RRB NTPC Graduate CBT 1 2025 PYQs

Concept: Drop all letters, write the remaining symbols in order, then count from the right.

1. Symbols in order: £, $, Ω, %, *, &, #, @, *, & (10 symbols).
2. From the right: 1st &, 2nd *, 3rd @, 4th #, 5th &, 6th *, 7th %.

Trick: "$n$-th from the right" in a list of $N$ items is the $(N-n+1)$-th from the left: $10-7+1=4$th → %.

Answer: (B) %

Concept: Find one operation on the whole number that turns the first number into the second, and confirm it on the second pair.

1. $8\to 41$: $8\times 5+1=41$.
2. $81\to 406$: $81\times 5+1=406$ ✓ (same rule).
3. $13\to 13\times 5+1=66$.

Trick: When the second number is roughly 5 times the first, try $5n\pm k$ first; the pair $(81,406)$ settles $k=+1$.

Answer: (B) 66

Concept: Replace each letter by its symbol and then apply BODMAS (÷ and × before + and −).

1. 13 B 2 C 6 D 8 A 4 becomes $13\times 2+6-8÷4$.
2. Division first: $8÷4=2$; multiplication: $13\times 2=26$.
3. $26+6-2=30$.

Trick: Write the decoded expression fully before calculating – most mistakes come from solving left-to-right instead of BODMAS.

Answer: (C) 30

Concept: Compare the two letters of each pair position-wise using alphabet positions.

1. ZC−AD: Z→A (+1, cyclic), C→D (+1).
2. DG−EH: D→E (+1), G→H (+1). RU−SV: R→S (+1), U→V (+1).
3. OR−OU: O→O (0), R→U (+3) – breaks the +1,+1 pattern.

Trick: Z→A counts as +1 (alphabet wraps around) – don't pick ZC−AD as odd just because it crosses Z.

Answer: (B) OR−OU

Concept: Identify the required numbers first, then pick the asked digits.

1. Numbers: 442, 405, 297, 384, 316. Highest = 442, lowest = 297.
2. Third digit of 442 = 2; second digit of 297 = 9.
3. Sum $=2+9=11$.

Trick: Underline the highest and lowest numbers as soon as you read the list – it saves re-scanning for every sub-question of this set.

Answer: (B) 11

Concept: Find one operation on the whole number that works for both given pairs.

1. $22\times 3-6=66-6=60$ ✓
2. $27\times 3-6=81-6=75$ ✓
3. $33\times 3-6=99-6=93$.

Trick: The second number is a little less than $3\times$ the first (66→60, 81→75, both −6) – test "×3 then −6" immediately.

Answer: (D) 93

Concept: Draw the Venn diagram with the minimum overlap allowed by the statements and test each conclusion.

1. All hands are legs → the "hands" circle lies inside "legs". Some hands are faces → the hands–faces overlap is non-empty, and it also lies inside legs. So some legs are faces – conclusion (I) follows.
2. Some faces are arms – the arms can overlap faces outside hands, so "some hands are arms" is not necessary. Conclusion (II) does not follow.

Trick: Two "Some" statements in a chain (Some hands are faces, Some faces are arms) never give a definite conclusion between the end terms.

Answer: (B) Only conclusion (I) follows

Concept: Treat letters and numbers as separate series.

1. Letters: DHG → CGF → BFE → AED – every letter moves back by 1. Next: A→Z, E→D, D→C = ZDC.
2. Numbers: 42, 38, 34, 30 – decreasing by 4. Next $=26$.
3. Combined: ZDC 26.

Trick: A −1 = Z (cyclic). Only option (A) starts with ZD and has 26.

Answer: (A) ZDC 26

Concept: Number the 7 seats 1–7 from left to right (all face north, so left/right are the usual).

1. Only two people left of C → C at seat 3. Three between C and F → F at seat 7.
2. B second to the right of C → B at seat 5. Two between B and D → D at seat 2 (seat 8 does not exist).
3. Seats 1, 4, 6 remain for G, E, A with G left of E left of A → G 1, E 4, A 6.
4. Left of D (seat 2): only G → 1 person.

Trick: Start with the clue that fixes a seat ("only two to the left of C") – it anchors the whole row.

Answer: (A) 1

Concept: Handle letters and numbers separately.

1. Letters: N(14)→K(11)→H(8)→E(5): each −3 → next B(2). V→S→P→M: −3 → J. A→X→U→R: −3 → O. Cluster = BJO.
2. Numbers: 7, 14, 21, 28 → +7 → 35.
3. Answer: BJO 35.

Trick: Check only the first letter and the number first: E −3 = B and 28 + 7 = 35 → option (C) is the only match.

Answer: (C) BJO 35

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

1. Only K below R → K at 1, R at 2.
2. Only A above E → E at 6, A at 7.
3. Only S between E and T → S at 5, T at 4.
4. Remaining M at 3. Between M (3) and K (1): only R → one box.

Trick: "Only X below/above Y" fixes Y at the second position from that end – use such clues first.

Answer: (C) One

Concept: Put A at the origin, East = +x, North = +y, and track the position after each leg (every left turn rotates the direction anticlockwise).

1. East 38 → (38, 0); left = North 35 → (38, 35); left = West 39 → (−1, 35).
2. Left = South 37 → (−1, −2); left = East 1 → P = (0, −2).
3. P is directly 2 km south of A, so he must drive 2 km to the North.

Trick: Net East–West: $38-39+1=0$; net North–South: $35-37=-2$ → 2 km south of A.

Answer: (D) 2 km to the North

Concept: Look at the differences between consecutive terms.

1. $111\to 110$: −1; $110\to 106$: −4; $106\to 97$: −9; $97\to 81$: −16.
2. Differences are $1^2,2^2,3^2,4^2$ → next difference $5^2=25$.
3. $81-25=56$.

Trick: Differences 1, 4, 9, 16 → perfect squares; subtract 25 next.

Answer: (A) 56

Concept: The same letter shift must relate # to POM and LKI to %.

1. Test option (D): TSQ → POM: T(20)→P(16), S(19)→O(15), Q(17)→M(13) – each −4.
2. LKI → HGE: L(12)→H(8), K(11)→G(7), I(9)→E(5) – each −4 ✓.
3. Other options: SRP→POM is −3,−3,−3 but GFE from LKI is −5,−5,−4; TJQ, RTP do not give a uniform shift.

Trick: Use the fully known pair LKI : % first – only HGE (option D) is a uniform −4 from LKI; then confirm # = TSQ.

Answer: (D) # = TSQ , % = HGE

Concept: Find one rule on the whole number that fits both pairs.

1. $23\times 6+1=138+1=139$ ✓
2. $28\times 6+1=168+1=169$ ✓
3. $25\times 6+1=150+1=151$.

Trick: $139÷23\approx 6.04$ and $169÷28\approx 6.04$ – both just above 6, so try "×6 + 1".

Answer: (A) 151

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

1. T + P: T is the sister of P. P × M: P is the son of M → T is the daughter of M.
2. M − V: M is the husband of V → V is the mother of T.
3. V ÷ S: V is the daughter of S.
4. So T is the daughter of V, and V is the daughter of S → T is S's daughter's daughter (granddaughter).

Trick: T's gender is fixed by "sister" (female), so options with "son" or "brother" as the link are eliminated at once.

Answer: (C) Daughter’s daughter

Concept: Fix the definite days first, then fit the ordered group into the gaps.

1. No one before N → N on Monday. Three between N and O → O on Friday.
2. One person between O and M → M on Wednesday or Sunday; Wednesday is L's → M on Sunday.
3. Tuesday, Thursday, Saturday remain for C, B, A in the order C → B → A.
4. B gets Thursday.

Trick: "No one has exam before N" = N is on the first day; such absolute clues anchor the whole schedule.

Answer: (C) Thursday

Concept: Check the alphabetical gap from the first cluster to the second.

1. AB → DE: A(+3)D, B(+3)E.
2. TU → WX: T(+3)W, U(+3)X.
3. KL → NO: K(+3)N, L(+3)O.
4. XY → YZ: X(+1)Y, Y(+1)Z – gap is +1, not +3.

Trick: Write the place numbers (A=1 … Z=26) and subtract – the pair with a different gap is the odd one.

Answer: (B) XY - YZ

Concept: Remove every digit and then count from the left in the remaining series.

1. Series: 5 # % 8 * $ % @ 1 3 @ $ 4 6 9 7 # $ 2 % 5 4.
2. After dropping numbers: # % * $ % @ @ $ # $ %.
3. Fourth from the left = $.

Trick: Strike the digits with a pencil first; never try to count and skip at the same time.

Answer: (A) $

Concept: Replace each symbol by its given meaning and then apply BODMAS.

1. 79 − 8 ÷ 55 + 423 × 3 becomes $79\times 8+55-423÷3$.
2. $423÷3=141$, $79\times 8=632$.
3. $632+55-141=546$.

Trick: Rewrite the whole expression with new symbols first, then solve – do not substitute and calculate in the same step.

Answer: (A) 546