RRB NTPC Undergraduate CBT 1 2025 PYQs

Concept: Vowels move +1, consonants move −1 in the alphabet.

1. N O T F E R N → N−1 = M, O+1 = P, T−1 = S, F−1 = E, E+1 = F, R−1 = Q, N−1 = M.
2. New group: M P S E F Q M.
3. Only M appears exactly twice → one letter.

Trick: The two N’s both become M; no other letter repeats.

Answer: (B) One

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

1. LOTS → 8 4 3 9; JOLS → 4 8 5 3.
2. Common letters O, L, S ↔ common digits {8, 4, 3}.
3. Leftover in JOLS is J ↔ leftover digit 5.

Trick: Digits need not be in letter order – just compare the sets.

Answer: (C) 5

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

1. TR − OM: T→O = −5, R→M = −5.
2. PN − KI: −5, −5. XV − SQ: −5, −5.
3. US − OP: U→O = −6, S→P = −3 – odd one.

Trick: Three pairs have a uniform −5 shift; check the first option last if it looks different.

Answer: (A) US−OP

Concept: Find the letter-wise shift in the given pair and apply it to the new cluster.

1. LOSM → QTXR: L+5 = Q, O+5 = T, S+5 = X, M+5 = R.
2. ILPJ → NQUO: each letter +5 (confirms rule).
3. FIMG: F+5 = K, I+5 = N, M+5 = R, G+5 = L → KNRL.

Trick: Uniform +5; options with the same letters in a different order (KNLR, NKRL) are traps.

Answer: (C) KNRL

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

1. A ÷ B: A is sister of B. B × C: B is husband of C.
2. C + D: C is mother of D; so B and C are parents of D.
3. D – E: D is brother of E; E ÷ F: E is sister of F → D, E, F are siblings, children of B and C.
4. A is the sister of B, F’s father → A is F’s father’s sister.

Trick: Siblings share parents – once B is D’s father, he is F’s father too.

Answer: (A) Father’s sister

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

1. Given: $12÷4+9\times 3-8$. After swapping (+ ↔ −, × ↔ ÷): $12\times 4-9÷3+8$.
2. Division first: $9÷3=3$; multiplication: $12\times 4=48$.
3. $48-3+8=53$.

Trick: Rewrite the whole expression with new signs before calculating – avoids mixing old and new.

Answer: (A) 53

Concept: Decode each symbol and build the tree generation by generation.

1. C ≠ D: C is wife of D. D ₹ E: D is father of E → C is E’s mother.
2. E × L: E is mother of L. L ÷ M: L is daughter of M → M is L’s father, i.e. E’s husband.
3. C is the mother of E, and E is M’s wife → C is M’s wife’s mother.

Trick: Mother of L + father of L = couple; then C (E’s mother) is M’s mother-in-law.

Answer: (C) Wife’s mother

Concept: Use coordinates: East = +x, North = +y; left turn rotates E→N→W→S→E.

1. A (0, 0) → East 17 → (17, 0) → left (North) 18 → (17, 18) → left (West) 30 → (−13, 18).
2. Left (South) 31 → (−13, −13) → left (East) 9 → (−4, −13) → left (North) 6 → (−4, −7).
3. Right turn → East 4 → P (0, −7).
4. P is 7 km directly south of A → he must drive 7 km North.

Trick: Net East–West: $17-30+9+4=0$; net North–South: $18-31+6=-7$ → 7 km due north.

Answer: (D) 7 km to the North

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

1. All Eagles are Seagulls → Eagles circle inside Seagulls. Some Seagulls are Peacocks → Peacocks overlaps Seagulls somewhere.
2. The overlap may or may not include Eagles → “Some Eagles are Peacocks” is not certain (I ✗).
3. Equally, “No Peacocks are Eagles” is not certain (II ✗).

Trick: All + Some (through the middle term) → no definite conclusion between the end terms.

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

Concept: Facing north, left and right are the same as ours. Number seats 1–7 from the left.

1. Two people to the left of N → N at 3. Three between N and G → G at 7.
2. One between G and E → E at 5.
3. I immediately left of R: free seats 1, 2, 4, 6 → only (1, 2) are adjacent → I = 1, R = 2.
4. A is not next to G → A ≠ 6 → A = 4, T = 6.
5. To the right of T (seat 6) there is only G → one person.

Trick: “Only two to the left of N” fixes N at seat 3 instantly – start there.

Answer: (C) One

Concept: Apply the same operation to each letter and to the number.

1. KL 9 → IH −2: K−2 = I, L−4 = H, $9-11=-2$.
2. PG 7 → NC −4: P−2 = N, G−4 = C, $7-11=-4$ – rule confirmed.
3. JQ 14: J−2 = H, Q−4 = M, $14-11=3$ → HM 3.

Trick: First letter −2, second letter −4, number −11; eliminate options whose letters are wrong first.

Answer: (D) HM 3

Concept: Look at the differences between consecutive terms.

1. $150-135=15,180-150=30,225-180=45,285-225=60$.
2. Differences are multiples of 15: 15, 30, 45, 60 → next 75.
3. $285+75=360$.

Trick: All terms are multiples of 15: $9,10,12,15,19,24\times 15$ (gaps 1, 2, 3, 4, 5) → $24\times 15=360$.

Answer: (A) 360

Concept: Replace the letters by their operators and apply BODMAS.

1. 3 A 42 B 14 C 29 D 22 → $3÷42\times 14+29-22$.
2. $3÷42=\frac{1}{14}$; $\frac{1}{14}\times 14=1$.
3. $1+29-22=8$.

Trick: $÷$ then $\times$ left to right: $\frac{3\times 14}{42}=1$ in one step.

Answer: (C) 8

Concept: A conclusion follows only if it holds in every Venn diagram consistent with the statements.

1. All apples are baskets → Apples inside Baskets. Some baskets are cups → Cups overlaps Baskets.
2. The Cups overlap need not reach Apples → “Some apples are cups” is only possible, not definite (I ✗).
3. “All cups are apples” is certainly not forced – cups may lie outside apples (II ✗).

Trick: “All A are B” + “Some B are C” never gives a definite A–C conclusion.

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

Concept: Track each position separately; here every letter moves +4.

1. 1st letters: T → X → B → F → ? : +4 → J.
2. 2nd letters: U → Y → C → G → ? : +4 → K.
3. 3rd letters: W → A → E → I → ? : +4 → M.

Trick: Within each cluster the pattern is (n, n+1, n+3) – JKM fits; JLM, JMK, KJL do not.

Answer: (B) JKM

Concept: Number 1 (top) to 7 (bottom); test the two cases a “between” clue creates.

1. K = 4. One box between H and K → H = 2 or 6.
2. Three between H and S with S above K: H = 2 gives S = 6 (below K ✗); so H = 6, S = 2. I immediately below S → I = 3.
3. Free: 1, 5, 7 for N, G, T. G not adjacent to H (6) → G ≠ 5, 7 → G = 1.
4. N above T → N = 5, T = 7. Below N: H, T → two boxes.

Trick: Use the “S above K” condition to kill one case immediately.

Answer: (B) Two

Concept: Substitute the operators, simplify both sides with BODMAS, then solve for ?.

1. 7 A 7 C 20 B 2 = ? C 20 → $7÷7+20\times 2=?+20$.
2. LHS: $1+40=41$.
3. $?=41-20=21$.

Trick: Do not forget the right side also has an operation (+ 20).

Answer: (B) 21

Concept: When differences are irregular, try “×2 ± constant”.

1. $4\times 2+2=10$, $10\times 2-2=18$, $18\times 2+2=38$, $38\times 2-2=74$.
2. Pattern alternates +2, −2; next is +2.
3. $74\times 2+2=150$.

Trick: Each term ≈ double the previous → check ×2 first, then fix the small adjustment.

Answer: (B) 150

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

1. RF – NB: R→N = −4, F→B = −4.
2. MJ – IF: −4, −4. UG – QC: −4, −4.
3. PT – LG: P→L = −4, T→G = −13 – odd one.

Trick: Check the second letter too – the first letter of PT–LG also drops by 4, which is the trap.

Answer: (D) PT – LG

Concept: Track each position separately; every letter moves +5.

1. 1st letters: Y → D → I → N → ? : +5 → S.
2. 2nd letters: B → G → L → Q → ? : +5 → V.
3. 3rd letters: Z → E → J → O → ? : +5 → T.

Trick: Within each cluster the pattern is (n, n+3, n+1) – SVT fits (S=19, V=22, T=20).

Answer: (B) SVT