RRB NTPC Undergraduate CBT 1 2025 PYQs

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

1. Only two boxes below G → G at 3. Only one box above B → B at 6. One box between B and A → A at 4 (8 is impossible).
2. Remaining places 1, 2, 5, 7 for C, D, E, F. F immediately above E needs consecutive places → E = 1, F = 2.
3. C and D take 5 and 7 with D below C → D = 5, C = 7.

Trick: Clues with 'only one/two' fix absolute positions – start with them, then fit the 'immediately' pair into the only consecutive gap.

Answer: (C) C

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

1. YI → BM: Y(25)+3 = 28 → B(2), I(9)+4 = M(13): shifts $+3,+4$.
2. GV → JZ: $+3,+4$; BU → EY: $+3,+4$.
3. OS → RV: O(15)+3 = R(18), S(19)+3 = V(22): shifts $+3,+3$ – different.

Trick: Write letter numbers (EJOTY: E=5, J=10, O=15, T=20, Y=25) and check both shifts; the odd one differs in the second shift.

Answer: (D) OS - RV

Concept: Facing north, 'right' is the higher position number. Number seats 1 to 8 from left.

1. Y is at an end and S is second to the left of Y → Y = 8, S = 6 (Y = 1 is impossible). S is third to the right of N → N = 3.
2. B–M are adjacent with Z two places right of M. Free seats 1, 2, 4, 5, 7: B = 4, M = 5, Z = 7 (B = 1, M = 2 leaves no adjacent pair for A–T).
3. T immediately right of A → A = 1, T = 2.

Trick: Fix the person at the extreme end first; every other clue then collapses to one option.

Answer: (B) A

Concept: Check the differences between consecutive terms for a repeating pattern.

1. Differences: $34-26=8$, $45-34=11$, $53-45=8$, $64-53=11$, $72-64=8$.
2. The pattern is $+8,+11$ alternating, so the next difference is $+11$.
3. $72+11=83$.

Trick: Two alternating differences → the series is really two interleaved series each growing by $19$: $26,45,64,83$.

Answer: (D) 83

Concept: Rearrange the letters alphabetically, then count positions from the required end.

1. NERVOUS in alphabetical order: E, N, O, R, S, U, V.
2. Fourth from the left = R; third from the right = S.
3. R(18) and S(19) are consecutive, so there are $0$ letters between them.

Trick: Letters between two positions $=|b-a|-1$; here $19-18-1=0$.

Answer: (C) 0

Concept: Swap the signs as instructed, then apply BODMAS.

1. After interchanging: $215+147\times 110-1604÷4$.
2. $147\times 110=16170$ and $1604÷4=401$.
3. $215+16170-401=15984$.

Trick: Rewrite the whole expression with new signs first – never swap and calculate at the same time.

Answer: (B) 15984

Concept: Number positions 1 (bottom) to 7 (top); fix the 'only' clues first.

1. Two boxes below A → A = 3. One box above E → E = 6. One box between E and B → B = 4.
2. Free places 1, 2, 5, 7 for C, D, F, G. C immediately above G needs consecutive places → G = 1, C = 2.
3. D below F → D = 5, F = 7. Between A (3) and D (5) there is only B → 1 box.

Trick: Boxes between positions $a$ and $b$ $=|a-b|-1=5-3-1=1$.

Answer: (B) 1

Concept: When the unknown is early in the series, work out the differences from the known terms at the end.

1. $110-80=30$, $80-56=24$, $56-38=18$ → differences fall by $6$ each step: next (earlier) differences are $12$ and $6$.
2. $?=38-12=26$; check: $26-20=6$ ✓.
3. Series: $20,26,38,56,80,110$ with differences $6,12,18,24,30$.

Trick: Differences that are multiples of one number ($6,12,18,\dots$) – spot the table of 6.

Answer: (D) 26

Concept: Find one operation on the whole first number that gives the second in both given pairs.

1. ${18}^2=324$, $324+1=325$ ✓; ${19}^2=361$, $361+1=362$ ✓. Rule: second $=$ first${}^2+1$.
2. Option C: ${22}^2+1=484+1=485$ ✓.
3. Others: ${14}^2+1=197\ne 196$; ${21}^2+1=442\ne 440$; ${20}^2+1=401\ne 399$.

Trick: When the second number is close to a perfect square, test $n^2\pm k$ first.

Answer: (C) 22, 485

Concept: Sort the digits in ascending order and compare position by position with the original.

1. Original: 3 1 4 7 8 5 6; ascending: 1 3 4 5 6 7 8.
2. Position-wise: 3→1, 1→3, 4→4, 7→5, 8→6, 5→7, 6→8.
3. Only the digit 4 (third place) stays where it was.

Trick: Write the two rows one below the other and tick the matching columns.

Answer: (A) One

Concept: Check the gaps between the letters inside each cluster.

1. GDH: G(7) → D(4) is $-3$, G → H is $+1$. MJN: $-3,+1$. JGK: $-3,+1$.
2. DAF: D(4) → A(1) is $-3$, but D → F(6) is $+2$ – breaks the pattern.

Trick: Compare the first and third letters first – in three clusters they are neighbours (G-H, M-N, J-K); D-F are not.

Answer: (B) DAF

Concept: Two particular ('Some') premises never give a definite conclusion; draw the Venn diagram to confirm.

1. Cups and jugs overlap, jugs and vases overlap – but cups and vases may or may not touch, so 'Some cups are vases' is not certain. (I) does not follow.
2. 'All vases are jugs' is a universal claim; the premise only says some jugs are vases. (II) does not follow.

Trick: Some + Some = No conclusion; a 'Some' premise can never prove an 'All' conclusion.

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

Concept: Treat letters and the number separately and find the rule from both given pairs.

1. IP 18 → KL 4: I$+2$ = K, P$-4$ = L, $18-14=4$.
2. KL 16 → MH 2: K$+2$ = M, L$-4$ = H, $16-14=2$ – same rule.
3. AT 9: A$+2$ = C, T$-4$ = P, $9-14=-5$ → CP $-5$.

Trick: Verify the rule on the second pair before applying it – here the number rule is 'subtract 14', not 'divide'.

Answer: (D) CP -5

Concept: 'Some A are B' + 'All B are C' gives 'Some A are C'; the converse of 'All' is only 'Some'.

1. The papers that are rains lie inside rains, and all rains lie inside trees → those papers are trees. (I) follows.
2. 'All rains are trees' does not mean all trees are rains – trees may exist outside rains. (II) does not follow.

Trick: Some + All (middle term shared) = Some; reversing 'All A are B' gives only 'Some B are A'.

Answer: (D) Only conclusion (I) follows

Concept: Track the position numbers of letters at each step; the same position-shuffle must apply in the option.

1. ROSE (1234) → RSEO = positions 1, 3, 4, 2. Then RSEO (1234) → OSRE = positions 4, 2, 1, 3. JUGS → JGSU (1,3,4,2) → UGJS (4,2,1,3) ✓.
2. PING → positions 1,3,4,2 = P N G I ✓; PNGI → positions 4,2,1,3 = I N P G ✓.
3. LADY → LDYA ✓ but 4,2,1,3 gives A D L Y, not ALYD; COLD and BOND fail at the first step.

Trick: Step 1 keeps the first letter fixed – options C and D are eliminated instantly.

Answer: (A) PING - PNGI - INPG

Concept: Compare the gaps from the first letter to the other two letters in each cluster.

1. XTW: X(24) → T(20) is $-4$, X → W(23) is $-1$. FBE: $-4,-1$. JFI: $-4,-1$.
2. PMO: P(16) → M(13) is $-3$, P → O(15) is $-1$ – the first gap differs.

Trick: First letter minus 4 = second letter in three clusters (X-T, F-B, J-F); P-M is only 3 apart.

Answer: (A) PMO

Concept: Sort the digits in ascending order and compare each position with the original number.

1. Original: 3 5 1 6 4 7 2; ascending: 1 2 3 4 5 6 7.
2. Position-wise: 3→1, 5→2, 1→3, 6→4, 4→5, 7→6, 2→7 – no column matches.
3. So zero digits keep their position.

Trick: With digits 1–7 the sorted number is always 1234567 – just check whether digit $d$ sits at place $d$ in the original.

Answer: (D) Zero

Concept: Letters common to both words get the codes common to both; the leftover letter gets the leftover digit.

1. PRIME = 37862 and REMAP = 87423 share the letters P, R, M, E and the digits 3, 7, 8, 2.
2. In PRIME the extra letter is I → extra digit 6. In REMAP the extra letter is A → extra digit 4.

Trick: Cancel common letters against common digits – whatever remains is the answer.

Answer: (B) 4

Concept: Find the pattern within each pair and between consecutive pairs.

1. JL – MO – PR: inside each pair the gap is $+2$ (J→L, M→O, P→R); first letters J, M, P go $+3$. EG – HJ – KM: same ($+2$ inside, $+3$ across).
2. MO – PR – SU: M→O $+2$, P→R $+2$, S→U $+2$; M, P, S go $+3$ ✓.
3. CH, DP, NF have internal gaps of 5, 12, −8 – they fail at once.

Trick: Check the internal gap of the first pair only – three options are out in one look.

Answer: (D) MO – PR – SU

Concept: Take A as origin; a right turn from east → south → west → north → east. Track east-west and north-south totals separately.

1. East 11 → right (south) 8 → right (west) 14 → right (north) 12 → right (east) 3.
2. East-west: $11-14+3=0$ km, so P is directly above/below A. North-south: $-8+12=+4$ km north.
3. P is 4 km north of A, so she must drive 4 km south to reach A.

Trick: Net E–W $=0$ means the answer is purely north or south – just net the N–S legs.

Answer: (A) 4 km to the south