In the relation $x > y + z$, $x + y > p$ and $z < p$, which of the following is necessarily true?

### Concept & Logic The transitive property of inequalities states that if $A > B$ and $B > C$, then $A > C$. Finding a common variable that bridges two inequalities is the key to determining a necessary truth. ### Step-by-Step Solution 1. Identify the given relations: $x > y + z$ (Equation 1), $x + y > p$ (Equation 2), and $z < p$ (Equation 3). 2. Rewrite Equation 3 so the inequality sign faces the same direction as Equation 2: $p > z$. 3. Analyze the relationship between Equation 2 and the rewritten Equation 3. 4. We have $x + y > p$ and $p > z$. 5. Applying the transitive property, we can chain these together: $x + y > p > z$. 6. This strictly implies that $x + y > z$. 7. The first relation ($x > y + z$) is extra information and is not required to prove this conclusion. ### Exam Strategy & Shortcut Scan the inequalities for a "bridge" variable. Here, $p$ appears in two separate simple relations. Linking $x + y > p$ directly to $p > z$ gives you $x + y > z$ in seconds without any complex manipulation or number substitution. ### Common Pitfall Getting distracted by the first equation ($x > y + z$) and attempting to add, subtract, or substitute all three equations together. This wastes time and can lead to mathematical dead-ends or incorrect assumptions. ### Final Answer **Therefore, the correct answer is $x + y > z$.**