition to make 10. Then A + 10 will end with A and there will be a carry over of 1 which will appear as the hundred's digit. Since the hundred's digit is A, it must be equal to 1.
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A test has 50 questions. A student scores 1 mark for a correct answer, –1/3 for a wrong
answer, and –1/6 for not attempting a question. If the net score of a student is 32, the number
of questions answered wrongly by that student cannot be less than
1. 6 2. 12 3. 3 4. 9
Soln. (3) — Let the number of correct answers be ‘x’, number of wrong answers be ‘y’ and
number of questions not attempted be ‘z’.
Thus, x + y + z = 50 … (i)
And  x – y – z 32
3 6
The second equation can be written as,
6x – 2y – z = 192 … (ii)
Adding the two equations we get,
7x – y = 242 or x = 242 + y
7
Since, x and y are both integers, y cannot be 1 or 2. The minimum value that y can
have is 3.

can u explain why Y can not be 1 or 2....i tried to make 242/7+Y integer but even 3 is not enough

The equation you get is 7x - y = 242
x = (242 + y)/7 (don't forget that the entire 242+y is divided by 7, not just 242)
Since x must be an integer, (242+y) must be divisible by 7. After 242, the closest multiple of 7 is 245 (if you are wondering how to get it, divide 242 by 7. You get 4 as remainder. So you need another 3 to go to the next multiple of 7). So y must be at least 3.
Think: Can y take other values? If yes, which ones?
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