RRB NTPC Graduate CBT 1 2025 PYQs

Concept: Number Monday–Sunday as 1–7 and start with the clue that pins positions.

1. 'Only G has exam before K' → G = Mon (1), K = Tue (2).
2. Three people between H and K → H = Sat (6).
3. Three between L and J → the free pair with gap 4 is (Wed, Sun). F < E < J forces J = Sun (7), L = Wed (3); F = Thu, E = Fri.
4. Before L (Wed): G and K → two people.

Trick: 'Only X before Y' fixes two slots at once – always the first clue to use.

Answer: (D) Two

Concept: Place the fixed person first, then use the 'gap' clue and the consecutive block.

1. C → Friday. Three people between C and F → F is on Monday (Tuesday–Thursday are the three days between).
2. E, D, B are on consecutive days in that order (E, then D, then B). The only three consecutive free days are Tue, Wed, Thu → E = Tuesday, D = Wednesday, B = Thursday.
3. A takes the remaining day, Saturday.

Trick: Fix the definite clue (C = Friday), then slot the three-in-a-row block E–D–B; the answer appears without trial and error.

Answer: (C) Tuesday

Concept: When two words share exactly one letter, the common digit in their codes is the code of that letter.

1. BUST → 3, 5, 1, 9 and LIDS → 6, 8, 9, 4.
2. The only common letter is S; the only common digit is 9.
3. So S = 9.

Trick: Match the common letters with the common digits – no need to decode the whole word.

Answer: (B) 9

Concept: Look at the differences between consecutive terms.

1. $145-141=4$, $141-133=8$, $133-121=12$, $121-105=16$.
2. The differences 4, 8, 12, 16 increase by 4, so the next difference is 20.
3. $105-20=85$.

Trick: When the series falls slowly then faster, check for differences in an arithmetic pattern (multiples of a number).

Answer: (D) 85

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

1. Letters: TRZ → RPX → PNV → NLT: each letter moves 2 steps back (T→R→P→N, R→P→N→L, Z→X→V→T). Next: N−2 = L, L−2 = J, T−2 = R → LJR.
2. Numbers: 2, 5, 8, 11 → +3 each → next 14.
3. Required cluster: LJR 14.

Trick: Note that the first letter of each cluster is the second letter of the previous one (T R Z → R P X → P N V…) – a quick confirmation that letters shift by −2.

Answer: (D) LJR 14

Concept: Letters common to both words get the digits common to both codes; the leftover letter gets the leftover digit.

1. BITE → 4, 5, 6, 8 and TIME → 4, 8, 2, 6.
2. Common letters: I, T, E → common digits 4, 6, 8.
3. The letter only in TIME is M; the digit only in its code is 2. So M = 2.

Trick: Elimination: strike out the shared digits (4, 6, 8) from TIME's code – what remains (2) is M.

Answer: (C) 2

Concept: Build the family tree generation by generation, then read the relation from E's side.

1. B is the mother of C and D (C is D's brother). A is B's sister → A is the maternal aunt of D.
2. E is D's husband → D is E's wife. A is the sister of E's wife's mother.
3. So A is E's wife's mother's sister.

Trick: Write 'E's wife = D' first; whatever A is to D, prefix it with 'wife's'. A = D's mother's sister → wife's mother's sister.

Answer: (B) Wife's mother's sister

Concept: Fix the couple first: Kc's mother is Ef and Kc's father is Cw, so Ef and Cw are husband and wife.

1. Qk is Ef's brother and Lu is Qk's father → Lu is also Ef's father.
2. Ef is Cw's wife, so Lu is Cw's wife's father (father-in-law).

Trick: Whenever two persons are mother and father of the same child, link them as a couple at once – the chain shortens immediately.

Answer: (A) Wife's father

Concept: Scan each symbol and check both neighbours: letter on the left AND letter on the right.

1. Series: K { J % I + H = G * F / D : - ! B C ? A #.
2. Qualifying symbols: { (K…J), % (J…I), + (I…H), = (H…G), * (G…F), / (F…D), ? (C…A) – 7 symbols.
3. Rejected: ':' and '!' are next to '-', '-' is between two symbols, and '#' has nothing on its right.

Trick: Underline every symbol, then tick only those sandwiched between two letters; symbols adjacent to another symbol or at the end are out.

Answer: (A) SEVEN

Concept: Start from the fixed floor and use the 'only n between' clues; six floors, 1 at the bottom.

1. D = 1. Only one person between D and C → C = 3. Only two between O and C → O = 6 (floor 0 does not exist).
2. Q is below C with two people between Q and P: Q = 2 and P = 5 (Q = 1 is taken by D).
3. The last person N takes floor 4. Above N are P (5) and O (6) → two people.

Trick: Build the tower bottom-up: 1-D, 2-Q, 3-C, 4-N, 5-P, 6-O. 'Above N' = floors higher than 4.

Answer: (A) Two

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

1. $63\times 3=189$, $189-30=159$ ✓
2. $44\times 3=132$, $132-30=102$ ✓ – rule: $3n-30$.
3. $57\times 3-30=171-30=141$.

Trick: Check the difference of the pairs: $159-102=57$ and $63-44=19$; $57=3\times 19$, so the multiplier is 3, then adjust the constant.

Answer: (C) 141

Concept: Find the shift from the right-hand pair where both terms are partly known, then apply it (in reverse) on the left.

1. Right side XHP : %. Testing option letters: X(24)+6 = 30 → D(4), H(8)+6 = N(14), P(16)+6 = V(22) → % = DNV (each letter +6, cyclic).
2. So the rule is 'first + 6 = second'. On the left, # + 6 = KUC → # = K−6 = E, U−6 = O, C−6 = W (3−6 = −3 → 23 = W).
3. # = EOW, % = DNV.

Trick: Only option (D) has both EOW and DNV; options with DNO or FOW break the uniform +6 shift.

Answer: (D) # = EOW % = DNV

Concept: Compare letter positions in the given pair to find the position-wise shift.

1. RILA → SKOE: R+1 = S, I+2 = K, L+3 = O, A+4 = E. Pattern +1, +2, +3, +4.
2. Check: KEPO → K+1 = L, E+2 = G, P+3 = S, O+4 = S → LGSS ✓
3. MJRT → M+1 = N, J+2 = L, R+3 = U, T+4 = X → NLUX.

Trick: When the first letters differ by 1 and the last by 4, suspect the +1, +2, +3, +4 staircase.

Answer: (B) NLUX

Concept: Facing the centre: 'left' means clockwise and 'right' means anticlockwise. Start with the clue linking two fixed people.

1. F is to the immediate right of C. C is third to the left of G, so G is third to the right of C. F is third to the right of E → E is third to the left of F.
2. Going clockwise from C: C, _, E, _, G, _, F (E is 3 places left of F, G is 3 right of C, F just right of C).
3. Only one person between D and B counted from the left of B → D is 2nd to the left of B. The empty seats are the 1st, 3rd and 5th clockwise from C; A is not next to C, so A is at the 5th seat (between G and F), B at the 1st and D at the 3rd.
4. From A's left (clockwise): F, then C → one person between A and C.

Trick: Draw 7 dots, mark C, and put F anticlockwise of C at once; the rest of the clues fill in without any trial.

Answer: (B) 1

Concept: Draw the sets: Spoons inside Forks inside Cups; Plates overlapping Spoons.

1. Some spoons are plates, and every spoon is a fork → those spoon-plates are forks → 'Some forks are plates' follows.
2. Every fork is a cup, so those same elements are cups too → 'Some cups are plates' follows.
3. Both conclusions follow.

Trick: 'All A are B' + 'Some A are C' always gives 'Some B are C'; chain it up through 'All B are D' to get 'Some D are C'.

Answer: (A) Both conclusions I and II follow.

Concept: Caps inside Helmets inside Hats; Crowns completely outside Hats.

1. Every cap is a helmet and every helmet is a hat → every cap is a hat. No hat is a crown → no cap is a crown. Conclusion I follows.
2. All caps are hats; hence at least those caps are hats that are caps → 'Some hats are caps' follows (conversion of 'All caps are hats').
3. Both conclusions follow.

Trick: 'All A are B' can always be converted to 'Some B are A'; and 'All A are B' + 'No B is C' gives 'No A is C'.

Answer: (A) Both conclusions (I) and (II) follow.

Concept: Remove the symbols and count positions from the left in the new series.

1. Symbols in the series: #, %, #, €, *, ©, &, €. Dropping them leaves: S 2 S 4 P Z 4 B 8 5 R 8 J N.
2. Counting from the left: 1st S, 2nd 2, 3rd S, 4th = 4.

Trick: In "drop" questions, write only the remaining elements in one line before counting – avoids skipping errors.

Answer: (C) 4

Concept: Check each letter whose immediate left neighbour and immediate right neighbour are both symbols.

1. Series: H F G J D J V D K P = % S O C B = % B + #.
2. Letters next to symbols: S (between % and O – O is a letter, fails), B (between C and =, fails), B (between % and +) – passes.
3. So only one such letter.

Trick: Look only around the symbols; a letter needs a symbol on BOTH sides, so scan each symbol-letter-symbol triplet.

Answer: (D) One

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

1. F is 2nd from bottom → F = 2. Three boxes between F and D → D = 6. G is below F → G = 1.
2. Only E between D and B → E = 5, B = 4 (position 8 does not exist).
3. A above D → A = 7. The remaining box C = 3.
4. Boxes above C: B, E, D, A = 4.

Trick: Always start with the clue that gives an exact position (F = 2) and the clue with a fixed gap (3 between F and D).

Answer: (A) 4

Concept: Test each statement alone; the question needs only the tallest tree.

1. Statement I: N > M > V, and M is shorter than only one tree → M is 2nd tallest, so the only tree above M, N, is the tallest. Sufficient.
2. Statement II: V > T gives no information about the tallest. Not sufficient.

Trick: "Shorter than only one" = second position; whoever is already known to be above that person is the top.

Answer: (D) Data in Statement I alone is sufficient to answer the question while data in statement II is not