RRB Technician Grade III 2024, 2025 PYQs

Concept: Find the gaps between letters using their positions.

1. FHK: $F(6)\to H(8)\to K(11)$, gaps $+2,+3$.
2. UWZ: $21\to 23\to 26$, gaps $+2,+3$; PRU: $16\to 18\to 21$, gaps $+2,+3$.
3. JLP: $10\to 12\to 16$, gaps $+2,+4$ – different.

Trick: Learn EJOTY ($5,10,15,20,25$) to get letter positions quickly.

Answer: (C) JLP

Concept: A word common to two sentences takes the code common to both.

1. 'someday in paradise' → yd pl ws.
2. 'in dense forest' → fc ws nm.
3. Common word 'in', common code 'ws'.

Trick: Match common words with common codes – no need to decode the rest.

Answer: (D) ws

Concept: Arrange the letters alphabetically and compare position by position.

1. Original: A R T I C L E.
2. Sorted: A C E I L R T.
3. Same position: A (1st) and I (4th) – 2 letters.

Trick: Write both words one under the other and tick the matching columns.

Answer: (B) 2

Concept: Draw the least-overlap Venn diagram and test each conclusion.

1. All shirts are pants – the shirts circle lies inside pants.
2. I: Some pants are shirts – true (conversion of 'All'). Follows.
3. II: Tie touches pants, not necessarily shirts – does not follow.

Trick: 'All A are B' always gives 'Some B are A'; a 'Some' link through B does not reach A.

Answer: (B) Only conclusion I follows

Concept: An effect must be a direct, logical response to the cause.

1. I: Recording entries/exits and verifying delivery staff directly targets thefts – valid.
2. II: Installing CCTV cameras is a direct anti-theft step – valid.
3. III: Appointing more women constables has no link to bicycle thefts – not valid.

Trick: Ask 'Would this happen because of the cause?' – keep only actions that address that exact problem.

Answer: (B) Both I and II can be the possible effects

Concept: Compare the moves from 1st to 2nd and 2nd to 3rd letter.

1. RYT: $18\to 25\to 20$, i.e. $+7,-5$.
2. JQL: $10\to 17\to 12$, $+7,-5$; FMH: $6\to 13\to 8$, $+7,-5$.
3. MTP: $13\to 20\to 16$, $+7,-4$ – odd one.

Trick: Shortcut: in the group, 3rd letter $=$ 1st letter $+2$ (R→T, J→L, F→H); M→P is $+3$.

Answer: (B) MTP

Concept: Each letter moves back by 9 positions (wrapping from A to Z).

1. $Z(26)\to Q(17)\to H(8)\to Y(25)\to P(16)$.
2. $S(19)\to J(10)\to A(1)\to R(18)\to I(9)$.
3. $N(14)\to E(5)\to V(22)\to M(13)\to D(4)$.
4. Next term: PID.

Trick: Moving back 9 is the same as moving forward 17 – use whichever is easier near A/Z.

Answer: (C) PID

Concept: Test options so that one rule fits both pairs.

1. Option (D): $47\times 2+2=96$.
2. $53\times 2+2=108$ – same rule.
3. Others fail, e.g. (A): $58\times 2-20=96$ but $53\times 2-20=86\ne 112$.

Trick: When the second number is about double, try $\times 2\pm k$ and check the same $k$ on both sides.

Answer: (D) X = 47, Y = 108

Concept: Vowels go one letter back, consonants one letter forward.

1. F→G, R→S, I→H, E→D, N→O, D→E, L→M, Y→Z.
2. New group: G S H D O E M Z.
3. All 8 letters are different – none appears twice.

Trick: Y is treated as a consonant here; check repeats only in the new group.

Answer: (C) 0

Concept: Number the letters of the first word 1 2 3 4 and track their order.

1. JAIL (1234) → ALIJ $=2431$ → LIAJ $=4321$ (reverse).
2. ZEAL → ELAZ (2431) → LAEZ (4321) – same.
3. BORE → OERB (2431) → EROB (4321) – matches.

Trick: The third word is just the first word reversed – check that first to eliminate options fast.

Answer: (D) BORE - OERB - EROB

Concept: Two 'Some' statements in a chain give no definite link between the end terms.

1. Rocks–Apples, Apples–Horses, Horses–Nails are each 'Some' links.
2. I: Nails and Apples are linked only through Horses (Some + Some) – no definite conclusion.
3. II: Horses and Rocks are linked only through Apples (Some + Some) – no definite conclusion.

Yaad rakhein: Some + Some = no conclusion. Draw a chain of just-touching circles to see it.

Answer: (C) Neither conclusion (I) nor (II) follows.

Concept: Trace each move on a grid with A at the origin (East = $+x$, North = $+y$).

1. North 7: $(0,7)$; right → East 2: $(2,7)$; right → South 1: $(2,6)$.
2. Left → East 3: $(5,6)$; right → South 3: $(5,3)$; right → West 3: $(2,3)$.
3. Left → South 3: B $=(2,0)$.
4. A $=(0,0)$ lies 2 km to the West of B.

Trick: Only net displacement matters: East $2+3-3=2$, North $7-1-3-3=0$.

Answer: (D) 2 km West

Concept: Decode each symbol from left to right and draw the family tree.

1. A + B: A is the mother of B.
2. B − C: B is the husband of C; C + D: C is the mother of D.
3. D × E: D is the sister of E, so E is a child of B and C.
4. B is E's father and A is B's mother – A is E's father's mother.

Trick: Mark gender as soon as a relation reveals it (mother ♀, husband ♂, sister ♀).

Answer: (B) Father’s mother

Concept: Draw the Schools circle inside the Colleges circle, with University completely separate.

1. All schools are colleges ⇒ every school is a college, so some colleges (at least the schools) are schools ⇒ I follows.
2. No college is a university and every school is a college ⇒ no school is a university, so 'Some schools may be universities' is impossible ⇒ II does not follow.

Yaad rakhein: 'All A are B' ⇒ 'Some B are A' always follows; 'All A are B' + 'No B is C' ⇒ 'No A is C', so 'A may be C' is false.

Answer: (C) Only conclusion I follows

Concept: Treat each letter position as its own series.

1. 1st letters: Y, Z, A, B, C ($+1$).
2. 2nd letters: E, G, I, K, M ($+2$).
3. 3rd letters: S, V, Y, B, E ($+3$, wrapping Z → A).
4. Next term: CME.

Trick: Split clusters column-wise; each column usually has a simple constant step.

Answer: (B) CME

Concept: Look at every symbol followed by an odd number, then check what is just before it.

1. Symbols followed by odd numbers: # (1), $ (7), & (3), Ω (5).
2. Before them: # ← 9, $ ← 8, & ← 7, Ω ← 3 – all numbers.
3. No symbol has a symbol before it and an odd number after it.

Trick: Scan for the rarer condition first (odd number after a symbol) – it gives a short list to check.

Answer: (A) None

Concept: All face north, so their left is our left. Fix A and B first.

1. Three people right of A: A at 4. B at 2.
2. F sits between C and G, so C, F, G are consecutive; D is second to the left of C.
3. Only D at 3, C at 5, F at 6, G at 7 fits; E takes seat 1.
4. Between E (1) and D (3): only B – one person.

Trick: 'Immediate neighbour of both C and G' means F sits exactly in the middle of them.

Answer: (C) One

Concept: Letters in a word are coded by digits in jumbled order; compare common letters and digits.

1. TREF → 3528, KERF → 5438.
2. Common letters R, E, F ↔ common digits 3, 5, 8.
3. Remaining: T ↔ 2 and K ↔ 4.

Trick: The letter unique to a word takes the digit unique to its code.

Answer: (C) 4

Concept: Each letter moves back by 11 positions.

1. $V(22)\to K(11)$, $S(19)\to H(8)$, $U(21)\to J(10)$, $R(18)\to G(7)$: all $-11$.
2. Check: $Y(25)\to N(14)$ – also $-11$.
3. SPRO: $S(19)\to H(8)$, $P(16)\to E(5)$, $R(18)\to G(7)$, $O(15)\to D(4)$.

Trick: Shifting all letters by the same amount keeps the pattern – only the 1st letter must be checked to pick an option starting with H, then verify the order.

Answer: (A) HEGD

Concept: Check each symbol: the element just before it and just after it must both be numbers.

1. Series: 6 @ $ 7 & 9 # 1 * £ 5 3 2 + % 8 4 Ω
2. 7 & 9 ✓ and 9 # 1 ✓
3. @, $, *, £, +, %, Ω fail (a symbol or the end is next to them)
4. Count $=2$

Trick: Look for the pattern "number–symbol–number" while scanning once left to right.

Answer: (C) 2