RRB NTPC Graduate CBT 1 2024 PYQs

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

1. Series: 6 @ $ 1 8 # 9 % ≺ 2 4 ⋟ 3 ^ 5 & 7.
2. Valid: 8 # 9, 4 ⋟ 3, 3 ^ 5, 5 & 7 → 4 symbols.
3. @, $, % and ≺ each touch another symbol, so they do not count.
4. 4 is more than three.

Trick: Scan only the symbols and check both neighbours of each.

Answer: (C) More than three

Concept: Each letter moves $+12$ positions (cycling after Z).

1. First letters: F(6) → R(18) → D(4) → P(16) → B(2) → N(14).
2. Second letters: R → D → P → B → N.
3. Third letters: O(15) → A(1) → M(13) → Y(25) → K(11).
4. Next term = BNK.

Trick: $+12$ and $-14$ give the same letter; also each term's 2nd letter becomes the next term's 1st.

Answer: (B) BNK

Concept: Decode each symbol from left to right.

1. E ÷ F: E is the father of F. F + G: F is the son of G, so G is F's mother, i.e. E's wife.
2. G – H: G is the sister of H. H × K: H is the wife of K.
3. So K's wife is H, H's sister is G, and G's husband is E.
4. E is K's wife's sister's husband.

Trick: Draw the couples with double lines – the relation reads straight off the chart.

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

Concept: For people facing the centre, 'right' is anticlockwise and 'left' is clockwise.

1. Fix Z. S is third to the right of Z; M is immediately left of S; T is third to the right of M.
2. Remaining seats for A, B, L: two seats beside Z and the seat between S and T.
3. B is second to the left of L. If A sat next to Z on M's side, A would neighbour M – not allowed. So A sits between S and T, L between Z and M, B between T and Z.
4. From A going right: T, B, Z, then L → 3 people.

Trick: Place the chain Z → M → S → T first; only three seats remain to test.

Answer: (C) 3

Concept: Find the rule from the complete pair and apply it to both sides of ::.

1. Option C: UOR → ZTW: U+5=Z, O+5=T, R+5=W.
2. NHK → SMP: N+5=S, H+5=M, K+5=P – same rule $+5$.
3. Other options do not give a single consistent rule on both sides.

Trick: Test each option on the left pair first; the moment it fails, drop it.

Answer: (C) # = UOR, % = SMP

Concept: Fill the definite clues first.

1. R lives on floor 1.
2. Only A above T → A = 6, T = 5.
3. Only two between R (1) and P → P = 4.
4. Y on an even floor → Y = 2 (4 and 6 are taken), so L = 3.
5. Between L (3) and A (6): T and P → two people.

Trick: 'Only X above Y' → X on the top floor, Y just below it.

Answer: (C) Two

Concept: Take A as origin; East = +x, North = +y.

1. North 9 → (0, 9); right = East 11 → (11, 9).
2. Right = South 17 → (11, −8); right = West 12 → (−1, −8).
3. Right = North 8 → P(−1, 0).
4. P is 1 km West of A, so he must drive 1 km to the East.

Trick: Net E–W: $11-12=-1$; net N–S: $9-17+8=0$.

Answer: (C) 1 km to the east

Concept: A conclusion follows only if it is true in every possible diagram.

1. All letters are messages; some messages are emails; no email is a post.
2. Letters may lie in the part of 'messages' outside 'emails', so 'Some emails are letters' is not certain – (II) fails.
3. Post is barred only from emails; letters may still overlap post, so 'No letter is a post' is not certain – (I) fails.

Trick: 'Some' linking through the middle term (messages) never gives a definite link between letters and emails.

Answer: (B) Neither conclusion (I) nor (II) follows

Concept: Treat the two letters and the number separately.

1. 1st letter: N, M, L, K → J ($-1$).
2. 2nd letter: R, Q, P, O → N ($-1$).
3. Number: 25, 36, 47, 58 → 69 ($+11$).
4. Next term = JN69.

Trick: Number +11 rules out JM68, JN68; letter −1 rules out JM69.

Answer: (B) JN69

Concept: Change each digit by its own rule: odd $+1$, even $-1$.

1. Number: 2 6 3 7 4 5 8.
2. New: 1 5 4 8 3 6 7.
3. Greatest digit $=8$, smallest $=1$; sum $=9$.

Trick: Odd digits rise and even digits fall, so check 7 → 8 (largest) and 2 → 1 (smallest) directly.

Answer: (A) 9

Concept: Track the position with East–West and North–South totals; a left turn while facing West means facing South, a right turn means facing North.

1. West 18, left (South) 6, right (West) 21 → he is 39 km West and 6 km South of A.
2. Right (North) 18 → 6 km South becomes 12 km North; right (East) 39 → East–West offset becomes 0.
3. So P is exactly 12 km North of A; to return he must drive 12 km towards the South.

Trick: Net E–W: $-18-21+39=0$; net N–S: $-6+18=12$ North → go back South.

Answer: (A) 12 km to the South

Concept: A conclusion follows only if it is true in every possible Venn diagram of the statements.

1. Induction lies inside Stove; Oven meets Stove only in “some” part – that part may lie outside Induction.
2. Tandoor meets Oven only in “some” part, so it need not touch Induction → (I) Some tandoor are induction: not certain.
3. Oven may overlap only the non-induction part of Stove → (II) Some oven are induction: not certain.

Trick: “Some + Some” or “All + Some” never gives a definite “some” link to the first term.

Answer: (A) Neither conclusion (I) nor (II) follows

Concept: Check the differences between consecutive terms.

1. Differences: $13-6=7$, $25-13=12$, $42-25=17$, $64-42=22$
2. The differences rise by 5: next difference $=27$
3. $64+27=91$

Trick: Second differences constant (5) → keep adding 5 to the last gap.

Answer: (C) $91$

Concept: Decode each symbol step by step and draw the family tree.

1. P + Q: P is the father of Q. Q × R: Q is the son of R, so R is Q’s mother (wife of P).
2. R – S: R is the daughter of S. S ÷ T: S is the wife of T, so T is R’s father.
3. R is T’s daughter and P is R’s husband → P is T’s daughter’s husband (son-in-law).

Trick: Go from T downwards: T → daughter R → her husband P.

Answer: (A) Daughter's husband

Concept: A word common to two sentences has the code common to both coded sentences.

1. ‘talk to him’ → ‘yd hp mu’; ‘when you talk’ → ‘nc mu bt’
2. Common word: talk; common code: mu
3. So talk = mu.

Trick: Match the common word with the common code – order of codes does not matter.

Answer: (D) mu

Concept: Treat the first letter, second letter and number as three separate series.

1. First letters: J, I, H, G → each −1 → F
2. Second letters: N, M, L, K → each −1 → J
3. Numbers: 28, 40, 52, 64 → each $+12$ → $76$

Trick: Both letters move back 1; number $+12$ → FJ76.

Answer: (D) FJ76

Concept: Swap the signs first, then solve using BODMAS.

1. Swap ÷ ↔ − and × ↔ +: $4320÷96+37\times 21-372$
2. $4320÷96=45$, $37\times 21=777$
3. $45+777-372=450$

Trick: Write the new expression fully before calculating – most errors come from half-swapped signs.

Answer: (B) $450$

Concept: Find the letter-shift that turns the left term into the right term in each pair.

1. GLI → %: try +5 on each letter: G+5 = L, L+5 = Q, I+5 = N → % = LQN.
2. Then # → QVS must also be +5: L+5 = Q, Q+5 = V, N+5 = S → # = LQN.
3. Only option (B) gives # = LQN and % = LQN.

Trick: Check one option at a time with the same +5 rule on both sides.

Answer: (B) # = LQN, % = LQN

Concept: Look for the pattern symbol – number – symbol.

1. Series: 9 # 2 6 7 © 6 8 & © $ & 3 9 1 7 $ 5 # @ * 4
2. # 2 6 (6 is a number), & 3 9 (9 is a number) – not valid.
3. Only $ 5 # fits → count = 1.

Trick: Scan only the numbers that come right after a symbol, then check the next element.

Answer: (D) 1

Concept: Add up East–West and North–South movements separately.

1. West 5, right (North) 11, right (East) 3, left (North) 4, right (East) 3, right (South) 15.
2. East–West: $-5+3+3=+1$ → 1 km East. North–South: $11+4-15=0$.
3. B is 1 km East of A, so he must drive 1 km towards the West.

Trick: North–South cancels ($15=11+4$) → only the East–West gap of 1 km remains.

Answer: (A) 1 km towards the West