RRB NTPC Graduate CBT 1 2024 PYQs

Concept: Total = rank from top + rank from bottom − 1 (Jon is counted twice).

1. Rank from top $=24$, from bottom $=16$.
2. Total $=24+16-1=39$.

Trick: Always subtract 1 because the person is in both counts.

Answer: (A) 39

Concept: Check the gap inside each cluster and between clusters.

1. PS–NQ: $16\to 19$ ($+3$), $14\to 17$ ($+3$), first letters $-2$.
2. EH–CF and KN–IL: the same pattern ($+3$, $+3$, $-2$).
3. GJ–FK: $7\to 10$ ($+3$) but $6\to 11$ ($+5$) ✗.

Trick: Find one rule from two pairs, then test the rest.

Answer: (D) GJ - FK

Concept: Check each symbol: the element just before it and just after it must both be numbers.

1. Series: € 3 © 3 $ % % 5 5 £ € 4 * € 3 7 5 & 8 2 5 *
2. 3 © 3 → number-symbol-number ✔
3. 5 & 8 → number-symbol-number ✔
4. All others fail ($ is followed by %, £ by €, * by €, etc.).

Trick: Scan only for the pattern "number – symbol – number"; ignore everything else.

Answer: (A) 2

Concept: Facing the centre, "right" means anticlockwise and "left" means clockwise.

1. Fix L; K is 4th to the right of L.
2. N sits between R and I. Only one placement works: L, R, N, I, K, J, M going anticlockwise (J is 3rd to the left of R; I is 3rd to the left of M).
3. From N, move to the left (clockwise): R, L, M, J, then K.
4. People between them: R, L, M, J = 4.

Trick: Place the fixed pair (L–K) first, then test the "N between R and I" block in the empty seats.

Answer: (B) Four

Concept: Track the position on a coordinate grid, taking B as origin.

1. South 4 → $(0,-4)$; left turn = east 7 → $(7,-4)$.
2. Right turn = south 7 → $(7,-11)$; left = east 9 → $(16,-11)$.
3. Left = north 11 → A $=(16,0)$.
4. A is 16 km east of B, so he must drive 16 km towards the west.

Trick: Net N–S: $-4-7+11=0$; net E–W: $7+9=16$ east → return 16 km west.

Answer: (B) 16 km towards the west

Concept: A person who is a plumber, a hairdresser and a judge lies in the common region of all three circles.

1. Plumber ∩ Hairdresser only = 13 (not judges).
2. Plumber ∩ Judge only = 12; Hairdresser ∩ Judge only = 11.
3. Region common to all three = 19.

Trick: "Also A and B" with three sets = the central region only.

Answer: (D) 19

Concept: Track each letter position separately.

1. 1st letters: G, I, K, M, O ($+2$ each).
2. 2nd letters: E, G, I, K, M ($+2$).
3. 3rd letters: B, D, F, H, J ($+2$).
4. Next term = OMJ.

Trick: Every letter of each term is +2 from the previous term: MKH + 2 → OMJ.

Answer: (A) OMJ

Concept: Apply the rule digit by digit, then read the required positions.

1. 2614759: even digits +2 (2→4, 6→8, 4→6); odd digits −1 (1→0, 7→6, 5→4, 9→8).
2. New number: 4 8 0 6 6 4 8.
3. 6th from left = 4; 1st from right = 8.
4. Sum $=4+8=12$

Trick: Transform only the two digits you need: 6th digit 5 → 4, last digit 9 → 8.

Answer: (C) $12$

Concept: Swap the signs first, then solve using BODMAS.

1. $127÷5+707\times 7-38$ becomes $127\times 5-707÷7+38$
2. $=635-101+38$
3. $=572$

Trick: Rewrite the whole expression with new signs before calculating — most errors come from swapping only some signs.

Answer: (A) $572$

Concept: Number floors 1 (bottom) to 6 (top) and place definite clues first.

1. Only one person above D → D on floor 5; X above D → X on 6.
2. Only A below P → A on 1, P on 2.
3. E on an even floor: only 4 is left → E = 4, so N = 3.
4. Below N: P and A = 2 people.

Trick: "Only one above" and "only one below" clues fix the top and bottom immediately.

Answer: (A) Two

Concept: Check every condition for the given group, apply only the one that fits.

1. Codes: E = @, P = 9, T = $, F = 7, U = %.
2. Cond. 1: last letter U is a vowel ✗. Cond. 2 and 4: 2nd letter P is a consonant ✗.
3. Cond. 3: 3rd T and 4th F are consonants ✔ → both get the code of T = $.
4. Code: @ 9 $ $ %

Trick: Check the vowel/consonant pattern of positions 1, 2, 3, 4 and last at once: E P T F U = V C C C V.

Answer: (C) @ 9 $ $ %

Concept: Find the change in each part: 1st letter, 2nd letter and the number.

1. UB 15 → SF 11: U $-2$ = S, B $+4$ = F, $15-4=11$.
2. IO 8 → GS 4: I $-2$ = G, O $+4$ = S, $8-4=4$ ✔.
3. EL 3 → E $-2$ = C, L $+4$ = P, $3-4=-1$ → CP −1.

Trick: Check the number first: $3-4=-1$ already leaves only option (A).

Answer: (A) CP −1

Concept: Look for the pattern number – number – symbol.

1. Series: 9 8 3 1 4 6 # 6 4 6 * % & $ @ 9 1 % 1 % # £
2. 4 6 # ✔
3. 4 6 * ✔
4. 9 1 % ✔ (the next 1 is preceded by %, so ✗)

Trick: Find each symbol and check the two elements just before it.

Answer: (D) 3

Concept: Decode each symbol and draw the family tree, then trace from Q to M.

1. M – N: M is the brother of N.
2. N + O: N is the mother of O.
3. O ÷ P: O is the father of P; P × Q: P is the wife of Q.
4. From Q: wife (P) → her father (O) → his mother (N) → her brother (M).

Trick: Read the chain backwards from Q and add one relation word per link.

Answer: (B) Wife's father's mother's brother

Concept: Look for even number – symbol – odd number.

1. Series: 8 $ 7 & 3 Ω 9 # 1 8 £ 6 @ & % 4 6 @ * 4 5
2. 8 $ 7 → even – symbol – odd ✔
3. Others fail: 7 & 3, 3 Ω 9, 9 # 1 have odd before; 8 £ 6 has even after; the @ after 6 is followed by a symbol.

Trick: First list only symbols preceded by an even number (only $, £, @, @), then check what follows.

Answer: (C) One

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

1. eat healthy food → xy zd uv
2. food is essential → uv aq bc
3. Common word "food" ↔ common code "uv".

Trick: Just find the code appearing in both lines.

Answer: (C) uv

Concept: Test each option: the rule from # to LQT must also take ZEH to %.

1. Option (C): OTW → LQT: O $-3$ = L, T $-3$ = Q, W $-3$ = T.
2. ZEH → WBE: Z $-3$ = W, E $-3$ = B, H $-3$ = E ✔.
3. Other options do not give one consistent rule.

Trick: Check only the first letter: Z $-3$ = W appears only in option (C).

Answer: (C) # = OTW, % = WBE

Concept: Look at the differences between consecutive terms.

1. $201-1=200$, $200-3=197$, $197-5=192$, $192-7=185$
2. Subtracted numbers are consecutive odd numbers; next is 9.
3. $185-9=176$

Trick: Term $=201-(n-1{)}^2$: $201-25=176$.

Answer: (A) $176$

Concept: Find the rule from # to KSM and apply it from OWQ to %.

1. Option (C): LTN → KSM: each letter $-1$ (L→K, T→S, N→M).
2. OWQ → NVP: each letter $-1$ ✔.
3. (A) LTM → KSL, not KSM ✗; (B) and (D) give NUP / NVO, not $-1$ of OWQ ✗.

Trick: # is KSM shifted $+1$ = LTN; % is OWQ shifted $-1$ = NVP.

Answer: (C) # = LTN, % = NVP

Concept: Facing the centre, right = anticlockwise, left = clockwise.

1. Fix Q; R is 3rd to the left of Q. U and P sit on either side of Q.
2. P is 2nd to the left of T: works only with P on Q's right; then T is two seats further right, and U is on Q's left.
3. Two seats remain for S and V; V is not next to R, so S sits beside R. Order (anticlockwise): Q, P, V, T, R, S, U.
4. From T moving left: V, P, Q, then U → 3 people between.

Trick: Try both positions of P beside Q; the wrong one makes T clash with U.

Answer: (A) 3