RRB Technician Grade I 20 Dec 2024 Shift 1 Paper

Concept:Change each vowel to the second letter after it and each consonant to the previous letter, then count the new clusters that have no vowel.Explanation:Original clusters: $\mathrm{CAB},\ \mathrm{BIT},\ \mathrm{YET},\ \mathrm{AFT}$.Apply the given transformation to each letter:$\mathrm{CAB} \to \mathrm{BCA}$ because $C \to B$, $A \to C$, $B \to A$.$\mathrm{BIT} \to \mathrm{AKS}$ because $B \to A$, $I \to K$, $T \to S$.$\mathrm{YET} \to \mathrm{XGS}$ because $Y \to X$, $E \to G$, $T \to S$.$\mathrm{AFT} \to \mathrm{CES}$ because $A \to C$, $F \to E$, $T \to S$.Now check vowels in the new clusters:$\mathrm{BCA}$ has $A$, so it contains a vowel.$\mathrm{AKS}$ has $A$, so it contains a vowel.$\mathrm{XGS}$ has no $A,\ E,\ I,\ O,\ U$, so it has no vowel.$\mathrm{CES}$ has $E$, so it contains a vowel.Therefore, exactly one cluster has no vowel.Answer:One cluster. Correct option: D.