RRB NTPC Undergraduate CBT 1 2025 PYQs

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

1. All vans are trailers → Vans circle lies inside Trailers.
2. No trailer is a scooter → Scooters circle is completely separate from Trailers.
3. Since vans are inside trailers, no van can touch scooters → Conclusion I follows.
4. Scooters and trailers never meet, so ‘Some scooters are trailers’ is false → II does not follow.

Trick: All A are B + No B is C ⇒ No A is C (always).

Answer: (C) Only conclusion (I) follows

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

1. P £ D → P is the daughter of D (P is female).
2. D ← M → D is the wife of M; so P is the daughter of M too.
3. M @ L → M is the father of L; L ≥ U → L is the brother of U, so U is also M's child.
4. P, L and U are children of M and D → P (female) is U's sister.

Trick: Children of the same parents: a female is ‘sister’, a male is ‘brother’.

Answer: (B) Sister

Concept: Apply the rule digit by digit, then count repeated digits.

1. Even digits ($-2$): $8\to 6,2\to 0,4\to 2,6\to 4$.
2. Odd digits ($+3$): $5\to 8,3\to 6,1\to 4$.
3. 8254316 → 6 0 8 2 6 4 4.
4. 6 appears twice and 4 appears twice → two digits repeat.

Trick: Write the new number under the old one in a table – mistakes vanish.

Answer: (A) Two

Concept: Check the differences between consecutive terms.

1. Known differences: $98-89=9,110-98=12,125-110=15$ → they rise by 3.
2. So the differences are 3, 6, 9, 12, 15.
3. $80+3=83$ and $83+6=89$ ✓.

Trick: Differences in steps of 3 → multiples of 3 series.

Answer: (C) 83

Concept: Mark the vowels first, then look only at the letters sitting between two vowels.

1. Vowels in the series: E (2nd), I (4th), I (10th), A (19th).
2. Only one consonant has a vowel on both sides: E S I (S is the 3rd letter).
3. Around the other vowels the neighbours are consonants (e.g. R I Y, S A P), so no other case.

Trick: Search for the pattern ‘vowel–consonant–vowel’ – it needs two vowels just two places apart.

Answer: (D) One

Concept: Find each vowel and check its left and right neighbours.

1. Vowels: I (3rd), E (6th), U (13th), A (19th), U (20th).
2. R I D ✓, P E Q ✓, Q U Y ✓ – consonants on both sides.
3. H A U ✗ (followed by a vowel) and A U Z ✗ (preceded by a vowel).
4. Count = 3.

Trick: Two vowels side by side (AU) can never count – skip them quickly.

Answer: (B) Three

Concept: Compare the shift from each letter of the first cluster to the matching letter of the second (A = 1 … Z = 26).

1. DP – ZL: $D-4=Z$ (looping back), $P-4=L$.
2. KS – GO: $K-4=G,S-4=O$. YH – UD: $Y-4=U,H-4=D$.
3. RM – IN: $R-9=I$ but $M+1=N$ → different rule.

Trick: Learn reverse jumps: D−4 = Z (wrap round the alphabet).

Answer: (B) RM – IN

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

1. FORT → RTFO = letters 3 4 1 2; third word TROF = FORT written backwards (4 3 2 1).
2. SIGN → GNSI (3 4 1 2) → NGIS (reverse) ✓.
3. RAFT → FTRA (3 4 1 2) → TFAR (reverse) ✓.
4. DIRT would give RTDI; DOVE gives VEDO; FILE gives LEFI – but its reverse is ELIF, not ELFI.

Trick: Middle = swap the two halves; last = full reverse of the first word.

Answer: (A) RAFT - FTRA - TFAR

Concept: Find the gaps between consecutive letters using their positions.

1. KMP: $K\stackrel{+2}{\to }M\stackrel{+3}{\to }P$; QSV: $+2,+3$; FHK: $+2,+3$.
2. DFJ: $D\stackrel{+2}{\to }F\stackrel{+4}{\to }J$ → the second gap is 4, not 3.

Trick: Write the gaps as a pair (+2, +3) – the different pair is the odd one.

Answer: (D) DFJ

Concept: Number the letters 1 2 3 4 and see where each moves.

1. WORD → RWOD = letters 3 1 2 4; DROW = WORD reversed (4 3 2 1).
2. JAMS → MJAS (3 1 2 4) → SMAJ (reverse) ✓.
3. YEAR → AYER (3 1 2 4) → RAEY (reverse) ✓.
4. AGES should give EAGS; VAST gives SVAT but its reverse is TSAV; POLE gives LPOE.

Trick: Step 1: 3rd letter jumps to the front; Step 2: write the original word backwards.

Answer: (C) YEAR - AYER - RAEY

Concept: Decode the symbols one by one and draw the tree.

1. H + I → H is the wife of I.
2. I * O → I is the father of O, so H is the mother of O.
3. O × D → O is the mother of D; D – J → D is the brother of J, so O is also J's mother.
4. H is the mother of J's mother → mother's mother.

Trick: Two generations up on the mother's side = mother's mother (maternal grandmother).

Answer: (B) Mother's mother

Concept: Look at the differences between consecutive terms.

1. Differences: 9, 14, 25, 30, 41.
2. Pattern of these differences: $+5,+11,+5,+11$ → next is $+5$.
3. Next difference $=41+5=46$, so $?=125+46=171$.

Trick: When the differences are irregular, take differences once more – an alternating pattern often appears.

Answer: (D) 171

Concept: Track each letter position and the number separately.

1. First letters: B, F, J, N → $+4$ each → next R.
2. Second letters: D, H, L, P → T; third letters: F, J, N, R → V.
3. Numbers: 2, 5, 8, 11 → $+3$ → 14.
4. Missing term = RTV 14.

Trick: Each cluster has letters 2 apart (B_D_F); just add 4 to the first letter and fill the rest.

Answer: (A) RTV 14

Concept: Draw the Venn diagram with the minimum overlap allowed by the statements.

1. Some pens are markers → Pens and Markers overlap.
2. All markers are stands → Markers circle lies inside Stands.
3. The pens that are markers are therefore also stands → Conclusion I follows.
4. Stands may contain many things that are not markers → II (All stands are markers) does not follow.

Trick: Some A are B + All B are C ⇒ Some A are C. ‘All B are C’ never reverses to ‘All C are B’.

Answer: (B) Only conclusion (I) follows

Concept: Take A as the origin, with east and north positive, and follow each move.

1. 23 km north → $(0,23)$; left turn = west 8 km → $(-8,23)$.
2. Left = south 28 km → $(-8,-5)$; left = east 17 km → $(9,-5)$.
3. Left = north 5 km → P at $(9,0)$ – on the same horizontal line as A.
4. P is 9 km east of A, so he must drive 9 km to the west.

Trick: East–west net: $17-8=9$ km east; north–south net: $23+5-28=0$.

Answer: (B) 9 km to the west

Concept: Track each letter position with its alphabet number (A = 1 … Z = 26) and the number separately.

1. First letters: B(2) → W(23) → R(18) → M(13): each step $-5$ (B − 5 wraps to W) → next $13-5=8$ = H.
2. Second letters: V → Q → L → G ($-5$) → B; third letters: Y → T → O → J ($-5$) → E.
3. Numbers: 32, 16, 8, 4 → $÷2$ → 2.
4. Missing term = HBE 2.

Trick: All three letters move back 5 together, the number halves.

Answer: (D) HBE 2

Concept: Write the original number and the ascending-order number one below the other and compare position by position.

1. Original: 3 7 4 9 6 8 5.
2. Ascending: 3 4 5 6 7 8 9.
3. Same digit at the same place: 1st (3) and 6th (8).
4. So two digits keep their positions.

Trick: Compare column by column – never count by memory.

Answer: (C) Two

Concept: Compare each letter of the first cluster with the matching letter of the second (A = 1 … Z = 26).

1. VB-TN: $V-2=T$, $B+12=N$.
2. SY-QK: $S-2=Q$, $Y+12=K$ (25 + 12 = 37, $37-26=11$). KQ-IC: $K-2=I$, $Q+12=C$.
3. PV-MH: $P-3=M$ (not $-2$), $V+12=H$ → the first rule breaks.
4. So PV – MH is the odd one.

Trick: Check the simple shift (first letters) first – the odd pair often fails there.

Answer: (D) PV – MH

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

1. ‘tale jaw bow’ → vl ud np.
2. ‘law bow fresh’ → pq re ud.
3. Common word = bow; common code = ud → bow = ud.

Trick: Match the common word with the common code – order of codes does not matter.

Answer: (A) ud

Concept: Find how each letter of the first cluster changes into the second (A = 1 … Z = 26).

1. VUT – PON: $V(22)-6=P(16)$, $U-6=O$, $T-6=N$; also each cluster is three consecutive letters going backward.
2. TSR – NML: again $-6$ for every letter.
3. YXW: consecutive backward ✓; $Y(25)-6=S(19)$, $X-6=R$, $W-6=Q$ → SRQ.
4. YWU is not consecutive; RRP and RQP give a wrong shift.

Trick: First-letter check: Y − 6 = S, so only options starting with S survive.

Answer: (C) YXW−SRQ