RRB NTPC Undergraduate CBT 1 2025 PYQs

Concept: Place towns on a grid step by step (N up, E right).

1. J – T – R lie on one west-to-east line.
2. K is south of R; M is west of K and south-west of T; J is north of M ⇒ J, T, R, K, M form a rectangle-like grid.
3. K is east and south of J ⇒ J is to the north-west of K.

Trick: Asked ‘J w.r.t. K’ – stand at K and look at J.

Answer: (D) North-west

Concept: Take first and second differences.

1. Differences: $14,42,70,98$
2. Second differences: $28,28,28$ ⇒ next difference $=98+28=126$.
3. Next term $=232+126=358$.

Trick: When first differences grow evenly, add the constant second difference.

Answer: (C) 358

Concept: Find the shift for each position: $(+5,-3,-1)$.

1. JTT → OQS: J+5=O, T−3=Q, T−1=S
2. NQS → SNR: N+5=S, Q−3=N, S−1=R
3. EOR → JLQ: E+5=J, O−3=L, R−1=Q ✓

Trick: Options B, C, D start with $-5$ – reject them from the first letter itself.

Answer: (A) EOR : JLQ

Concept: Facing north, left/right are the solver’s left/right; 7 seats numbered from left.

1. Five between I and L ⇒ they occupy seats 1 and 7. S is immediately left of L ⇒ L $=7$, S $=6$, I $=1$.
2. One between S and N ⇒ N $=4$.
3. T, U, E in seats 2, 3, 5 with T left of U left of E ⇒ T $=2$, U $=3$, E $=5$.
4. Third from left = U.

Trick: ‘Five between’ in a row of 7 means the two people sit at the ends.

Answer: (A) U

Concept: Find one rule linking the first number to the second.

1. $19\times 12+1=229$; $22\times 12+1=265$
2. Rule: second $=$ first $\times 12+1$.
3. $28\times 12+1=337$ ✓ (others: $24\to 289$, $23\to 277$, $26\to 313$).

Trick: If the second number is about 12 times the first, test $\times 12\pm$ small number.

Answer: (C) 28, 337

Concept: Facing the centre, moving to a person’s left = clockwise.

1. Fix Ojas; Neha is 2nd to his left; Upendra is 2nd to Neha’s left.
2. Tina–Shina must be adjacent; the only free adjacent pair is the two seats after Upendra ⇒ Shina, then Tina to her left.
3. Manu is not next to Upendra ⇒ Manu is to the immediate left of Ojas; Pooja sits between Neha and Upendra.
4. From Manu moving left: Neha, Pooja, then Upendra ⇒ two people.

Trick: Draw the circle and always move clockwise for ‘left’ when facing the centre.

Answer: (B) Two

Concept: Check every even digit: previous digit even AND next digit odd.

1. Even digits with neighbours: 4 (start), 9-6-5, 3-6-2, 6-2-1, 1-2-1, 1-6-9, 7-4-7.
2. Only 6-2-1 fits (6 even before, 1 odd after).
3. Count $=1$.

Trick: First find even-even pairs (here only 6 2), then check the next digit.

Answer: (B) One

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

1. ‘know city routes’ → ha mi ko
2. ‘city lights beautiful’ → ly ha jh
3. Common word: city; common code: ha.

Trick: Common word ↔ common code.

Answer: (D) ha

Concept: Look at the differences between consecutive terms.

1. Differences: $12,13,14,15,16$
2. Next difference $=17$.
3. Next term $=111+17=128$.

Trick: Consecutive increasing differences ⇒ just add the next natural number.

Answer: (C) 128

Concept: Track each letter position separately.

1. 1st letters: J, K, L, M → N ($+1$)
2. 2nd letters: K, N, Q, T → W ($+3$)
3. 3rd letters: A, F, K, P → U ($+5$)

Trick: Only option C starts with N – the first letter alone gives the answer.

Answer: (C) NWU

Concept: Scan the series left to right and test each even digit: the digit before it must be a perfect square (1, 4 or 9) and the digit after it must be odd.

1. Series: 9 6 3 1 8 7 2 9 5 6 1 4 7 5 3 4 5 6 2 6 8 7 2 3 7 6 5 8.
2. Matches: 9 6 3 (6 between 9 and 3), 1 8 7 (8 between 1 and 7), 1 4 7 (4 between 1 and 7).
3. Check the rest: 4 5 6 → 6 is followed by 2 (even) ✗; 9 5 → 5 is odd ✗. So exactly three.

Trick: Underline every 1, 4, 9 first and look only at the digit right after each – the search shrinks to a few spots.

Answer: (C) प

Concept: Write the place values (A = 1 … Z = 26) and compare how the second cluster is obtained from the first.

1. OR–KM: O(15)→K(11) is $-4$, R(18)→M(13) is $-5$.
2. QT–MO: Q(17)→M(13) $-4$, T(20)→O(15) $-5$. SV–OQ: S(19)→O(15) $-4$, V(22)→Q(17) $-5$.
3. UX–QR: U(21)→Q(17) $-4$, but X(24)→R(18) is $-6$ – breaks the $(-4,-5)$ pattern.

Trick: Also see the gap inside each cluster: OR, QT, SV, UX all $+3$; KM, MO, OQ are $+2$ but QR is $+1$ – the odd one jumps out.

Answer: (A) ह

Concept: Decode each symbol step by step and build the family tree, then read L’s relation to T.

1. L # I → L is the wife of I; I $ G → I is the father of G. So L is the mother of G.
2. G @ H → G is the sister of H; H & T → H is the brother of T. So G, H and T are all children of I and L.
3. Hence L is the mother of T.

Trick: In a chain, fix the generation of the first and last person only – here L is one level above G, H and T, and she is female, so “mother”.

Answer: (D) Mother

Concept: Look at the differences: $+4,+9,+4,+22,+4,?$ – the $+4$ steps alternate with a second rule.

1. $7+4=11,11\times 2-2=20$
2. $20+4=24,24\times 2-2=46$
3. $46+4=50,50\times 2-2=98$

Trick: When differences alternate small–big–small, test the big jumps as “$\times 2\pm k$”: $11\to 20$ and $24\to 46$ are both $\times 2-2$.

Answer: (A) ज

Concept: Position from the left $=$ Total $-$ position from the right $+1$. Moving to the left decreases the left-end position.

1. Ccd from the left $=122-54+1=69$th.
2. Vks is 17th to the left of Ccd: $69-17=52$nd from the left.

Trick: “Left of” → subtract from the left-position; “right of” → add. First convert everyone to the same end.

Answer: (A) “

Concept: Each term is obtained by multiplying by an increasing number and subtracting the next number: $\times 1-2,\times 2-3,\times 3-4,\dots$

1. $8\times 1-2=6,6\times 2-3=9$
2. $9\times 3-4=23,23\times 4-5=87$
3. $87\times 5-6=435-6=429$

Trick: A series that grows faster and faster (6, 9, 23, 87) usually hides a multiplier that increases by 1 each step.

Answer: (B) त

Concept: For people facing the centre, “right” means anticlockwise and “left” means clockwise. Fix Av and place the others.

1. Bg is 2nd to the right (anticlockwise) of Av; Gp is 2nd to the left (clockwise) of Av.
2. Dq is 3rd to the right of Gp – going anticlockwise from Gp past Av, Dq lands just before Bg. Three seats remain: one between Av and Gp, and two between Gp and Bg.
3. Ck is immediately right of Fx, so Fx and Ck take the two adjacent seats between Gp and Bg (Ck next to Gp); Et takes the seat between Av and Gp.
4. Counting from the left (clockwise) of Et: Gp, then Ck – one person between.

Trick: Draw the circle first and mark arrows “L ↻, R ↺” – half the mistakes in such questions are direction slips.

Answer: (D) प

Concept: Track the first letters of the clusters and the internal pattern of each cluster.

1. First letters: U(21) → G(7) → S(19) → E(5) → ? : alternately $-14$ and $+12$, so next $=5+12=17=$ Q.
2. Inside every cluster: 2nd letter $=$ 1st $-3$, 3rd letter $=$ 1st $+2$ (U R W: 21, 18, 23).
3. For Q(17): $17-3=14=$ N, $17+2=19=$ S → QNS.

Trick: Check the internal pattern against the options: only QNS has the shape (x, x−3, x+2).

Answer: (A) भ

Concept: Number the positions from the top (1) to the bottom (7) and place the boxes using the definite clues first.

1. Only one box above E → E is 2nd; B is immediately above E → B is 1st (top).
2. Two boxes between C and E → C is 5th.
3. “Only Q is below S” → S is 6th and Q is 7th (bottom). The remaining D and R go to 3rd and 4th with D above R.
4. Order (top→bottom): B, E, D, R, C, S, Q. Between B (1st) and S (6th): E, D, R, C = four boxes.

Trick: Start from clues that fix an exact slot (“only one above”, “immediately above”) – they anchor the stack in seconds.

Answer: (B) ठ

Concept: Find how each letter and the number change in the given pairs, then apply the same change.

1. MJ 8 → OM $-6$: M$+2=$O, J$+3=$M, $8-14=-6$.
2. RW 12 → TZ $-2$: R$+2=$T, W$+3=$Z, $12-14=-2$ ✓ same rule.
3. UR 1 → U$+2=$W, R$+3=$U, $1-14=-13$ → WU $-13$.

Trick: Numbers going negative almost always mean a fixed subtraction – find it from the first pair ($8\to -6$ is $-14$) and confirm with the second.

Answer: (C) ऋ