RRB NTPC Undergraduate CBT 1 2025 PYQs

Concept: Decode each symbol step by step and draw the family tree.

1. G * K: G is the daughter of K. K & V: K is the husband of V → G is the daughter of K and V.
2. V @ P: V is the mother of P → P is G's brother (son of K and V).
3. P # W: P is the father of W. W % J: W is the son of J → J is W's mother, i.e. P's wife.
4. So J is the wife of G's brother → brother's wife.

Trick: Two parents named for the same child (P father of W, W son of J) ⇒ P and J are husband and wife.

Answer: (B) Brother's wife

Concept: Find the position-shift from the first cluster to the second, letter by letter, in each pair.

1. RL–IQ: R(18)→I(9) = $-9$; L(12)→Q(17) = $+5$.
2. VP–MU: V(22)→M(13) = $-9$; P(16)→U(21) = $+5$. CW–TB: C(3)→T(20) = $-9$ (3−9 = −6 → 20); W(23)→B(2) = $+5$ (28 → 2).
3. OI–EN: O(15)→E(5) = $-10$; I(9)→N(14) = $+5$. The first-letter shift differs.

Trick: Pattern (−9, +5) in three pairs; OI–EN breaks it with −10.

Answer: (B) OI – EN

Concept: Take successive differences and look for a pattern in them.

1. 217 → 190 → 165 → 142 → 121 → 102: differences $-27,-25,-23,-21,-19$.
2. Differences decrease by 2 each time, so the next difference is $-17$.
3. $102-17=85$.

Trick: Consecutive odd numbers subtracted (27, 25, 23 …) – a very common second-level pattern.

Answer: (D) 85

Concept: Within each term the letters are alternate (A, C, E), and each term shifts by a fixed amount.

1. A(1) → D(4) → G(7) → J(10): $+3$ each time.
2. C → F → I → L: $+3$; E → H → K → N: $+3$.
3. Next: J+3 = M, L+3 = O, N+3 = Q → MOQ.

Trick: The series is just ACE, DFH, GIK, JLN, MOQ – every letter +3; option MPQ is a trap (P breaks the +2 gap inside the term).

Answer: (C) MOQ

Concept: Find the shift applied to each letter position in the given pairs and apply the same shifts.

1. BGKM → HKMN: B+6 = H, G+4 = K, K+2 = M, M+1 = N → shifts (+6, +4, +2, +1).
2. Check FHNP → LLPQ: F+6 = L, H+4 = L, N+2 = P, P+1 = Q ✓.
3. NRTX: N(14)+6 = T(20), R(18)+4 = V(22), T(20)+2 = V(22), X(24)+1 = Y(25) → TVVY.

Trick: Shifts need not be equal – write them as a list (+6, +4, +2, +1) and reuse.

Answer: (B) TVVY

Concept: Look at the gaps between consecutive letters inside each cluster.

1. CEH: C→E $+2$, E→H $+3$. PRU: P→R $+2$, R→U $+3$.
2. JLO: J→L $+2$, L→O $+3$.
3. UWX: U→W $+2$, W→X $+1$ – the second gap is different.

Trick: Pattern (+2, +3) in three clusters; UWX has (+2, +1), so it is the odd one.

Answer: (C) UWX

Concept: All face north, so left and right are the same as ours. Number the seats 1 (left) to 7 (right).

1. No one to the right of I → I = 7. Two people between I and K → K = 4.
2. K is to the immediate left of W → W = 5. Three people between H and W → H = 1.
3. Z is immediately right of J → J, Z need adjacent seats among 2, 3, 6 → J = 2, Z = 3; Y = 6.
4. Between Y (6) and Z (3): K and W → two people.

Trick: Fix the end person first ("no one to the right of I"), then work inward with "n people between" clues.

Answer: (D) Two

Concept: Replace the signs as instructed, then apply BODMAS (÷ and × before + and −).

1. After interchange: $2814÷7+3\times 75-5$.
2. $2814÷7=402$; $3\times 75=225$.
3. $402+225-5=622$.

Trick: Do the division and multiplication first; the big number divided by 7 gives a clean 402, which is the signal you are on track.

Answer: (B) 622

Concept: Apply the rule to each digit separately: odd digit $+2$, even digit $-2$.

1. 7213546 → 7+2 = 9, 2−2 = 0, 1+2 = 3, 3+2 = 5, 5+2 = 7, 4−2 = 2, 6−2 = 4.
2. New number: 9 0 3 5 7 2 4.
3. Third digit from the left = 3.

Trick: Only the asked position matters – third digit is 1 (odd) → 1+2 = 3; no need to convert the whole number.

Answer: (D) 3

Concept: Arrange the letters alphabetically, pick the required letters and count the letters strictly between them in the alphabet.

1. GUILDER → D, E, G, I, L, R, U.
2. Fourth from left = I (9); second from right = R (18).
3. Letters between I and R in the alphabet: J K L M N O P Q = $18-9-1=8$.

Trick: Letters between two letters = difference of positions $-1$.

Answer: (A) 8

Concept: Draw a Venn diagram: 'All eggs are squares' puts the eggs circle inside squares; 'Some crows are eggs' makes crows overlap eggs.

1. The crows that are eggs lie inside squares, so some squares are crows – Conclusion (I) follows.
2. 'All eggs are crows' is not forced: eggs may exist outside the crows circle – Conclusion (II) does not follow.

Trick: All A are B + Some C are A ⇒ Some B are C (definite); a 'Some' statement never yields an 'All' conclusion.

Answer: (B) Only conclusion (I) follows

Concept: In sentence coding, a word common to two sentences has the code common to both.

1. 'eagle gum novel' → fq ko hz; 'novel fresh zebra' → ko ef ni.
2. Common word: novel; common code: ko.
3. Hence novel = ko.

Trick: Match the overlap of words with the overlap of codes – no other step needed.

Answer: (D) ko

Concept: Substitute A = ÷, B = ×, C = +, D = − and evaluate each option with BODMAS.

1. Option D: $404÷4-15+50\times 8=101-15+400=486$ ✓.
2. Option A: $404÷4+15-50\times 8=101+15-400=-284$.
3. Option B: $404-4-15+50÷8=385+6.25=391.25$; Option C: $404\times 4-15+50÷8=1607.25$.

Trick: $404÷4=101$ and $50\times 8=400$ add up near 486 – test the option that has A with 4 and B with 8 first.

Answer: (D) 404 A 4 D 15 C 50 B 8

Concept: Track the position with coordinates, taking A as origin, North as $+y$ and East as $+x$. A left turn rotates N → W → S → E → N.

1. North 9: (0, 9). Left → West 13: (−13, 9). Left → South 19: (−13, −10).
2. Left → East 44: (31, −10). Left → North 10: (31, 0) = P.
3. P is 31 km due east of A, so he must drive 31 km towards the West to reach A.

Trick: Net N–S: $9-19+10=0$; net E–W: $-13+44=31$ km east → go back 31 km west.

Answer: (C) 31 km to the West

Concept: When neighbouring terms rise and fall alternately, split the series into two alternate sub-series.

1. Odd places: 43, 31, 19 → each $-12$.
2. Even places: ?, 67, 79 → each $+12$, so ? $=67-12=55$.
3. Check: 43, 55, 31, 67, 19, 79 – alternate differences $-12$ and $+12$ hold.

Trick: Zig-zag series → read alternate terms; both chains here use the same step 12.

Answer: (D) 55

Concept: Compare the shift of each letter from the first cluster to the second.

1. HC–GB: H→G $-1$, C→B $-1$. NI–MH: N→M $-1$, I→H $-1$. LG–KF: L→K $-1$, G→F $-1$.
2. JE–MK: J→M $+3$, E→K $+6$ – different pattern.
3. Also, within every first cluster the gap is 5 (H–C, J–E, N–I, L–G), but in MK it is only 2.

Trick: Three pairs are simply "each letter one step back"; the one that moves forward is the odd one.

Answer: (B) JE - MK

Concept: For people facing the centre, 'left' means clockwise and 'right' means anticlockwise. Fix one person and place the others.

1. Fix Ej. Bg is third to the right (3 places anticlockwise); Gv is third to the left (3 places clockwise) – in a circle of 7 they sit next to each other opposite Ej.
2. Fm is a neighbour of Bg (the seat not taken by Gv), and Cf is Fm's other neighbour → going clockwise from Ej: Ej, _, _, Gv, Bg, Fm, Cf.
3. Dn is not Ej's neighbour → Dn takes the seat two places from Ej, Ar sits immediately left of Ej.
4. From the left of Fm (clockwise) to Ar: Cf, Ej, then Ar → two people between.

Trick: Facing centre: left = clockwise. 'Third to right' and 'third to left' of the same person in a 7-seat circle are adjacent seats.

Answer: (C) Two

Concept: Compare words letter by letter from the left; the first differing letter decides the order.

1. All start with 'Br'. Third letters: Brain (a), Bread (e), Break (e), Bright (i), Bring (i).
2. a < e < i → Brain first; among Bread/Break the fourth letters match (a), fifth: d < k → Bread, Break.
3. Bright vs Bring: fourth letter g = g, fifth: h < n → Bright, Bring.
4. Order: Brain (5), Bread (3), Break (4), Bright (1), Bring (2).

Trick: Move to the next letter only when the current letters tie; vowels order a < e < i < o < u helps quick sorting.

Answer: (A) 5, 3, 4, 1, 2

Concept: Find the shift of each letter in the given pairs and apply it to the new cluster.

1. LHLP → TPTX: L(12)+8 = T(20), H(8)+8 = P(16), L+8 = T, P(16)+8 = X(24) → every letter $+8$.
2. Check DZDH → LHLP: D+8 = L, Z(26)+8 = 34−26 = 8 = H, D+8 = L, H+8 = P ✓.
3. JFJN: J(10)+8 = R(18), F(6)+8 = N(14), J+8 = R, N(14)+8 = V(22) → RNRV.

Trick: The 1st and 3rd letters are identical in every cluster, so the answer must have identical 1st and 3rd letters – only RNRV does.

Answer: (D) RNRV

Concept: 'All carrots are onions' and 'All onions are peas' give nested circles: carrots inside onions inside peas.

1. Conclusion (I) 'All peas are carrots' reverses the inclusion – peas may exist that are not carrots, so it does not follow.
2. Conclusion (II) 'Some onions are carrots': since all carrots are onions, the carrots are onions too, so some onions are definitely carrots – follows.

Trick: All A are B ⇒ Some B are A (always true); but All A are B never gives All B are A.

Answer: (B) Only conclusion (II) follows