RRB ALP CBT 1 2025 PYQs

Concept: Adding the three pair-rates counts each person twice.

1. $2(A+B+C)=\frac{1}{10}+\frac{1}{12}+\frac{1}{15}=\frac{6+5+4}{60}=\frac{1}{4}$
2. $A+B+C=\frac{1}{8}$ → 8 days.

Short trick: LCM method: work = 60 units; pair rates 6, 5, 4 → sum 15 → each-person total 7.5/day → 60/7.5 = 8 days.

Answer: (A) 8 days

Concept: Subtract the reductions from the total, then share in the ratio.

1. Remaining = 6820 − (470 + 110 + 540) = 6820 − 1120 = 5700.
2. 1 part = 5700 ÷ 20 = 285 → Tina's reduced share = 9 × 285 = 2565.
3. Original = 2565 + 110 = ₹2,675.

Short trick: Tina − 110 must be a multiple of 9 × 285 — check: 2675 − 110 = 2565 = 9 × 285 ✔.

Answer: (A) ₹2,675

Concept: Form one equation: new SP = 1.425 × new CP.

1. Let CP = x → SP = 1.64x.
2. $1.64x-65=1.425(x+60)$ ⇒ $0.215x=65+85.5=150.5$
3. $x=700$

Short trick: Option check: x = 700 → SP = 1148 → 1083 vs 1.425 × 760 = 1083 ✔.

Answer: (D) 700

Concept: For constant area, % decrease = $\frac{r}{100+r}\times 100$.

1. $\frac{40}{140}\times 100=\frac{200}{7}=28.57\mathrm{\%}$

Short trick: 40% = 2/5 increase → decrease = 2/(5 + 2) = 2/7 = 28.57%.

Answer: (D) 28.57%

Concept: After 2 years the sum of ages becomes 40 + 4 = 44.

1. After 2 years, K : P = 3 : 1 → K = $\frac{3}{4}\times 44=33$.
2. Present age of Kamlesh = 33 − 2 = 31.

Short trick: Option check: 31 + 9 = 40 and 33 = 3 × 11 ✔.

Answer: (C) 31

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

1. Fix L at the top. O is immediate right of L (anticlockwise), so S (other neighbour of L) is immediate left.
2. K is immediate right of O.
3. V is second to the left of L (two seats clockwise); J takes the last seat.
4. Clockwise from L: L, S, V, J, K, O. Immediate right of J (anticlockwise) is V.

Short trick: Place the 'immediate' clues first (O, K), then the 'second to' clue — the empty seat goes to J automatically.

Answer: (C) V

Concept: Check each symbol's immediate neighbours: pattern Number – Symbol – Number.

1. 5 % 3 ✔
2. 3 + 8 ✔
3. 8 \ / , / % , 6 ... 1 + # , # % 3 – all fail.
4. Count = 2.

Short trick: Scan only symbols whose left neighbour is a digit; then just check the right neighbour.

Answer: (A) Two

Concept: Fix the definite positions first, then place the pairs.

1. Only L above Z → L = 1, Z = 2 (from top).
2. Only two above Q → Q = 3, so A = 4.
3. O–W pair fits in 5-6 or 6-7; O is not 3rd from bottom (position 5) → O = 6, W = 7, P = 5.
4. Between P (5) and L (1): Z, Q, A = 3 boxes.

Short trick: Boxes between positions p and q = |p − q| − 1 = 5 − 1 − 1 = 3.

Answer: (B) 3

Concept: Look for the pattern Letter – Symbol – Symbol.

1. O * © ✔
2. V € £ ✔
3. D * # ✔
4. I % X, X © V, B $ D, L © A – fail.
5. Count = 3.

Short trick: Find every pair of adjacent symbols; then check if the first one has a letter before it.

Answer: (A) 3

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

1. S - M: S is the son of M.
2. M @ H: M is the sister of H.
3. H # Y: H is the daughter of Y; Y % U: Y is the father of U.
4. So M, H and U are siblings; S is the son of M, i.e. U's sister's son.

Short trick: M is female (sister) and S is her son → 'sister's son' is the only option with a female link.

Answer: (B) Sister's son

Concept: Pattern required: Letter – Symbol – Symbol.

1. H £ # ✔
2. B $ + ✔
3. U * % ✔
4. E @ H, K > X, M $ Z – fail.
5. Count = 3.

Short trick: Only symbol pairs (£#, $+, *%) can qualify — check the letter before each pair.

Answer: (A) Three

Concept: Check differences; they form an AP.

1. Differences: 3, 10, 17, 24 (each +7).
2. Next difference = 31 → 136 + 31 = 167.

Short trick: Unit digit check: 136 + 31 ends in 7 → only 167 fits.

Answer: (B) 167

Concept: Substitute, then apply BODMAS.

1. 104 ÷ 4 − 3 × 9 + 12
2. = 26 − 27 + 12 = 11

Short trick: Do ÷ and × first: 26 and 27; then 26 − 27 = −1, −1 + 12 = 11.

Answer: (C) 11

Concept: Code = (number of letters)².

1. PEN: 3 letters → 3² = 9.
2. WOLF: 4 letters → 4² = 16.
3. ISLAND: 6 letters → 6² = 36.

Short trick: When codes are perfect squares (9, 16), test the letter count first.

Answer: (C) 36

Concept: Facing the centre: left = clockwise.

1. F at top; B third to the left of F → B opposite F.
2. C second to the left of B.
3. Remaining seats (F's left side) hold A, D and E; D immediately left of A → order clockwise from F: F, A, D, B, E, C.
4. From E, moving left (clockwise): C, F, A → A is third to the left of E.

Short trick: In 6 seats, 'third to the left' = 'third to the right' = opposite seat. A and E are opposite, so only 'Third to the left' fits.

Answer: (B) Third to the left

Concept: Find the shift of each letter position.

1. GNP → KRU: +4, +4, +5.
2. COK → GSP: +4, +4, +5.
3. VAM → ZER: V+4 = Z, A+4 = E, M+5 = R ✔ (others give +4, +5/+6 patterns).

Short trick: Check only the 2nd letter (+4): VIQ→N (+5), HDL→J (+6), HNR→T (+6) fail at once.

Answer: (D) VAM : ZER

Concept: Pattern: Number – Symbol – Number.

1. 3 ^ 5 ✔, 5 # 6 ✔, 6 £ 1 ✔, 8 % 7 ✔, 7 * 4 ✔.
2. 1 $ @, 4 ≺ Ω – fail.
3. Count = 5, i.e. more than three.

Short trick: Stop counting once you reach 4 — the answer is already 'More than three'.

Answer: (D) More than three

Concept: Identify the chain of operations from the given sets.

1. 4 ×2 = 8, +2 = 10, +2 = 12; 3 ×2 = 6, +2 = 8, +2 = 10.
2. Rule: ×2, +2, +2.
3. 1 ×2 = 2, +2 = 4, +2 = 6 ✔ (others end with +4).

Short trick: All options satisfy ×2 and the first +2; just check the last step (+2).

Answer: (D) 1 - 2 - 4 - 6

Concept: Replace the signs, then apply BODMAS.

1. 188 + 11 × 16 − 234 ÷ 18
2. = 188 + 176 − 13 = 351

Short trick: Unit digit: 8 + 6 − 3 = 11 → ends in 1 → only 351.

Answer: (A) 351

Concept: Compare letter gaps within and across each pair.

1. KG - HD: K−4 = G, H−4 = D, K−3 = H.
2. PL - MI and HD - EA follow the same (−4, −4, −3).
3. JF - HT: J−4 = F but H−4 ≠ T → odd one.

Short trick: Second cluster's letters should both be 3 behind the first's: T is ahead of F, so JF-HT is out instantly.

Answer: (D) JF - HT