2025 CSAT — Q19

A relation is “always” forced only if an unbroken chain of the same-direction inequalities links the two letters. Hunt for the break.

Statement I: $A \le B > C < D > E > F \ge G = H$. To compare $B$ and $E$, trace from $B$: $B > C$, but then $C < D$ — the direction reverses at $C$. The chain $B>C<D>E$ is broken, so $B$ and $E$ are not linked. Counterexample: $B=2,\ C=1,\ D=10,\ E=5$ (with $A\le2$, $F<5$, etc.). Then $B=2 \not> E=5$. So I is FALSE.

Statement II: $P > Q = R \le S = T \le U = V > W$. Trace $S$ to $V$: $S = T \le U = V$, giving $S \le V$. The relation that is forced is $S \le V$ — equality is allowed (take $T = U$). So ”$S$ is always less than $V$” overstates it: $S = V$ is possible. Counterexample: $S=T=U=V=3$ gives $S = V$, not $S < V$. So II is FALSE.

Both fail. Answer: (d) Neither I nor II.

Two distinct traps live in one item. I tests whether you spot a direction reversal ($>$ then $<$) that severs the chain; II tests whether you respect the $\le$ vs $<$ distinction — a single non-strict link makes “always strictly less” false. The scoring discipline for inequality-chain claims is: a strict conclusion needs an unbroken run of strict inequalities; one $\le$ anywhere, or one reversal, kills the “always.” Build the equality/reversal counterexample rather than arguing in words.