RRB NTPC Graduate CBT 1 2024 PYQs

Concept: Interchange the signs as given, then apply BODMAS.

1. $171+95÷5\times 4-33$
2. $=171+19\times 4-33$
3. $=171+76-33=214$

Trick: Write the new expression fully before calculating — most mistakes come from half-replaced signs.

Answer: (D) 214

Concept: Change each digit by the rule and read the required position.

1. 6127385: odd digits 1, 7, 3, 5 → 3, 9, 5, 7; even digits 6, 2, 8 → 4, 0, 6.
2. New number: 4 3 0 9 5 6 7.
3. Third from the right: 7, 6, 5.

Trick: Only the 3rd digit from the right is needed: original 3 is odd ⇒ $3+2=5$.

Answer: (B) 5

Concept: Check the shift of each letter position across the terms.

1. S → Q → O → M → K (−2 each)
2. U → S → Q → O → M (−2 each)
3. X → V → T → R → P (−2 each)

Trick: Every letter moves 2 steps back: MOR − 2 = KMP.

Answer: (C) KMP

Concept: Take A as origin and track the coordinates after each turn (all left turns).

1. 8 km N → (0, 8); left = W 9 km → (−9, 8); left = S 12 km → (−9, −4).
2. Left = E 13 km → (4, −4); left = N 4 km → P (4, 0).
3. P is 4 km east of A, so he must drive 4 km towards the west.

Trick: Net N–S: $8-12+4=0$; net E–W: $13-9=4$ km east ⇒ go back west 4 km.

Answer: (D) 4 km to the west

Concept: A letter common to both words gets the digit common to both codes.

1. GIFT → 4268 and TURN → 3527.
2. Common letter: T. Common digit: 2.
3. So T = 2.

Trick: Common letter ↔ common digit.

Answer: (A) 2

Concept: Look for the pattern Number – Symbol – Number.

1. Check each symbol's neighbours: 6 # % ✗, 4 £ * ✗, 1 £ © ✗, 3 % % ✗, % % 2 ✗, 2 $ £ ✗.
2. Only 7 # 4 has a number on both sides ✓.
3. Count = 1.

Trick: Scan only the symbols that have a number just before them, then check the next element.

Answer: (B) 1

Concept: Facing north, the right end is the east end. Fix the people with fixed positions first.

1. No one to the right of A ⇒ A at position 7 (right end).
2. Three between A and D ⇒ D at position 3.
3. Two between D and B ⇒ B at position 6.
4. C third to the left of P and N immediately right of P with free seats 1, 2, 4, 5 ⇒ C = 1, P = 4, N = 5, so O = 2.
5. Between O (2) and B (6): D, P, N = 3 people.

Trick: People between positions $i$ and $j$ $=|i-j|-1=6-2-1=3$.

Answer: (C) Three

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

1. After swapping: $131\times 6-1818÷18+23$
2. $=786-101+23$
3. $=708$

Trick: Rewrite the whole expression first, then solve — never swap signs mentally while calculating.

Answer: (C) 708

Concept: Take Y as origin (0, 0); East = +x, North = +y.

1. 19 E → (19, 0); right (south) 28 → (19, −28); left (east) 17 → (36, −28).
2. Left (north) 59 → (36, 31); left (west) 36 → (0, 31); right (north) 9 → Z (0, 40).
3. Z is 40 km directly north of Y, so he must drive 40 km towards the south.

Trick: Net east–west: $19+17-36=0$; net north–south: $-28+59+9=40$ north.

Answer: (C) 40 km towards the south

Concept: Check the gap between letters within each cluster using positions (A = 1 … Z = 26).

1. KP: $11\to 16$ (+5); DW: $4\to 23$ (+19).
2. UZ: +5; NG: $14\to 7$ (+19, wrapping).
3. CH: +5; VO: $22\to 15$ (+19, wrapping).
4. RW: +5; but LD: $12\to 4$ is +18 — breaks the pattern.

Trick: Moving +19 is the same as moving −7: D→W, N→G, V→O all go back 7, but L→D goes back 8.

Answer: (B) RW – LD

Concept: Check differences between consecutive terms.

1. $832-830=2$, $830-826=4$, $826-820=6$, $820-812=8$
2. Next difference $=10$
3. $812-10=802$

Trick: Differences rise by 2: −2, −4, −6, −8, −10.

Answer: (D) 802

Concept: Find the rule inside each pair and between the pairs (A = 1 … Z = 26).

1. EB-GD-IF: in each pair second letter = first − 3 (E→B, G→D, I→F); first letters go +2 (E, G, I).
2. BY-DA-FC: B→Y, D→A, F→C (−3 with wrap); first letters B, D, F (+2).
3. UR-WT-YV: U→R, W→T, Y→V (−3); first letters U, W, Y (+2). ✓
4. GD-JF-KH fails (G→J is +3); PM-RO-SQ fails (R→S is +1); KH-MI-OL fails (M→I is −4).

Trick: Check the easy “−3 inside each pair” rule first; it removes (A)’s JF and (D)’s MI immediately.

Answer: (B) UR-WT-YV

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

1. M + N: M is the son of N.
2. N – O: N is the brother of O.
3. O ÷ P: O is the father of P.
4. P × Q: P is the wife of Q.
5. From Q: wife (P) → her father (O) → his brother (N) → his son (M).

Trick: Read the chain backwards starting from Q to build the relation.

Answer: (A) Wife's Father's Brother's Son

Concept: Find the rule from the left pair and apply it to the right pair: # → DFJ means LNR → %.

1. ACG → DFJ: A+3 = D, C+3 = F, G+3 = J, so # = ACG.
2. Apply +3 to LNR: L+3 = O, N+3 = Q, R+3 = U, so % = OQU.

Trick: Both pairs must follow the same direction of change (here +3 to each letter).

Answer: (C) # = ACG , % = OQU

Concept: Compare each letter of the first cluster with the corresponding letter of the second.

1. MGP → NIS: M+1, G+2, P+3.
2. SVB → TXE: S+1, V+2, B+3.
3. CTR → DVU: C+1, T+2, R+3. ✓

Trick: Pattern +1, +2, +3 — check only the first letter (+1) to eliminate (B) and (C) quickly.

Answer: (A) CTR : DVU

Concept: Look for the pattern Number – Symbol – Symbol, and count the middle symbol.

1. 3 £ $ → £ ✓
2. 8 * % → * ✓
3. 6 € $ → € ✓
4. 8 & $ → & ✓
5. Other symbols fail (e.g. 3 $ 9 — followed by a number). Total = 4.

Trick: Scan only symbols that come right after a digit, then check the next item.

Answer: (B) 4

Concept: Track each position of the clusters separately.

1. 1st letters: V, R, N, J, F (−4 each).
2. 2nd letters: D, G, J, M, P (+3 each).
3. 3rd letters: U, S, Q, O, M (−2 each).
4. Missing term = JMO.

Trick: Use the last term FPM to check: J−4 = F, M+3 = P, O−2 = M ✓.

Answer: (D) JMO

Concept: The series is made of two interleaved series.

1. Odd places: 86, 73, 60, 47, ? (−13 each).
2. Even places: 97, 82, 67, 52 (−15 each).
3. Next odd-place term $=47-13=34$.

Trick: When terms rise and fall alternately, split into odd/even positions.

Answer: (B) 34

Concept: Check the gaps inside each cluster (A = 1 … Z = 26, with wrap-around).

1. XV: X−2 = V; FZ: F−6 = Z.
2. BZ: B−2 = Z; JD: J−6 = D.
3. HF: H−2 = F; PJ: P−6 = J.
4. YX: Y−1 = X (not −2); GA: G−6 = A. So YX – GA is odd.

Trick: Also the first letters jump +8 between clusters (X→F, B→J, H→P, Y→G); the odd gap is inside YX.

Answer: (D) YX – GA

Concept: Code each element from the table, then check which condition applies.

1. Direct codes: 9 → R, 7 → F, 8 → A, 6 → L.
2. Condition 1: first element is not a symbol — not applicable.
3. Condition 2: first = 9 (odd), last = 6 (even) — applies: first and last become ©.
4. Condition 3: 7 and 8 are not perfect squares — not applicable.
5. Code = © F A ©.

Trick: Check the conditions in order before writing direct codes — options (C) and (D) are the traps.

Answer: (B) ©FA©