RRB NTPC Undergraduate CBT 1 2025 PYQs

Concept: Vowels move one letter forward; consonants move one letter back.

1. P → O, E → F, R → Q, B → A, L → K, A → B, D → C.
2. New group: O F Q A K B C.
3. All seven letters are different, so no letter appears twice → zero.

Trick: Write the new letters in alphabetical order (A B C F K O Q) – repeats become obvious at once.

Answer: (D) Zero

Concept: Try each option, then solve using BODMAS.

1. As given: $(6\times 14)÷2+18\times 4-28÷7+34=42+72-4+34=144\ne 87$.
2. Interchange 4 and 2: $(6\times 14)÷4+18\times 2-28÷7+34$.
3. $=21+36-4+34=87$ ✓.

Trick: The value must fall from 144 to 87 – swap numbers so that a big term (18 × 4) shrinks.

Answer: (D) 4 and 2

Concept: All face north, so left and right are the solver's left and right. Fix the sure positions first.

1. Only K is to the right of E → K at the right end (7), E at 6.
2. One person between I and E → I at 4. Two people between C and I → C at 1 (7 is taken).
3. R, S, T fill 2, 3, 5 in the order R – S – T (S between R and T) → R 2, S 3, T 5.
4. Row: C R S I T E K → third from the right is T.

Trick: Clues with ‘only’ (only K to the right) lock the end seats – start there.

Answer: (D) T

Concept: Facing north, left–right is the same as ours. Place the people fixed by count clues first.

1. Five people to the right of D → D is 2nd from the left. Four people to the left of H → H is 5th.
2. R is the neighbour of both A and D → R at 3 and A at 4 (R at 1 would leave no seat for A).
3. Seats 1, 6, 7 remain; J is immediately right of B → B 6, J 7, so U 1.
4. Row: U D R A H B J → U and D are adjacent, no one between them.

Trick: ‘n people to the right’ in a row of 7 means position $7-n$ from the left.

Answer: (C) None

Concept: Number the positions 1 (top) to 7 (bottom) and place the fixed boxes first.

1. Three boxes above C → C at 4. One box between D and C → D at 2 or 6.
2. Three boxes between D and E, and E above C → D at 6, E at 2. G just below E → G at 3.
3. Positions 1, 5, 7 remain for A, B, F. A not next to D (5 or 7) → A at 1; F above B → F 5, B 7.
4. Stack: A, E, G, C, F, D, B → box just above F is C.

Trick: Test both cases of an ‘either–or’ clue; the one that breaks a later rule is dropped.

Answer: (D) C

Concept: Follow each letter position and the number separately.

1. First letters: A, E, I, M → $+4$ → Q.
2. Second letters: D, H, L, P → T; third letters: G, K, O, S → W.
3. Numbers: 3, 6, 9, 12 → $+3$ → 15.
4. Missing term = QTW 15.

Trick: Inside each cluster letters are 3 apart (A_ _D_ _G) – Q, T, W keeps this gap.

Answer: (C) QTW 15

Concept: Number positions 1 (top) to 6 (bottom).

1. No box above G → G at 1. Three boxes below A → A at 3.
2. Two boxes between A and E → E at 6.
3. Positions 2, 4, 5 remain; F above H above D → F 2, H 4, D 5.
4. H and D are adjacent → zero boxes between them.

Trick: ‘No box above X’ means X is on top – the easiest anchor.

Answer: (A) Zero

Concept: Find the operation that turns the first number into the second in both given pairs.

1. $104-89=15$ and $172-157=15$ → rule: add 15.
2. $138+15=153$ ✓.
3. Others: $56+15=71\ne 72$, $97+15=112\ne 111$, $39+15=54\ne 64$.

Trick: Small, equal differences → simple addition rule; check the difference first.

Answer: (D) 138, 153

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

1. ‘bat view month’ → nd wz nh.
2. ‘oven month gum’ → on ju wz.
3. Common word = month; common code = wz → month = wz.

Trick: Common word ↔ common code – ignore positions.

Answer: (C) wz

Concept: Number the positions 1 (top) to 7 (bottom).

1. Only F above G → F at 1, G at 2. Three boxes below D → D at 4.
2. Two boxes between D and C → C at 1 or 7; 1 is F, so C at 7.
3. Positions 3, 5, 6 remain; A above B above E → A 3, B 5, E 6.
4. Stack: F, G, A, D, B, E, C → box just above B is D.

Trick: ‘Only F above G’ fixes the top two boxes at once.

Answer: (D) D

Concept: Compare letters position-wise between the two clusters.

1. DCF → GFI: D+3 = G, C+3 = F, F+3 = I.
2. YXA → BAD: Y+3 = B, X+3 = A, A+3 = D (wrapping after Z).
3. RQT → UTW: R+3 = U, Q+3 = T, T+3 = W ✓.
4. TRT → WVY fails (R+3 = U, not V); TSV → VVX and TRT → VUX fail at the first letter.

Trick: Check only the first letter of each option first – T+3 = W, T+2 = V – it eliminates fast.

Answer: (B) RQT − UTW

Concept: Each term = previous term $\times 2$ minus an increasing number.

1. $8\times 2-1=15$, $15\times 2-2=28$, $28\times 2-3=53$
2. $53\times 2-4=102$
3. Check: $102\times 2-5=199$ ✓

Trick: Terms roughly double each time → test $\times 2$ and look at the small leftover.

Answer: (A) $102$

Concept: Find the alphabet-position gap between consecutive letters (A = 1 … Z = 26, wrapping around).

1. ICW: I(9) − 6 = C(3); C(3) − 6 → W(23).
2. NHB: N(14) − 6 = H(8); H(8) − 6 = B(2).
3. GAU: G(7) − 6 = A(1); A(1) − 6 → U(21).
4. FZO: F(6) − 6 → Z(26), but Z − 6 = T, not O. So FZO is odd.

Trick: Backward −6 with wrap: from A go to U, from C go to W.

Answer: (A) FZO

Concept: Each step moves the first letter of the word to the end.

1. FORK → ORKF (F to end) → RKFO (O to end).
2. TUCK → UCKT → CKTU – same rule.
3. BAKE → AKEB → KEBA ✓.
4. THEM → MTHE moves the last letter to the front – opposite rule; (C) and (D) break the order.

Trick: Read the 2nd word: it should start with the 2nd letter of the 1st word.

Answer: (A) BAKE - AKEB - KEBA

Concept: Look for the pattern Odd – Even – Even and count the middle digit.

1. Scan: 3 8 4 ✓ (3 odd, 4 even).
2. 9 4 6 ✓ (9 odd, 6 even).
3. 7 2 4 ✓ (7 odd, 4 even).
4. Others fail: 9 8 9 (followed by odd), 3 6 3 (followed by odd). Total = 3.

Trick: First mark every odd digit, then check the next two digits are both even.

Answer: (C) Three

Concept: Decode each symbol and build the tree from H.

1. Y ÷ H: Y is the son of H.
2. R # Y: R is the wife of Y (H's son's wife).
3. D & R: D is the father of R; S + D: S is the brother of D.
4. So S is H's son's wife's father's brother.

Trick: Read the chain backwards from H: son → wife → father → brother.

Answer: (C) Son's wife's father's brother

Concept: Number positions 1 (bottom) to 7 (top) and fix the definite clues first.

1. No box below L → L = 1. Four boxes between M and L → M = 6. H above M → H = 7.
2. Left: positions 2–5 for G, I, N, O. O and N need two boxes between them → 2 and 5.
3. I immediately above G → G = 3, I = 4.
4. N not immediately above I → N ≠ 5, so N = 2, O = 5. Fourth from bottom = I.

Trick: Start with words like 'no box below' – they fix an exact position.

Answer: (D) I

Concept: Take A as origin; east and north positive.

1. North 2 → (0, 2); right = east 5 → (5, 2).
2. Right = south 7 → (5, −5); right = west 8 → (−3, −5).
3. Right = north 5 → (−3, 0) = P.
4. P is 3 km west of A, so she must drive 3 km to the east.

Trick: Net east-west: $5-8=-3$ (3 km west); net north-south: $2-7+5=0$.

Answer: (A) 3 km to the east

Concept: Vowel → next letter; consonant → previous letter.

1. C→B, O→P, N→M, T→S, A→B, C→B, T→S.
2. New group: B P M S B B S.
3. First from the right = S.

Trick: Only the last letter matters: T is a consonant → S.

Answer: (C) S

Concept: Compare the first letters (+7) and the second letters of the two clusters.

1. QM – XJ: Q+7 = X, M−3 = J.
2. HP – OM: H+7 = O, P−3 = M.
3. RD – YA: R+7 = Y, D−3 = A.
4. LG – SC: L+7 = S but G−4 = C (not −3). Odd one.

Trick: When all first letters fit, the odd one hides in the second letters.

Answer: (D) LG – SC