RRB NTPC Graduate CBT 2 2024 PYQs

Concept: Common words in two sentences share the common code.

1. ‘dogs’ in 1 and 2 → common code xl.
2. ‘bark’ in 1 and 3 → mn; ‘and’ in 2 and 3 → rs.
3. Sentence 2: cats and dogs = rs tp xl → cats = tp.

Trick: Strike out known codes; the leftover is the answer.

Answer: (C) tp

Concept: Check every number with its left and right neighbours.

1. Numbers: 1, 9, 7, 0.
2. 1: preceded by T (letter) ✗; 9: preceded by & (symbol), followed by Z (letter) ✓.
3. 7: preceded by Z ✗; 0: preceded by A ✗.
4. Only one such number (9).

Trick: Pattern to find: Symbol – Number – Letter.

Answer: (B) One

Concept: Draw the minimum (possible-case) Venn diagram.

1. Cutter lies inside Knife; Nail and Scissor are both outside Cutter.
2. Scissor may or may not overlap the Knife area outside Cutter → I (No scissor is a knife) is not certain.
3. Nail may or may not overlap Knife → II (Some nails are knives) is not certain.

Trick: "No A is B" + "All B are C" gives no definite relation between A and C.

Answer: (D) Neither conclusion (I) nor (II) follows.

Concept: Track each letter position separately.

1. 1st letters: W, A, E, I, M, Q (+4 each).
2. 2nd letters: O, S, W, A, E, I (+4 each).
3. 3rd letters: G, K, O, S, ?, A → should be W (+4), but MEV has V.
4. Correct cluster: MEW, so MEV is incorrect.

Trick: All three positions move +4: W(23)+4 = A(27−26).

Answer: (B) MEV

Concept: Put A at origin (east = +x, north = +y).

1. Laksh: 7 E → (7, 0); right (south) 12 → (7, −12); left (east) 6 → B (13, −12).
2. P = (16, 0); Vinod: 4 E → (20, 0); right (south) 12 → (20, −12); right (west) 3 → Q (17, −12).
3. B and Q on same line: $17-13=4\text{km}$.

Trick: Both end 12 km south, so only east–west positions matter.

Answer: (D) 4 km

Concept: Number positions 1–4 from west. Row 2 (north-facing): left = west; Row 1 (south-facing): left = east.

1. Z third to left of R → Z at 1, R at 4 (row 2).
2. Facing F is neighbour of R → position 3, so F at 3.
3. D third to right of G → G at 4, D at 1; A at 2 (right of F ✓).
4. A not facing P → position 2 of row 2 is N, P at 3. A faces N.

Trick: In opposite rows, "right" of a south-facing person is the viewer’s left.

Answer: (C) A

Concept: Look at the differences.

1. Differences: 16, 32, 48, 64 (multiples of 16).
2. Next difference $=80$.
3. $169+80=249$.

Trick: Differences = table of 16.

Answer: (C) 249

Concept: A valid course of action must solve the stated cause practically.

1. Cause: lack of trained teachers and poor infrastructure → recruiting and training teachers (I) directly addresses it.
2. Expelling failing students (II) is harsh and does not remove the cause.

Trick: Reject extreme / punitive actions like "expel all", "ban all".

Answer: (D) Only I follows

Concept: Find the rule from the complete pair and apply it.

1. Try (A): MUR → NVS: M+1 = N, U+1 = V, R+1 = S.
2. Same rule: JRO → J+1 = K, R+1 = S, O+1 = P → KSP.
3. Both # = MUR and % = KSP fit.

Trick: Each letter +1 on both sides of ::.

Answer: (A) # = MUR, % = KSP

Concept: Swap the signs, then use BODMAS.

1. After swapping: $50+8-2\times 6÷4$.
2. $2\times 6÷4=3$.
3. $50+8-3=55$.

Trick: Rewrite the whole expression first, then solve – avoids sign slips.

Answer: (D) 55

Concept: Draw the possible-case diagram: Park ⊂ Meadow ⊂ Forest; Garden overlaps only Forest.

1. "Some forests are gardens" – the common part may lie outside Meadow and Park.
2. So no definite link between Park and Garden → I does not follow.
3. Similarly no definite link between Meadow and Garden → II does not follow.

Trick: "Some" with the bigger set (Forest) never passes down to its subsets.

Answer: (B) Neither conclusion I nor II follows

Concept: Look at the multipliers.

1. $4\times 1=4$, $4\times 3=12$, $12\times 5=60$, $60\times 7=420$.
2. Multipliers are odd numbers 1, 3, 5, 7 → next 9.
3. $420\times 9=3780$.

Trick: Ratios growing by 2 → multiply by consecutive odd numbers.

Answer: (C) 3780

Concept: S must be female and a child of P (or P’s husband).

1. (A) P × Q: P is wife of Q. Q ÷ R: Q is father of R.
2. R – S: R is brother of S → S is also child of Q and P.
3. S + T: S is mother of T → S is female.
4. So S is the daughter of P.

Trick: For "daughter", the person must later appear as mother (+) or wife (×) to fix gender.

Answer: (A) P x Q ÷ R – S + T

Concept: Find a rule that fits both given pairs.

1. $9\times 6+1=55$ ✓; $12\times 6+1=73$ ✓.
2. $16\times 6+1=97$.

Trick: Check the gap: $73-55=18=6\times (12-9)$ → multiplier is 6.

Answer: (D) 97

Concept: Swap the signs, then use BODMAS.

1. After swapping: $60-15\times 3+10÷2$.
2. $15\times 3=45$, $10÷2=5$.
3. $60-45+5=20$.

Trick: Rewrite first, then solve ÷/× before +/−.

Answer: (B) 20

Concept: To find an age we need at least one actual value (present age of someone, or a year), not just age differences.

1. Statement I gives only the father's age at Rahul's birth (a difference of 33 years).
2. Statement II gives only a difference of 29 years with the mother.
3. Together we know two differences but no actual age of anyone, so Rahul's age cannot be found.

Trick: Only differences (no absolute age/year) → never sufficient.

Answer: (B) Both statements I and II together are NOT sufficient

Concept: Compare the shift from the first cluster to the second cluster letter by letter.

1. RX – MS: $R\to M=-5$, $X\to S=-5$
2. MF – HA: $-5,-5$; IO – DJ: $-5,-5$
3. NR – GL: $N\to G=-7$, $R\to L=-6$ → different

Trick: Write positions: R18→M13, X24→S19; the odd pair breaks the uniform $-5$.

Answer: (B) NR – GL

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

1. 'is' in sentences 1 & 2 → common code rt, so is = rt.
2. 'tasty' in 1 & 3 → common code pj; 'good' in 2 & 3 → common code bn.
3. Sentence 3: tasty good meal = pj bn dv → meal = dv.

Trick: Strike out codes already decoded in the third sentence; the leftover is the answer.

Answer: (B) dv

Concept: Check the gaps between consecutive letters of each cluster.

1. LPQ: $L\to P=+4$, $P\to Q=+1$
2. XBC: $X\to B=+4$ (cyclic), $B\to C=+1$; SWX: $+4,+1$
3. NSS: $N\to S=+5$, $S\to S=0$ → odd

Trick: Pattern $+4,+1$: last two letters are consecutive in three clusters; NSS has a repeated letter.

Answer: (A) NSS

Concept: Draw the least-overlap Venn diagram; a conclusion follows only if it is true in every possible diagram.

1. All calculators are inside Pens; some calculators overlap Stones.
2. Octopus is outside Calculators, but it may or may not overlap Stones or Pens.
3. I ('No octopus is a stone') and II ('Some pens are octopuses') are only possibilities, not certain.

Trick: 'No A is B' with no direct link between Octopus and Stones/Pens → neither definite.

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