2024 CSAT — Q76

Decode each symbol, then test whether the conclusion is forced or can be broken by a counterexample.

Statement 1. Premises: $2x \ge 3y$ and $3x = 4z$. Claimed conclusion: $9y > 8z$.

From $3x = 4z$, $z = \tfrac{3x}{4}$, so $8z = 6x$. From $2x \ge 3y$, $3y \le 2x \Rightarrow 9y \le 6x$. Hence $9y \le 6x = 8z$, i.e. $9y \le 8z$. The conclusion $9y > 8z$ is therefore never true — the derivation gives the opposite inequality. Statement 1 is false.

Statement 2. Premises: $x < 2y$ and $y \le z$. Claimed conclusion: $x \le z$.

From $y \le z$, $2y \le 2z$, so $x < 2y \le 2z$ gives only $x < 2z$ — not $x \le z$. Counterexample: $x = 1.9,\ y = 1,\ z = 1$ satisfies $x < 2y\ (1.9 < 2)$ and $y \le z\ (1 \le 1)$, yet $x \le z$ would need $1.9 \le 1$, which is false. Statement 2 is false.

Both fail, so neither is correct.

Answer: (d) Neither 1 nor 2.

Coded-inequality items reward substituting the symbol meanings and then doing genuine inequality algebra, not pattern-matching the letters. In Statement 1 the equality $3x = 4z$ lets you eliminate a variable and discover the conclusion is exactly reversed — a $\le$, not the claimed $>$. In Statement 2 the factor of $2$ in "$x < 2y$" is the trap: it survives into $x < 2z$, which is strictly weaker than $x \le z$, so a single fractional counterexample ($x$ just under $2$, $y = z = 1$) demolishes it. The general discipline: to validate ”$A$ relation $B$” claims, either prove the chain or kill it with one explicit counterexample — do not assume transitivity when a coefficient distorts the scale.