RRB NTPC Graduate CBT 1 2025 PYQs

Concept: Decode each symbol from left to right and build the family tree.

1. W % A: W is the mother of A.
2. A $ C: A is the father of C, so C is a son of A (C is male because C + J means C is J's brother).
3. C + J: C is the brother of J. So C is the son of W's son → W's grandson (son's son).

Trick: Two downward steps from W (mother → father → child) always mean grandchild; "brother" fixes C as male, so son's son.

Answer: (D) Son's son

Concept: Find one operation that converts the first number into the second in both given pairs.

1. $158-107=51$ and $173-122=51$; the rule is "add 51".
2. $154+51=205$.

Trick: When both pairs have the same difference, the rule is simply addition; no multiplication check needed.

Answer: (A) 205

Concept: Check each statement alone, then together, for whether the tallest person is fixed.

1. Statement I: Prateek > Mohit > Arjun, Kunal. Saurav's position is unknown, so not sufficient alone.
2. Statement II: Prateek > Saurav > Mohit. Arjun and Kunal are unknown, so not sufficient alone.
3. Together: Prateek > Saurav > Mohit > Arjun, Kunal → Prateek is tallest.

Trick: In "who is tallest" questions, a statement fails if even one person is missing from it.

Answer: (C) Both statements I and II together are sufficient

Concept: Remove the dropped elements, rewrite the series, then count from the required end.

1. Series without letters: $ % & # @ ! + ? > (9 symbols).
2. From the right: >(1), ?(2), +(3), !(4), @(5), #(6), &(7).

Trick: 7th from right of 9 items = 3rd from left → &.

Answer: (C) &

Concept: The same letter-shift must map # to JOP and YDE to %.

1. Try option C: OTU → JOP: O$-5$=J, T$-5$=O, U$-5$=P. Rule: each letter $-5$.
2. Apply to YDE: Y$-5$=T, D$-5$=Y, E$-5$=Z → TYZ, which matches % in option C.
3. Other options break the uniform shift (e.g. option A: L$+4$=P, not $-5$).

Trick: Work backwards from the known cluster: JOP $+5$ = OTU; only option C starts with OTU.

Answer: (C) # = OTU, % = TYZ

Concept: Find the uniform shift that turns # into PZH and apply it to MWE.

1. Option A: FPX → PZH: F(6)$+10$=P(16), P(16)$+10$=Z(26), X(24)$+10$=34→34$-$26=8=H. Rule: $+10$.
2. MWE $+10$: M(13)→W(23), W(23)→33$-$26=7=G, E(5)→O(15) → WGO.
3. Option B uses Y as first letter (Y$+10$=I, not P); C and D change WGO.

Trick: Compute % first (MWE$+10$=WGO); then only A and B remain, and F$+10$=P picks A.

Answer: (A) # = FPX, % = WGO

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

1. OMIT → 4295 and TIME → 2458 share the letters T, I, M and the digits 4, 2, 5.
2. In OMIT the extra letter O takes the extra digit 9; in TIME the extra letter E takes the extra digit 8.

Trick: Cancel common letters and common digits; what remains in TIME is E ↔ 8.

Answer: (B) 8

Concept: Test "multiply then add/subtract" when the second number is roughly a fixed multiple of the first.

1. $51\times 5=255$, $255-2=253$ ✓; $38\times 5=190$, $190-2=188$ ✓. Rule: $\times 5-2$.
2. $32\times 5-2=160-2=158$.

Trick: $253÷51\approx 5$ and $188÷38\approx 5$ – the ratio tells you the multiplier at once.

Answer: (B) 158

Concept: Draw the sets: pillows inside towels, towels inside cushions, beds overlapping pillows.

1. Conclusion I "All cushions are pillows": cushions is the biggest set; it can contain things that are not pillows → does not follow.
2. Conclusion II "Some towels are beds": the beds that are pillows are also towels, so some towels are definitely beds → follows.

Trick: Some A are B + All B are C ⇒ Some A are C (and its converse Some C are A). "All C are B" is never a valid reverse of "All B are C".

Answer: (A) Only conclusion (II) follows

Concept: Drop the numbers, rewrite the remaining series and count from the right.

1. Without numbers: E % Y & T K # C * S D $ R @ G (15 elements).
2. From the right: G(1), @(2), R(3), $(4), D(5), S(6), *(7).

Trick: 7th from right of 15 = 9th from left; counting E % Y & T K # C * gives * as well.

Answer: (B) *

Concept: Fix the most restrictive pair first (R and S with three floors between), then use the odd/even clue.

1. Three people between R and S with R below S: (R, S) = (1, 5) or (2, 6).
2. J is immediately above R on an odd floor → R must be even → R = 2, J = 3, S = 6.
3. Remaining floors 1, 4, 5 for T, L, K with T < L < K → T = 1, L = 4, K = 5. Floor 4 → L.

Trick: "Immediately above on an odd floor" forces the lower person onto an even floor; that single clue kills the (1, 5) case.

Answer: (C) L

Concept: Relate the middle number to the two outer numbers of each set.

1. $(34,1088,16)$: $34\times 16=544$, $544\times 2=1088$. $(21,420,10)$: $21\times 10\times 2=420$. Rule: middle $=$ first $\times$ third $\times 2$.
2. Option A: $28\times 4\times 2=224$ ✓.
3. B: $8\times 17\times 2=272\ne 279$; C: $4\times 9\times 2=72\ne 79$; D: $23\times 11\times 2=506\ne 513$.

Trick: Wrong options are all exactly 7 more than $2\times$product – spot the +7 and reject them.

Answer: (A) (28, 224, 4)

Concept: Start from the definite clue (T on floor 1) and place the rest.

1. T = 1. Two people between T and Y → Y = 4.
2. Only S lives above U → U = 5, S = 6.
3. Z is on an even floor; only floor 2 is left among even floors → Z = 2, so X = 3.
4. Between X (3) and T (1) there is one person, Z.

Trick: "Only S lives above U" always means U is second from top and S is on top.

Answer: (A) One

Concept: When everyone faces the centre, "right" means anticlockwise and "left" means clockwise.

1. Place E; F is immediate right (anticlockwise) of E. Three between D and E from the right of E: E → F → B/A/? … counting anticlockwise, D is the 4th seat from E.
2. Two between G and F from the right of G: G is 3 seats clockwise from F, i.e. G sits two seats clockwise of E.
3. Remaining three seats between F and D (anticlockwise) take B, A, with A immediate right of B; the seat between E and G goes to C. Order clockwise: E, C, G, D, A, B, F.
4. From the left of G (clockwise): D, A, then B → 2 people.

Trick: Fix the direction rule first (right = anticlockwise) and place the pair with the most clues (E, F, D) before anything else.

Answer: (B) 2

Concept: Draw lakes inside roads, lakes overlapping oceans, and planets inside oceans.

1. Conclusion I "Some oceans are roads": the lakes that are oceans are also roads (all lakes are roads) → follows.
2. Conclusion II "Some planets are lakes": planets and lakes are both inside oceans but need not touch each other → does not follow.

Trick: Some A are B + All A are C ⇒ Some B are C. Two subsets of the same set (planets, lakes in oceans) give no definite relation.

Answer: (A) Only conclusion (I) follows

Concept: Fix the box whose position is certain (U at the top), then place the others by counting gaps.

1. No box above U → U is at position 1 (top). Three boxes between U and P → P is 5th.
2. T is immediately above P → T is 4th. One box between R and T → R is 2nd or 6th.
3. Four boxes between R and V → R = 2nd, V = 7th (R = 6th would need V at position 1, which is U).
4. Q is above S, so Q = 3rd and S = 6th. Final stack (top to bottom): U, R, Q, T, P, S, V.
5. Between S (6th) and U (1st) lie R, Q, T, P → 4 boxes.

Trick: "Only n boxes between X and Y" means the positions differ by n + 1; start from the box with a fixed end position.

Answer: (C) 4

Concept: Count positions from the top (1 = top, 7 = bottom) and fill the fixed boxes first.

1. Two boxes below B → B is 5th. One box above G → G is 2nd.
2. One box between G and A → A is 4th (position 0 does not exist).
3. Vacant places: 1, 3, 6, 7. C immediately above F needs two consecutive places → C = 6th, F = 7th.
4. D and E take 1 and 3; E is below D → D = 1st, E = 3rd. Stack: D, G, E, A, B, C, F.
5. Below E are A, B, C, F → 4 boxes.

Trick: After fixing the definite boxes, look for the only pair of adjacent empty slots – that is where the "immediately above" pair goes.

Answer: (C) 4

Concept: Draw the Venn diagram: "All heaters are fans" puts the Heater circle inside Fans; "Some heaters are lights" makes Lights overlap Heaters; "No heater is a bulb" keeps Bulbs apart from Heaters.

1. Conclusion I: the heaters that are lights lie inside Fans too, so some fans are definitely lights → follows.
2. Conclusion II: "No heater is a bulb" can always be reversed to "No bulb is a heater" → follows.

Yaad rakhein: All A are B + Some A are C ⇒ Some B are C (definite). "No A is B" ⇔ "No B is A".

Answer: (C) Both conclusions (I) and (II) follow

Concept: For people facing the centre, "left" is clockwise and "right" is anticlockwise. Start with the pair whose relative position is fixed.

1. E is immediately right of F, and F is third to the left of D → going clockwise: D … F with two seats between, E just anticlockwise of F.
2. E is third to the right of A → A is third to the left (clockwise) of E; this places A two seats clockwise after F.
3. Only one person between G and C counting from the left of C → G is 2nd clockwise from C. The only arrangement that fits all this (clockwise): A, G, D, B, E, F, C.
4. Check: B's neighbours are D and E, not F ✓. Third to the right (anticlockwise) of G: A, C, F → F.

Trick: Facing centre → right = anticlockwise. Put the two "immediate" clues on the circle first, then test the "third to…" clues.

Answer: (A) F

Concept: Both unknowns must follow one rule; test each option's # against ZVT and apply the same rule to JFD.

1. Option (D): # = WSQ. W + 3 = Z, S + 3 = V, Q + 3 = T → each letter moves +3 to give ZVT ✓.
2. Apply +3 to JFD: J + 3 = M, F + 3 = I, D + 3 = G → MIG, which matches % in option (D).
3. Other options fail: e.g. (A) WTR → +3, +2, +2 (not uniform); (B) XSR → +2, +3, +2.

Trick: Write place values (W=23, S=19, Q=17 → Z=26, V=22, T=20); a constant gap confirms the rule instantly.

Answer: (D) # = WSQ, % = MIG