RRB Technician Grade III 2024, 2025 PYQs

Concept: $1\text{km/h}=\frac{5}{18}\text{m/s}$; $\text{Time}=\frac{\text{Distance}}{\text{Speed}}$.

1. $10\text{km/h}=10\times \frac{5}{18}=\frac{25}{9}\text{m/s}$
2. Time $=200÷\frac{25}{9}=\frac{200\times 9}{25}=72$ s

Trick: 10 km/h = 10,000 m per 3600 s → 200 m takes $\frac{200}{10000}\times 3600=72$ s.

Answer: (D) 72 seconds

Concept: Net discount for $a\%$ and $b\%$ $=a+b-\frac{ab}{100}$.

1. $25+8-\frac{25\times 8}{100}$
2. $=33-2=31\%$

Trick: On ₹100: after 25% → 75; 8% of 75 = 6 → pays 69; discount 31%.

Answer: (D) 31%

Concept: Form an equation linking the present age and the age at marriage.

1. Age at marriage $=x$; present $=x+10=\frac{6}{5}x$
2. $\frac{x}{5}=10\Rightarrow x=50$; present age $=60$
3. Son $=\frac{60}{10}=6$ years

Trick: The extra $\frac{1}{5}$ part equals the 10 years → $x=50$.

Answer: (A) 6 years

Concept: Percent less $=\frac{\text{difference}}{\text{base}}\times 100$, base = the number compared with (first).

1. Third $=100$: first $=70$, second $=63$
2. Difference $=7$
3. $\frac{7}{70}\times 100=10\%$

Trick: Base is the first number (70), not 100 — so not 7%.

Answer: (D) 10%

Concept: Read the data labels on the chart for each company and month.

1. Company X, February $=35$ crore
2. Company Y, April $=55$ crore
3. Difference $=55-35=20$ crore

Trick: Use the printed labels, not the axis height — the Y line sits above X on the scale.

Answer: (B) 20 crore

Concept: Find the work done in one 2-hour cycle, then finish the remainder.

1. C: $\frac{1}{6}$/h, D: $\frac{1}{9}$/h; one cycle $=\frac{1}{6}+\frac{1}{9}=\frac{5}{18}$
2. 3 cycles (6 h) $=\frac{15}{18}$; left $=\frac{3}{18}=\frac{1}{6}$
3. C (7th hour) fills $\frac{1}{6}$ exactly → total 7 h

Trick: LCM 18 units: C = 3, D = 2 per hour; 15 in 6 h, C adds the last 3 in hour 7.

Answer: (D) 7 hours

Concept: Letters common to both words share the common digits; the leftover letter takes the leftover digit.

1. ROAM = 2835, MORE = 4538
2. Common letters R, O, M ↔ common digits 3, 5, 8
3. ROAM leftover: A ↔ 2 (and E ↔ 4)

Trick: Strike out common letters and common digits — what remains pairs up directly.

Answer: (D) 2

Concept: Replace letters by signs (A = +, B = −, C = ×, D = ÷) and apply BODMAS.

1. $21÷3-4\times 5+22$
2. $=7-20+22$
3. $=9$

Trick: Do ÷ and × first: 7 and 20, then $7+22-20$.

Answer: (C) 9

Concept: Check the position change of each letter (A = 1 … Z = 26).

1. OJ-MP: O − 2 = M, J + 6 = P
2. QL-OR: Q − 2 = O, L + 6 = R; UP-SV: U − 2 = S, P + 6 = V
3. SN-QU: S − 2 = Q, but N + 7 = U

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

Answer: (B) SN-QU

Concept: A conclusion follows only if it is true in every possible Venn diagram.

1. All Biscuits are inside both Flours and Curds; Pulses only partly overlap Flours.
2. (I) All flours are curds: part of Flours can lie outside Curds → does not follow.
3. (II) Some biscuits are pulses: Biscuits can lie away from Pulses → does not follow.

Trick: 'Some' with 'All' in different circles gives no definite link — draw one case that breaks each conclusion.

Answer: (A) Neither conclusion (I) nor (II) follows.

Concept: Check each odd digit: the one before must be a perfect square (1, 4, 9) and the one after must be even.

1. Odd digits present: 7, 5, 3, 7, 1, 3, 3
2. 5 has 4 before it but 3 (odd) after; 3 (after 1) has 3 (odd) after
3. All other odd digits are preceded by 6, 5, 8, 7 or 3 (not squares) → count $=0$

Trick: Scan only the odd digits — there are just 7 of them here.

Answer: (B) 0

Concept: Track first and second letters separately through the triad.

1. UQ-VR-WY: first letters U, V, W (+1, +1); second Q, R, Y (+1, +7)
2. RN-SO-TV: R, S, T (+1, +1); N, O, V (+1, +7)
3. OK-PL-QS: O, P, Q (+1, +1); K, L, S (+1, +7) ✓

Trick: The +7 jump in the last second-letter decides it quickly: K → L → S.

Answer: (B) OK-PL-QS

Concept: Find a single operation linking the three numbers in both given sets.

1. $38-12=26$ and $75-50=25$: second − first = third
2. $(14,42,28)$: $42-14=28$ ✓
3. Others: $32-8=24\ne 7$, $40-21=19\ne 15$, $28-9=19\ne 16$

Trick: Equivalently first + third = second: $14+28=42$.

Answer: (D) (14, 42, 28)

Concept: Find the operation common to both sets.

1. $(25,5,20)$: $\frac{25}{5}=5$, $25-5=20$
2. $(40,8,32)$: $\frac{40}{5}=8$, $40-8=32$
3. $(80,16,64)$: $\frac{80}{5}=16$, $80-16=64$ ✓

Trick: Third $=\frac{4}{5}$ of the first: $\frac{4}{5}\times 80=64$.

Answer: (C) (80, 16, 64)

Concept: Pick each even digit and check that both neighbours are odd.

1. 5 2 7 ✓, 7 6 3 ✓, 3 2 9 ✓
2. 3 4 7 ✓ (near the end)
3. No other even digit has odd digits on both sides → count $=4$

Trick: Look for the pattern odd–even–odd in one left-to-right scan.

Answer: (B) 4

Concept: Common letters of the two words get the common digits; the extra letter takes the extra digit.

1. BEAM = 9483, DAME = 9683
2. Common letters E, A, M ↔ common digits 9, 8, 3
3. Leftover in BEAM: B ↔ 4 (and D ↔ 6)

Trick: Remove common letters/digits; the leftovers match directly.

Answer: (A) 4

Concept: Check the gaps between letters using positions (A = 1 … Z = 26).

1. AHO: 1, 8, 15 (+7, +7); HOV: 8, 15, 22 (+7, +7)
2. VCJ: 22 → 3 → 10 (+7, +7 cyclic)
3. EMS: 5, 13, 19 (+8, +6) — odd one

Trick: Remember the cyclic count: after Z comes A again.

Answer: (A) EMS

Concept: Each letter of the cluster moves forward by the same step.

1. H → K → N → Q (+3)
2. L → O → R → U (+3); G → J → M → P (+3)
3. Next: Q + 3 = T, U + 3 = X, P + 3 = S → TXS

Trick: Only option starting with T and ending with S is TXS.

Answer: (A) TXS

Concept: Fix the definite positions first (positions 1–7 from the left, all facing north).

1. Two to the left of B → B at 3; three between B and R → R at 7
2. P second to the right of B → P at 5; two between P and A → A at 2
3. E, G, Q fill 1, 4, 6 in that order (E left of G left of Q) → G at 4
4. Right of G: P, Q, R = 3

Trick: Facing north, your right is the viewer's right — count positions directly.

Answer: (A) Three

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

1. Fix L; F is third to the left of L
2. L's neighbours are D and N; N second to the left of E works only with N on L's right, D on L's left, E two seats right of N
3. G not beside F → G between E and N, M between D and F
4. From M going right: D, L, then N → 2 people between

Trick: Place the person linked to most clues (L) first.

Answer: (D) Two