RRB NTPC Undergraduate CBT 1 2025 PYQs

Concept: Convert to positions and compare the gaps.

1. EGF: 5, 7, 6 → $+2,-1$; BDC: 2, 4, 3 → $+2,-1$; HJI: 8, 10, 9 → $+2,-1$.
2. VYW: 22, 25, 23 → $+3,-2$ – different.
3. So VYW is the odd one.

Trick: Pattern 'x, x+2, x+1' = three consecutive letters with the last two swapped (E F G → E G F); V W X would give VXW, not VYW.

Answer: (C) VYW

Concept: Apply the rule digit by digit: odd digit $+1$, even digit $-1$.

1. 4 7 3 2 1 5 6 8 → 3 8 4 1 2 6 5 7.
2. New number = 38412657; its first digit from the left is 3.
3. (Only the first digit was really needed: 4 is even → $4-1=3$.)

Trick: Read the question before working: when only the first digit is asked, transform just that digit.

Answer: (C) 3

Concept: Treat the two letters and the number as three independent series.

1. 1st letters O(15) → M(13) → K(11) → ? : $-2$ → I(9).
2. 2nd letters N(14) → P(16) → R(18) → ? : $+2$ → T(20).
3. Numbers 35 → 44 → 53 → ? : $+9$ → 62. Answer IT62.

Trick: Opposite-moving letter pairs (one $-2$, other $+2$) are a common NTPC pattern.

Answer: (D) IT62

Concept: The word common to both sentences has the code common to both.

1. 'unit flash zebra' → mi de cz; 'zebra dance king' → de uv id.
2. Only 'zebra' appears in both sentences; only 'de' appears in both codes.
3. So zebra = de.

Trick: Common word ↔ common code; eliminate codes that appear once.

Answer: (D) de

Concept: Take successive differences; here the differences grow by an alternating amount.

1. Differences: $3,9,19,25,35$.
2. Second differences: $+6,+10,+6,+10$ → next $+6$ → difference $=35+6=41$.
3. Missing term $=101+41=142$.

Trick: If first differences look random, take the second differences – an alternating $6,10,6,10$ shows up.

Answer: (C) 142

Concept: Find the uniform shift from the first cluster to the second.

1. HLJN → JNLP: H→J, L→N, J→L, N→P – every letter $+2$.
2. EIGK → GKIM: also $+2$ each.
3. FJHL: F→H, J→L, H→J, L→N → HLJN.

Trick: Check the first letter in each option: only options starting with H survive, then L (J+2) fixes option C.

Answer: (C) HLJN

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

1. I is at 1; only one box between I and P → P at 3.
2. Three boxes between K and P → K at 7.
3. N immediately above Q with two boxes between N and M: N=5, Q=6, M=2 (N=4, Q=5 would need M at 7 or 1 – occupied).
4. J fills the last gap, position 4. Stack: I, M, P, J, N, Q, K → 4th from top is J.

Trick: Place the pair clue (N over Q) last; test each free slot pair and reject those that force M onto an occupied place.

Answer: (B) J

Concept: Compare consecutive gaps in each cluster.

1. LMN: 12, 13, 14; JKL: 10, 11, 12; FGH: 6, 7, 8 – three consecutive letters ($+1,+1$).
2. BDD: 2, 4, 4 → $+2,0$ – not consecutive.
3. So BDD is the odd one.

Trick: Repeated letter inside a cluster (DD) is an instant giveaway when the others are consecutive triplets.

Answer: (C) BDD

Concept: Draw the Venn diagram: 'All chairs are bats' puts Chairs inside Bats; 'No bat is a hat' keeps Bats and Hats apart.

1. Since every chair lies inside Bats and Bats never overlaps Hats, no chair can be a hat → (I) follows.
2. 'Some hats are bats' contradicts 'No bat is a hat' → (II) does not follow.
3. Only conclusion (I) follows.

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

Answer: (B) Only conclusion (I) follows

Concept: Total = rank from top + rank from bottom $-1$ (the person is counted in both).

1. Jay: 9th from top, 12th from bottom.
2. Total students $=9+12-1=20$.

Trick: Formula: $\text{Total}=T+B-1$; if two different people are given, use the overlap/gap rules instead.

Answer: (B) 20

Concept: The multiplier increases by 0.5 each step and 2 is subtracted every time.

1. $28\times 0.5-2=12$; $12\times 1-2=10$; $10\times 1.5-2=13$; $13\times 2-2=24$.
2. Next: $24\times 2.5-2=60-2=58$.

Trick: A series that first falls then rises sharply usually has a growing multiplier ($\times 0.5,\times 1,\times 1.5\dots$).

Answer: (D) 58

Concept: Draw Clouds inside Water; Vapor overlaps Water somewhere – it may or may not touch Clouds.

1. (I) 'Some clouds are vapor' is possible but not certain (Vapor may overlap only the non-cloud part of Water) → does not follow.
2. (II) 'Some vapor is water' is just the converse of 'Some water is vapor' → always true → follows.
3. Only conclusion (II) follows.

Trick: 'Some A are B' ⇔ 'Some B are A' (I-type converts); 'All' + 'Some' never gives a definite 'Some' about the first term.

Answer: (A) Only conclusion (II) follows

Concept: Track the position with coordinates: east = $+x$, north = $+y$; every right turn rotates N→E→S→W→N.

1. A(0, 0) → north 12 → (0, 12) → right (east) 15 → (15, 12).
2. Right (south) 17 → (15, −5) → right (west) 18 → (−3, −5).
3. Right (north) 5 → P(−3, 0).
4. P is 3 km west of A, on the same line → he must drive 3 km east.

Trick: Net E–W: $15-18=-3$ (3 km west); net N–S: $12-17+5=0$. The answer is just the net displacement reversed.

Answer: (D) 3 km to the east

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

1. $805÷2+26-1203\times 3$ becomes $805\times 2-26+1203÷3$.
2. $805\times 2=1610$; $1203÷3=401$.
3. $1610-26+401=1985$.

Trick: Do ÷ and × first, then + and − from left to right; $1203÷3$ is exact (1+2+0+3=6), a hint the swap is right.

Answer: (B) 1985

Concept: Facing the centre, a person's right is anticlockwise and left is clockwise. Number seats 1–7 clockwise.

1. Put Q at 1. One person between Q and L from Q's right (anticlockwise) → L at 6.
2. One between L and M from M's left (clockwise from M) → M at 4. One between Q and G from G's right → G at 3.
3. Two between H and M from M's left → H at 7. One between F and G from F's right → F at 5. O takes seat 2.
4. O's immediate right (seat 1) = Q; immediate left (seat 3) = G.

Trick: Draw the circle with arrows pointing inward; 'from the right of X' means move anticlockwise from X.

Answer: (A) Q and G

Concept: Replace each symbol by its meaning, then solve with BODMAS.

1. $18-3÷9+15\times 3$ → $18\times 3+9-15÷3$.
2. $18\times 3=54$; $15÷3=5$.
3. $54+9-5=58$.

Trick: Write the converted expression fully before calculating – most errors come from converting and computing in the same step.

Answer: (C) 58

Concept: Decode pairwise and build the tree around the couple P–J.

1. K & V → K is husband of V. V * P → V is daughter of P.
2. P # W → P is father of W; W % J → W is son of J → J is W's mother, i.e. P's wife.
3. So J is the mother of V, and V is K's wife.
4. J is K's wife's mother (mother-in-law). (G @ K – G is K's mother – is extra information.)

Trick: Find the chain that links the two asked persons (K → V → J) and ignore the rest.

Answer: (A) Wife's mother

Concept: Compare the first letters across the triad, and the second letters across the triad.

1. RN-TP-VY: first letters R→T→V ($+2,+2$); second letters N→P ($+2$), P→Y ($+9$).
2. PL-RN-TW: P→R→T ($+2,+2$); L→N ($+2$), N→W ($+9$). Same pattern.
3. NJ-PL-RU: N→P→R ($+2,+2$); J→L ($+2$), L→U ($12+9=21$) ✓.
4. Option D ends in V (22), option B/C break the $+2$ at the second letter.

Trick: Options A and D differ only in the last letter – compute just $L+9$ to decide.

Answer: (A) NJ-PL-RU

Concept: Each position of the cluster forms its own series.

1. 1st letters W, B, G, L → $+5$ each (W→B wraps past Z) → Q.
2. 2nd letters A, F, K, P → $+5$ → U.
3. 3rd letters X, C, H, M → $+5$ → R. Term = QUR.

Trick: All three series share the same jump ($+5$); after finding the 1st letter Q, the 2nd letter U eliminates every other option.

Answer: (C) QUR

Concept: Number positions 1 (top) to 7 (bottom); start from the box whose position is fixed by a count.

1. Two boxes below A → A at 5; R immediately above A → R at 4.
2. Three boxes between T and A → T at 1. One box between S and R → S at 2 or 6.
3. Four boxes between S and F: S=2 → F=7 ✓ (S=6 would need F at 1 = T). So S=2, F=7.
4. O above C in the free slots 3 and 6 → O=3, C=6. Stack: T S O R A C F → boxes below O = 4.

Trick: 'Only n boxes below X' fixes X at position $7-n$ immediately; build everything from there.

Answer: (C) Four