Finding relation between elements I have seen the following type of problems in reasoning tests:
If $A \le B = C > F < G = L$ , then which of the following is true? 
a. $A < G$ and $A >F$
b. $C >F$ 
c. $G =B$
[Answers are given by me just as examples.]
Can anybody explain a nice general method of solving these problems?
 A: Here's how I would do it. This isn't 100% maximally efficient, but it will get you the correct answer consistently.
Firstly, gather equal terms.
$$A \leq \{B,C\} > F < \{L,G\}$$
Then reorganize the available information.


*

*$A \leq \{B,C\}$

*$F < \{B,C\}$

*$F < \{L,G\}$


Then reorganize the available answers.
a. $A < G$ and $F < A$
b. $F < C$
c. $G = B$
Now just read off the answer. In this case, we can see that 2 implies b. (Thus b is the correct answer).
Its also important to remember that if $x < y$, then $x \leq y.$
This kind of reasoning works in quasi-ordered sets, but note that more sophisticated arguments are valid in partially ordered sets, and especially in a totally ordered sets.
A: I assume that $A,B,C,F,G,L$ are real numbers. I think it is better to rewrite the assumption this way :
$\begin{eqnarray}
A & \leq & B \tag{1}\\
B & = & C \tag{2}\\
F & < & C \tag{3}\\
F & < & G \tag{4}\\
G & = & L \tag{5}\\
\end{eqnarray}
$
Using $(2)$, inequalities $(3)$ and $(4)$ give $F < B$ and $F < G$. So, we know that $B$ and $G$ are both greater than $F$ but not necessarily equal. (For example, take $A=0$, $B=C=1$, $F=0$ and $G=L=2$. In this example, we don't have $G=B$ but $(1)$ to $(5)$ hold.
From $(3)$, the statment $C>F$ is obviously true.
The statment $A<G$ is false. Consider $A=-1$, $B=C=2$, $G=L=1$ and $F=0$. $(1)$ to $(5)$ still hold but we don't have $A<G$. The statment $A > F$ is also false.
To find such counter-examples, you could draw a line (which represents $\mathbb{R}$) and place $A,B,C,F,G,L$ on this line according to $(1)$,...,$(5)$. It might also help you to see which statement is true or not.
A: One can apply the principle:

No conclusion can be drawn about the ordering of items separated by at least one $\lt$ or $\leqslant$ and at least one $\gt$ or $\geqslant$.

In the present case, this means that A can only be compared to B and C, hence anything comparing A to F, G or L fails. Thus, 1. fails. Likewise, G can only be compared to L and F, hence anything comparing G to A, B or C fails. Thus, 3. fails.
One is left with 2., which is a part of the hypothesis.
A: I would usually start by cancelling out answers that are definitely not right... But it's obvious it's b because it's in the question -_-
