Choose the group of letters which is different from others.

Question

UnlockIAS · Accepted Answer

Correct — A, PSVX. Convert every letter to its position in the alphabet and the question answers itself. Option (b) ORUX is O = 15, R = 18, U = 21, X = 24, so the steps are +3, +3, +3. Option (c) JMPS is J = 10, M = 13, P = 16, S = 19 — again +3, +3, +3. Option (d) CFIL is C = 3, F = 6, I = 9, L = 12 — +3, +3, +3 once more. Option (a) PSVX is P = 16, S = 19, V = 22, X = 24, and its steps are +3, +3, +2. The final jump is one letter short: to hold the pattern the group would have had to end at Y, position 25, giving PSVY. That broken step is the only property that isolates exactly one of the four groups, which is what an odd-one-out question requires, so PSVX is the odd group. The same conclusion falls out of a faster check worth keeping as a habit: measure the span from the first letter to the last and skip the intermediate arithmetic altogether. ORUX spans 24 − 15 = 9, JMPS spans 19 − 10 = 9, CFIL spans 12 − 3 = 9, and PSVX spans 24 − 16 = 8. One span differs from the other three and you are done in four subtractions. It is equally important to see what does not separate them, because discarding a candidate classifier is the real work in this item type. Vowel content splits the set unevenly rather than three against one — ORUX carries two vowels, CFIL one, and JMPS and PSVX none. No group forms a word, no group repeats a letter, and all four begin with a consonant. Only the step size gives a 3–1 split, and it points at PSVX.

Why the other options are wrong — (b) A clean arithmetic chain: O = 15, R = 18, U = 21, X = 24, stepping +3 each time and spanning 9. It shares its final letter X with PSVX, which is the deliberate distraction — the shared endpoint tempts you to compare these two groups against each other rather than testing all four by the same rule.  (c) J = 10, M = 13, P = 16, S = 19: +3, +3, +3, span 9. It is the group most people convert last because its letters sit in the middle of the alphabet and are the hardest to place by eye, but it follows exactly the same rule as ORUX and CFIL and so cannot be the odd one.  (d) C = 3, F = 6, I = 9, L = 12 — the easiest chain to verify, since its positions are simply the first four multiples of three, and again +3, +3, +3 with a span of 9. Starting at the front of the alphabet makes it look unlike the others superficially, but position of the starting letter is not a property the question is testing.

Concept: Every letter-group question rests on one move: replace each letter by its position in the alphabet, so that a pattern in symbols becomes arithmetic you can actually compute. A = 1 through Z = 26, and the EJOTY mnemonic fixes five anchors — E = 5, J = 10, O = 15, T = 20, Y = 25 — so any letter is at most two counts away from a number you already know. Once the groups are numbers, only a handful of patterns ever appear in this exam. A constant step, as here. Alternating steps, such as +6, +3, +2 repeated across every group. Complementary or mirror letters, where a letter and its opposite from the far end of the alphabet sum to 27 (A + Z, B + Y, C + X and so on). Skip patterns that count forward through consonants only, or through the alphabet in reverse. In an odd-one-out item you are not asked to name the pattern the setter had in mind; you are asked to find a single test under which exactly one group behaves differently, which means a valid classifier must produce a 3–1 split and nothing else will do.

State PCS 2024 Q44 — Science & Technology PYQ with Answer

State PCS 2024 Q44 — Science & Technology PYQ with Answer