RRB NTPC Graduate CBT 1 2024 PYQs

Concept: Change every digit by the given rule, then read the required position.

1. Number: 8 3 1 5 4 7 6.
2. Odd digits +1: 3→4, 1→2, 5→6, 7→8. Even digits −2: 8→6, 4→2, 6→4.
3. New number: 6 4 2 6 2 8 4.
4. From the right: 4 (1st), 8 (2nd), 2 (3rd).

Trick: For 'third from right' you only need the last three digits: 5476 → …2 8 4.

Answer: (D) 2

Concept: Find the rule from the complete pair LHF : FBZ, then apply it backwards to get #.

1. L(12)−6=F(6), H(8)−6=B(2), F(6)−6=Z (wrap-around). So the rule is: each letter −6.
2. # → XTR means #−6 = XTR, so # = XTR + 6 each: X+6=D, T+6=Z, R+6=X → DZX.
3. Check: D−6=X, Z−6=T, X−6=R ✓.

Trick: Backward 6 in the alphabet = forward 20; going past A, continue from Z.

Answer: (B) # = DZX, % = FBZ

Concept: Check differences, and then the differences of the differences.

1. Differences: 7−4=3, 13−7=6, 24−13=11, 42−24=18.
2. Second differences: 3, 5, 7 → next 9, 11.
3. Next differences: 18+9=27, 27+11=38.
4. ? = 42+27 = 69; check 69+38 = 107 ✓.

Trick: Second differences forming odd numbers 3, 5, 7, 9… is a very common RRB pattern.

Answer: (C) 69

Concept: Apply the rule to each digit and count the digits that repeat.

1. Number: 7 1 5 6 4 2 3.
2. Odd +2: 7→9, 1→3, 5→7, 3→5. Even −1: 6→5, 4→3, 2→1.
3. New number: 9 3 7 5 3 1 5.
4. 3 appears twice and 5 appears twice; 9, 7, 1 appear once → two digits repeat.

Trick: Count distinct digits that repeat, not total repetitions.

Answer: (A) Two

Concept: Track the facing direction after each turn and add/subtract along each axis.

1. South 21 (facing S); right → West 19; right → North 26; right → East 19.
2. Net so far: 26−21 = 5 km north of A, 19−19 = 0 east-west.
3. Right → South 2 (now 3 km N); left → East 4; right → South 3 (now 0 km N).
4. P is exactly 4 km east of A, so he must drive 4 km to the west.

Trick: Right turn from S = W, from W = N, from N = E, from E = S — the clockwise order N→E→S→W.

Answer: (D) 4 km to the west

Concept: Check the conditions in order; apply the one that fits, else use direct codes.

1. Group: # 6 @ 2 → first is a symbol (#), last is a number (2): condition (i) applies.
2. (ii) fails (first is not odd number); (iii) fails (6 and @ are not perfect squares).
3. Direct codes: # = D, 6 = L, @ = Z, 2 = E.
4. Interchange first and last codes: E L Z D.

Trick: Write the direct code first (DLZE), then just swap the ends → ELZD.

Answer: (C) ELZD

Concept: Draw the minimum Venn diagram: 'Animals' lies inside Deer, Bears and Cows together.

1. Animals are both cows and bears, so the region of cows and bears overlaps → 'Some cows are bears' is definite. (II) follows.
2. Deer and Cows only share the animals; some deer may lie outside Cows → 'All deer are cows' is not certain. (I) does not follow.

Trick: All A are B + All A are C ⇒ Some B are C (always). 'All B are C' never follows.

Answer: (D) Only conclusion (II) follows.

Concept: Check each condition on the group @ 9 # 7.

1. First element @ is a symbol, last 7 is a number → condition (i) applies.
2. (ii) fails; (iii) fails because # is not a perfect square (only 9 is).
3. Direct codes: @ = A, 9 = D, # = G, 7 = P → A D G P.
4. Interchange first and last: P D G A.

Trick: In condition (iii) BOTH the 2nd and 3rd elements must be squares; one square is not enough.

Answer: (C) PDGA

Concept: Write the original and the arranged numbers one below the other and compare place by place.

1. Original: 1 6 7 3 4 5 9.
2. Ascending: 1 3 4 5 6 7 9.
3. Same place: 1st (1) and 7th (9); all others differ.

Trick: Smallest digit already first and largest already last — check these two quickly.

Answer: (D) Two

Concept: Draw the family tree with gender signs.

1. D is A's father and B is A's mother, so B is D's wife.
2. C is D's sister and E is C's brother, so E is D's brother.
3. B is the wife of E's brother.

Trick: Sister's brother of a father = father's brother (or the father himself); here E ≠ D, so E is the uncle.

Answer: (A) Brother’s wife

Concept: Check each condition on the group 8 4 3 %.

1. First element 8 is a number, last % is a symbol → condition (I) applies.
2. (II) fails (8 is even); (III) fails (4 is a square but 3 is not).
3. Direct codes: 8 = F, 4 = B, 3 = D, % = G → F B D G.
4. Interchange first and last: G B D F.

Trick: Write the direct code, then swap only the two ends.

Answer: (A) GBDF

Concept: Compare the positions of corresponding letters.

1. Y(25)→Q(17), R(18)→J(10), U(21)→M(13), O(15)→G(7): each letter −8.
2. Check SLOI → KDGA: S−8=K, L−8=D, O−8=G, I−8=A ✓.
3. UNQK: U(21)−8=M, N(14)−8=F, Q(17)−8=I, K(11)−8=C → MFIC.

Trick: Use EJOTY (5, 10, 15, 20, 25) to get letter positions fast.

Answer: (D) MFIC

Concept: Find the gaps between consecutive letters in each cluster.

1. TXW: T+4=X, X−1=W.
2. IML: I+4=M, M−1=L.
3. UYX: U+4=Y, Y−1=X.
4. PRS: P+2=R, R+1=S — a different pattern.

Trick: Write the pattern as (+4, −1) and test every option against it.

Answer: (D) PRS

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

1. Only F is above E → F = 1, E = 2.
2. Four boxes between E and D → D = 7.
3. J and K with two between in positions 3–6 → (3, 6). G immediately above I → G, I = 4, 5.
4. K is not immediately above G, so K ≠ 3 → J = 3, K = 6.
5. Order: F, E, J, G, I, K, D → 4th from top = G.

Trick: 'Only X is above Y' fixes both at once: X top, Y second.

Answer: (C) G

Concept: Treat each part of the cluster as its own series.

1. 1st letters: E, H, K, N → +3 → Q.
2. 2nd letters: F, I, L, O → +3 → R.
3. 3rd letters: R, U, X, A → +3 (wrap) → D.
4. Numbers: 7, 12, 17, 22 → +5 → 27. Answer QRD27.

Trick: First two letters are always consecutive (EF, HI, KL, NO) → next must be QR; this removes options (A) and (D).

Answer: (B) QRD27

Concept: Follow each letter position separately.

1. 1st letters: L, N, P, R → +2 → T.
2. 2nd letters: I, K, M, O → +2 → Q.
3. 3rd letters: P, R, T, V → +2 → X.
4. Next term: TQX.

Trick: Every letter moves +2 — just add 2 to each letter of ROV: TQX.

Answer: (B) TQX

Concept: 7 people sit around the table facing the centre, so for each person left = clockwise and right = anticlockwise.

1. Fix E. Two people between E and J counted from E's right, so J is the 3rd person to E's right.
2. Three people between E and I: going right from E the 4th seat is already J, so I is 4th to E's left. Two between J and F places F immediately to E's left.
3. Remaining seats: one between F and I, and two between J and E. D is immediate left of K, so K and D take the two seats next to E's right side (K, then D next to E); H sits between F and I.
4. Order clockwise: E, F, H, I, J, K, D. Counting 3 to the left (clockwise) of H: I, J, K.

Trick: Facing centre: left = clockwise. Draw the circle once and count seats instead of guessing.

Answer: (C) K

Concept: Find the position-value gaps inside each cluster and between the two clusters.

1. OJ – HC: O(15)→J(10) is $-5$, H(8)→C(3) is $-5$, O→H is $-7$.
2. NI – GB: N(14)→I(9) $-5$, G(7)→B(2) $-5$, N→G $-7$.
3. KF – DY: K(11)→F(6) $-5$, D(4)→Y(25) $-5$ (cyclic), K→D $-7$.
4. YT – RN: Y(25)→T(20) $-5$ but R(18)→N(14) is $-4$, so it is odd.

Trick: Check the second cluster's gap first; the one that breaks the $-5$ pattern is the answer.

Answer: (C) YT – RN

Concept: Apply the letter-wise shift found from the examples (cyclic after Z).

1. DIOR → MQXZ: D+9=M, I+8=Q, O+9=X, R+8=Z, i.e. $+9,+8,+9,+8$.
2. Check NOYX → WWHF: N+9=W, O+8=W, Y+9=H, X+8=F ✔
3. OMZV: O(15)+9=X, M(13)+8=U, Z(26)+9=35−26=9=I, V(22)+8=30−26=4=D.

Trick: Alternate $+9,+8$; subtract 26 when the value crosses Z.

Answer: (C) XUID

Concept: Swap the given pair in each option and evaluate using BODMAS.

1. Original: $4\times 36+4\times 5-17=147\ne 55$.
2. Interchange 56 and 28: $4\times (28-20)+(56÷7)\times 5-17$
3. $=4\times 8+8\times 5-17=32+40-17=55$ ✔

Trick: LHS is far too large, so the big bracket (56 − 20) must shrink – try swapping 56 first.

Answer: (D) 56 and 28