RRB NTPC Undergraduate CBT 1 2025 PYQs

Concept: Compare the shift of the first letters and of the second letters (use place values, wrapping past Z back to A).

1. SU–KA: S(19)→K(11) $-8$; U(21)→A(1) $-20$.
2. UW–MC: U(21)→M(13) $-8$; W(23)→C(3) $-20$. MO–EU: M(13)→E(5) $-8$; O(15)→U(21) is $-20$ after wrapping ($15-20=-5\Rightarrow -5+26=21$).
3. ZB–PH: Z(26)→P(16) is $-10$, not $-8$ – the odd pair (B→H is $-20$ with wrap, consistent, but the first letter breaks).

Trick: Check first letters alone first: 19→11, 21→13, 13→5 are all $-8$; 26→16 is $-10$ – answer found without touching the second letters.

Answer: (A) प

Concept: When two words share letters, the common letters share the common digits; the leftover letter gets the leftover digit.

1. DOLT → 4 9 6 3 and LOTS → 6 4 3 5.
2. Common letters O, L, T ↔ common digits 4, 6, 3.
3. The extra letter in LOTS is S and the extra digit is 5, so S = 5 (and D = 9).

Trick: Don’t try to match positions – just cancel the common letters and common digits; what is left pairs up.

Answer: (B) स

Concept: Find the letter-wise shift in the given pairs and apply it to the options.

1. QVI : KZJ → Q(17)→K(11) $-6$, V(22)→Z(26) $+4$, I(9)→J(10) $+1$.
2. USH : OWI → U(21)→O(15) $-6$, S(19)→W(23) $+4$, H(8)→I(9) $+1$ ✓ pattern $(-6,+4,+1)$.
3. LQG : FUH → L(12)→F(6) $-6$, Q(17)→U(21) $+4$, G(7)→H(8) $+1$ ✓.

Trick: Test only the third letter first ($+1$): LQG→FUH (G→H ✓), CNK→WRJ (K→J ✗), BRH→VVF ✗, AHL→UMK ✗ – one check settles it.

Answer: (A) प

Concept: Compare corresponding letters of consecutive clusters.

1. First letters F, G, H, I → next J; second letters H, I, J, K → next L; third letters D, E, F, G → next H.
2. Every letter moves $+1$ each step, so the next cluster is JLH.

Trick: When all three positions step uniformly, just add 1 to each letter of the last cluster: IKG → JLH.

Answer: (B) ज

Concept: Facing the centre: right = anticlockwise, left = clockwise. Begin with the pair that is fixed relative to each other (J and Z).

1. Z is immediately right (anticlockwise) of J. J is 3rd to the left (clockwise) of K, so K is 3 seats anticlockwise from J (one seat beyond Z... i.e. Z, then a gap, then K). Z is 3rd to the right of I, so I is 3 seats clockwise from Z, i.e. 2 seats clockwise from J.
2. Going clockwise from J: J, _, I, _, K, _, Z. The gaps are for H, W, Y.
3. Y is not a neighbour of J, so Y is not in the seat between J and I; one person between H and W counted from the left (clockwise) of W → W between J and I, H between I and K, Y between K and Z.
4. From the right (anticlockwise) of Y: K, H, I, then W → three people between Y and W.

Trick: Fill the circle with blanks first, then use “not a neighbour” clues to kill the wrong blanks.

Answer: (C) प

Concept: Replace the letters by the given operators and apply BODMAS (division and multiplication before addition and subtraction).

1. $81\text{A}3\text{D}7\text{B}2\text{C}16\Rightarrow 81÷3-7\times 2+16$
2. $=27-14+16$
3. $=29$

Trick: Write the substituted expression fully before computing – mixing substitution and calculation causes sign slips.

Answer: (C) ग

Concept: Treat the letters and the numbers as two separate series.

1. Letters: G→E→C→A (each $-2$), A→Y→W→U ($-2$), O→M→K→I ($-2$). Next: A$-2=$Y, U$-2=$S, I$-2=$G → YSG.
2. Numbers: $81,75,69,63$ – each $-6$, so next $=57$.
3. Combined: YSG 57.

Trick: Numbers alone eliminate three options ($63-6=57$); only option D has 57.

Answer: (D) क

Concept: Substitute the operators and follow BODMAS.

1. $25\text{A}51\text{D}3\text{B}6\text{C}5\Rightarrow 25+51÷3-6\times 5$
2. $=25+17-30$
3. $=12$

Trick: Do all $÷$ and $\times$ first ($51÷3=17$, $6\times 5=30$), then run left to right.

Answer: (A) प

Concept: Decode from the end (X) backwards so that each person’s generation relative to X is clear.

1. P = X → P is the father of X. O € P → O is the son of P, so O is X’s brother.
2. C ? O → C is the daughter of O. B % C → B is the sister of C, so B is also a daughter of O.
3. Therefore B is the daughter of X’s brother.

Trick: “Son of X’s father” = X’s brother (gender known!). The child of a brother is a niece/nephew – here a daughter, so “brother’s daughter”.

Answer: (D) “

Concept: Draw Venn circles: “All A are B” puts circle A inside circle B.

1. All tigers are chairs and all chairs are monks → Tigers ⊂ Chairs ⊂ Monks.
2. Conclusion (I) “Some monks are chairs” is the valid conversion of “All chairs are monks” – follows.
3. Conclusion (II) “All tigers are monks” follows by the chain (two universal affirmatives give a universal affirmative).

Trick: All + All = All (A→B, B→C gives A→C), and every “All B are C” also yields “Some C are B”.

Answer: (D) A

Concept: Look for an odd digit whose left neighbour is a perfect cube (1 or 8) and whose right neighbour is even.

1. Series: 3 5 8 9 7 4 1 5 9 6 2 3 6 5 8 7 4 1 5 9 8 6 3 2 1 5 4 7 8 4 1 2 5 9 8 7 4 1 5 6 2 3 9 8 7 8 9 7 5 1.
2. Matches: 8 7 4, 1 5 4, 8 7 4, 1 5 6, 8 7 8.
3. Near-misses: 8 9 7 (followed by odd) ✗, 1 5 9 ✗, 1 2 5 (2 is even) ✗. Total = 5.

Trick: Circle every 1 and 8, then read the two digits after each – odd then even is a hit.

Answer: (B) ह

Concept: Find an odd digit with an odd digit immediately before it and an even digit immediately after it.

1. Series: 7 5 9 2 7 9 4 1 5 1 7 8 3 2 8 4 1 2 3 8 2 7 9.
2. Matches: 5 9 2, 7 9 4, 1 7 8.
3. Others fail: 7 5 9 (followed by odd), 1 5 1 ✗, 7 8 → 8 is even itself, 1 2 3 → 1 not odd-preceded... So three.

Trick: Scan for the pattern “odd, odd, even” as a sliding window of three – the middle digit is the one you count.

Answer: (A) त

Concept: Find the rule from the first pair and verify with the second.

1. FR 6 → KN $-6$: F$+5=$K, R$-4=$N, $6-12=-6$.
2. LU 9 → QQ $-3$: L$+5=$Q, U$-4=$Q, $9-12=-3$ ✓.
3. LS 14 → L$+5=$Q, S$-4=$O, $14-12=2$ → QO 2.

Trick: The first letter rule ($+5$: L→Q) alone leaves only option A starting with Q.

Answer: (A) क

Concept: Take A as the origin and track the position after every turn (facing south, a left turn faces east; facing east, left faces north, and so on).

1. A → 23 km south → (0, −23); left (east) 19 km → (19, −23); left (north) 36 km → (19, 13).
2. Left (west) 19 km → (0, 13); left (south) 9 km → (0, 4); right (west) 5 km → (−5, 4); left (south) 4 km → P (−5, 0).
3. P is 5 km due west of A, so he must drive 5 km to the east to reach A.

Trick: Add up the east–west moves and north–south moves separately: E $19-19-5=-5$ (5 west); N $-23+36-9-4=0$ – P is 5 km west of A.

Answer: (D) प

Concept: Common letters of the two words take the common digits; the remaining letter takes the remaining digit.

1. VIBE → 4 3 9 5 and BITE → 5 9 1 4.
2. Common letters I, B, E ↔ common digits 4, 9, 5.
3. Extra letter in BITE is T, extra digit is 1 → T = 1 (and V = 3).

Trick: Cancel what is common on both sides; the leftovers pair up instantly.

Answer: (A) द

Concept: Find a rule linking the three numbers of each given set and test the options with the same rule.

1. (14, 7, 21): $7=14÷2$ and $21=7+14$. (56, 28, 42): $28=56÷2$ and $42=28+14$ ✓.
2. Rule: 2nd $=$ 1st $÷2$, 3rd $=$ 2nd $+14$.
3. (62, 31, 45): $31=62÷2$ ✓, $45=31+14$ ✓. Others fail: $57+14=71\ne 72$, $121+14=135\ne 134$, $173+14=187\ne 188$.

Trick: All options have 2nd $=$ 1st $÷2$, so the 3rd number decides: just check “2nd $+14$”.

Answer: (D) स

Concept: Position from the left $=$ Total $-$ position from the right $+1$; moving left reduces it.

1. Yusuf from the left $=15-10+1=6$th.
2. Suman is 4th to the left of Yusuf: $6-4=2$nd from the left.

Trick: Convert to one end first, then add for “right of” and subtract for “left of”.

Answer: (B) प

Concept: Compare the shift from the first cluster to the second for both letters.

1. RF–NB: R(18)→N(14) $-4$, F(6)→B(2) $-4$. MJ–IF: M(13)→I(9) $-4$, J(10)→F(6) $-4$. PL–LH: P(16)→L(12) $-4$, L(12)→H(8) $-4$.
2. XA–TU: X(24)→T(20) $-4$, but A(1)→U(21) is $+20$ (or $-6$ with wrap-around), not $-4$.
3. So XA–TU is the odd one.

Trick: Three pairs show a clean $(-4,-4)$; the pair that needs “wrapping past A” to explain is the odd one.

Answer: (A) त

Concept: Evaluate the equation as given, then try each interchange and recompute with BODMAS until the LHS equals 67.

1. Given: $(28÷4)\times 6+14\times 3-7\times 9+(66÷2)+38=42+42-63+33+38=92\ne 67$.
2. Interchange 2 and 3: $(28÷4)\times 6+14\times 2-7\times 9+(66÷3)+38$
3. $=42+28-63+22+38=67$ ✓.

Trick: LHS is 25 too big; swapping 3 and 2 lowers $14\times 3$ by 14 and $66÷2$ by 11 – exactly 25.

Answer: (C) L

Concept: Draw Venn circles: “Some poets are writers” makes the Poets and Writers circles overlap; “All writers are dreamers” puts Writers inside Dreamers.

1. The overlap of Poets and Writers lies inside Dreamers, so some poets are dreamers – hence “Some dreamers are poets” (I) follows.
2. Poets outside the Writers circle may or may not be dreamers, so “All poets are dreamers” (II) does not follow.

Trick: Some + All = Some (particular + universal gives only a particular conclusion); “Some A are B” always converts to “Some B are A”.

Answer: (D) S