RRB NTPC Undergraduate CBT 1 2025 PYQs

Concept: The multiplier increases by 0.5 each step and 2 is subtracted every time.

1. $28\times 0.5-2=12$; $12\times 1-2=10$; $10\times 1.5-2=13$; $13\times 2-2=24$.
2. Next: $24\times 2.5-2=60-2=58$.

Trick: A series that first falls then rises sharply usually has a growing multiplier ($\times 0.5,\times 1,\times 1.5\dots$).

Answer: (D) 58

Concept: Draw Clouds inside Water; Vapor overlaps Water somewhere – it may or may not touch Clouds.

1. (I) 'Some clouds are vapor' is possible but not certain (Vapor may overlap only the non-cloud part of Water) → does not follow.
2. (II) 'Some vapor is water' is just the converse of 'Some water is vapor' → always true → follows.
3. Only conclusion (II) follows.

Trick: 'Some A are B' ⇔ 'Some B are A' (I-type converts); 'All' + 'Some' never gives a definite 'Some' about the first term.

Answer: (A) Only conclusion (II) follows

Concept: Track the position with coordinates: east = $+x$, north = $+y$; every right turn rotates N→E→S→W→N.

1. A(0, 0) → north 12 → (0, 12) → right (east) 15 → (15, 12).
2. Right (south) 17 → (15, −5) → right (west) 18 → (−3, −5).
3. Right (north) 5 → P(−3, 0).
4. P is 3 km west of A, on the same line → he must drive 3 km east.

Trick: Net E–W: $15-18=-3$ (3 km west); net N–S: $12-17+5=0$. The answer is just the net displacement reversed.

Answer: (D) 3 km to the east

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

1. $805÷2+26-1203\times 3$ becomes $805\times 2-26+1203÷3$.
2. $805\times 2=1610$; $1203÷3=401$.
3. $1610-26+401=1985$.

Trick: Do ÷ and × first, then + and − from left to right; $1203÷3$ is exact (1+2+0+3=6), a hint the swap is right.

Answer: (B) 1985

Concept: Facing the centre, a person's right is anticlockwise and left is clockwise. Number seats 1–7 clockwise.

1. Put Q at 1. One person between Q and L from Q's right (anticlockwise) → L at 6.
2. One between L and M from M's left (clockwise from M) → M at 4. One between Q and G from G's right → G at 3.
3. Two between H and M from M's left → H at 7. One between F and G from F's right → F at 5. O takes seat 2.
4. O's immediate right (seat 1) = Q; immediate left (seat 3) = G.

Trick: Draw the circle with arrows pointing inward; 'from the right of X' means move anticlockwise from X.

Answer: (A) Q and G

Concept: Replace each symbol by its meaning, then solve with BODMAS.

1. $18-3÷9+15\times 3$ → $18\times 3+9-15÷3$.
2. $18\times 3=54$; $15÷3=5$.
3. $54+9-5=58$.

Trick: Write the converted expression fully before calculating – most errors come from converting and computing in the same step.

Answer: (C) 58

Concept: Decode pairwise and build the tree around the couple P–J.

1. K & V → K is husband of V. V * P → V is daughter of P.
2. P # W → P is father of W; W % J → W is son of J → J is W's mother, i.e. P's wife.
3. So J is the mother of V, and V is K's wife.
4. J is K's wife's mother (mother-in-law). (G @ K – G is K's mother – is extra information.)

Trick: Find the chain that links the two asked persons (K → V → J) and ignore the rest.

Answer: (A) Wife's mother

Concept: Compare the first letters across the triad, and the second letters across the triad.

1. RN-TP-VY: first letters R→T→V ($+2,+2$); second letters N→P ($+2$), P→Y ($+9$).
2. PL-RN-TW: P→R→T ($+2,+2$); L→N ($+2$), N→W ($+9$). Same pattern.
3. NJ-PL-RU: N→P→R ($+2,+2$); J→L ($+2$), L→U ($12+9=21$) ✓.
4. Option D ends in V (22), option B/C break the $+2$ at the second letter.

Trick: Options A and D differ only in the last letter – compute just $L+9$ to decide.

Answer: (A) NJ-PL-RU

Concept: Each position of the cluster forms its own series.

1. 1st letters W, B, G, L → $+5$ each (W→B wraps past Z) → Q.
2. 2nd letters A, F, K, P → $+5$ → U.
3. 3rd letters X, C, H, M → $+5$ → R. Term = QUR.

Trick: All three series share the same jump ($+5$); after finding the 1st letter Q, the 2nd letter U eliminates every other option.

Answer: (C) QUR

Concept: Number positions 1 (top) to 7 (bottom); start from the box whose position is fixed by a count.

1. Two boxes below A → A at 5; R immediately above A → R at 4.
2. Three boxes between T and A → T at 1. One box between S and R → S at 2 or 6.
3. Four boxes between S and F: S=2 → F=7 ✓ (S=6 would need F at 1 = T). So S=2, F=7.
4. O above C in the free slots 3 and 6 → O=3, C=6. Stack: T S O R A C F → boxes below O = 4.

Trick: 'Only n boxes below X' fixes X at position $7-n$ immediately; build everything from there.

Answer: (C) Four

Concept: Check each position of the clusters separately for a constant jump in the alphabet (A=1 … Z=26, cyclic).

1. 1st letters: G → Z → S → L : each $-7$ (G(7)−7 = 0 → Z(26)). Next: L(12)−7 = 5 → E.
2. 2nd letters: J → C → V → O : each $-7$. Next: O(15)−7 = 8 → H.
3. 3rd letters: E → X → Q → J : each $-7$. Next: J(10)−7 = 3 → C.

Trick: Subtracting 7 is the same as adding 19; when you go below A, add 26 and continue.

Answer: (A) EHC

Concept: When two coded sentences share exactly one word, the code common to both sentences is the code of that word.

1. Sentence 1: steak tooth badge → qj hc sv.
2. Sentence 2: badge poet lemon → tz qj wv.
3. Only "badge" is common and only "qj" is common → badge = qj.

Trick: Match the common word with the common code; order of codes in a sentence does not matter.

Answer: (C) qj

Concept: Decode each symbol step by step and build the family tree, then read the relation from P's side.

1. A @ D : A is the son of D. D £ J : D is the father of J → A and J are brothers.
2. J ≠ L : J is the brother of L → L is also a child of D, hence A's sister.
3. L ≤ P : L is the wife of P → A is the brother of P's wife.

Trick: Children of the same father are siblings; the sister's husband sees you as "wife's brother" (saala).

Answer: (B) Wife's brother

Concept: If a person is $m^{th}$ from the top and $n^{th}$ from the bottom, total students $=m+n-1$ (the person is counted twice).

1. $m=27,n=64$.
2. Total $=27+64-1=90$.

Trick: Top rank + bottom rank − 1 = total (subtract 1 for the double-counted person).

Answer: (C) 90

Concept: Number the seats 1–7 from left to right (facing north, left = west). Place the people with fixed positions first.

1. Six people right of E → E at 1. Five people left of Q → Q at 6. Three between M and Q → M at 2.
2. Seats 3, 4, 5, 7 remain for B, L, G, F. B is between L and G, so B at 4 with L, G at 3 and 5; F takes 7.
3. Three between F(7) and L → L at 3, so G at 5. Order: E M L B G Q F.
4. Between G(5) and F(7) sits only Q → one person.

Trick: "Only n people sit to the right/left" pins a person to an exact seat – start from such clues.

Answer: (A) One

Concept: Total students $=(\text{rank from top})+(\text{rank from bottom})-1$.

1. Rank from top $=44$, from bottom $=25$.
2. Total $=44+25-1=68$.

Trick: Add the two ranks and subtract 1 (Ayush himself is counted in both).

Answer: (A) 68

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

1. Letters: N(14) → I(9) → D(4) → Y(25) : each $-5$ (D−5 = −1 → wrap to 25 = Y). Next: Y(25)−5 = 20 → T.
2. Numbers: 32 → 25 → 18 → 11 : each $-7$. Next: $11-7=4$.
3. So the next term is T 4.

Trick: Going backwards past A, add 26: $4-5=-1\Rightarrow -1+26=25$ (Y).

Answer: (A) T 4

Concept: Compare corresponding letters of the two clusters in each pair and find the jump.

1. FE−IH : F→I (+3), E→H (+3).
2. LK−ON : L→O (+3), K→N (+3). CB−FE : C→F (+3), B→E (+3).
3. KJ−MO : K→M (+2), J→O (+5) – breaks the pattern.

Trick: In three pairs the second cluster is simply "each letter + 3"; the odd pair has unequal jumps.

Answer: (D) KJ−MO

Concept: Vowel → +2 in the alphabet, consonant → −2. Convert every letter of REFERTILIZES and count.

1. R→P, E→G, F→D, E→G, R→P, T→R, I→K, L→J, I→K, Z→X, E→G, S→Q.
2. New group: P G D G P R K J K X G Q.
3. Counts: G = 3, P = 2, K = 2, R = 1, X = 1. Among the options only K appears exactly twice.

Trick: Convert letter by letter in order and tally only the letters given in the options.

Answer: (B) K

Concept: Find the rule for letters and number separately from both given pairs, then apply it.

1. KY 1 → MU 17: K→M (+2), Y→U (−4), $1\to 17$ (+16).
2. FI 8 → HE 24: F→H (+2), I→E (−4), $8\to 24$ (+16). Rule confirmed.
3. RF 3: R+2 = T, F−4 = B, $3+16=19$ → TB 19.

Trick: Verify the rule on both examples before applying; here (+2, −4, +16).

Answer: (C) TB 19