Series Codes and Operations Module Review

How to use this review

Attempt these 20 original questions after the four Series Codes and Operations lessons. The review is free and untimed: +1 for correct, 0 for incorrect or unattempted, no negative marking and no pass cutoff. These learning settings are not the RRB exam's marking scheme or a full CBT simulation. Five letter, six number, five coding and four operator questions sample this module's selected skills; those counts are not predicted exam weightage.

Every finite-series question explicitly states its allowed rule family. Keep the stated English alphabet, cycle permission, word/token order, code domain and operator convention. Literal English code-bearing words and tokens are unchanged in Hindi. For new-symbol arithmetic, replace original signs once before calculating; for defined two-input operations, retain the exact left/right inputs and brackets.

Write a short working row, mapping or translated expression before choosing an option. In an insufficient-information question, the explanation must show different consistent possibilities, not simply say the pattern was difficult. After checking, revisit the skill behind each error. One correct response on a skill is not proof of mastery, and this short review does not establish readiness for the entire exam.

Questions

1. The English-letter sequence W, ?, O, K, G uses one constant backward shift, with no wraparound. Find the missing letter.

A. T B. S C. R D. Q

2. Consecutive terms have the same signed difference: ?, −3, 4, 11, 18. Find the first term.

A. 10 B. −4 C. −17 D. −10

3. Every English letter uses one fixed forward shift of 1–25 positions, returning to A after Z. WAVE→ZDYH and ZONE→CRQH. Encode FUZZ.

A. IXCC B. HWBB C. IXCF D. IWCC

4. Here # means + and @ means ×. Replace each original sign once, retain brackets, and evaluate the bracket before multiplication. Find (4 # 5) @ 2.

A. 14 B. 11 C. 18 D. 40

5. In B, D, G, ?, P, V, each forward English-alphabet gap is one larger than the previous gap. No wraparound is used. Fill the internal blank.

A. J B. L C. K D. I

6. Each term is multiplied by the same positive factor to obtain the next: ?, 96, 48, 24, 12. What is the first term?

A. 192 B. 96 C. 48 D. 384

7. Encoding uses the same forward English-alphabet shift of 1–25 positions for every letter, with Z→A wraparound: YARD→ACTF and DOVE→FQXG. Decode DQQM.

A. DQQM B. FSSO C. KOOB D. BOOK

8. In the original expression, + means ÷, ÷ means −, and − means +. Replace original signs once, simultaneously, then use ordinary precedence (×/÷ before +/−, left to right within each pair). Which rewrite and value correctly represent 18 + 3 − 4 ÷ 2?

A. 18 − 3 + 4 − 2 = 17 B. 18 ÷ 3 + 4 − 2 = 8 C. 18 ÷ 3 − 4 + 2 = 4 D. 18 + 3 − 4 ÷ 2 = 19

9. Odd and even term positions each form a separate constant-shift English-letter sequence, without wraparound: ?, B, S, F, O, J, K, N, G, R. Find the first term.

A. W B. V C. X D. U

10. The gaps between consecutive numbers increase by a fixed amount: 3, 9, 18, ?, 45, 63. Find the missing term.

A. 27 B. 33 C. 30 D. 36

11. Encode a four-letter word by reversing it, then shifting new positions 1 and 3 forward by one and new positions 2 and 4 backward by one, with no wraparound. GLOW→XNMF and TURN→OQVS. For SHED, which intermediate word must be placed under the new position labels immediately BEFORE the shifts?

A. SHED B. DEHS C. EDIR D. DEIR

12. Here @ means − and # means ×. Replace original signs once and preserve the brackets; then do the grouped multiplication before subtraction. For 16 @ (3 # 2), a learner writes: line 1, 16−(3×2); line 2, (16−3)×2; line 3, 13×2; line 4, 26. Which is the FIRST line that is no longer equal to the original under the stated rule?

A. Line 1 B. Line 3 C. Line 4 D. Line 2

13. In VC, TF, RI, ?, NO, each first-letter column has one fixed backward English-alphabet shift and each second-letter column one fixed forward shift. No wraparound occurs. Find the missing block.

A. OL B. PM C. LP D. PL

14. Odd term positions and even term positions each form their own constant-difference sequence: 11, 70, 17, ?, 23, 56, 29, 49. Find the term at position 4.

A. 64 B. 63 C. 35 D. 62

15. Among the letters used in this puzzle, each distinct letter has its own distinct single-digit code from 0–9, used consistently; letter order is preserved. ROAD→6248 and DORA→8264. Which option is justified for the code of OARX?

A. 2460 B. 2461 C. Cannot be determined uniquely D. 2468

16. Define a ★ b = a+2b, retaining left/right input order. Let A=(2 ★ 3) ★ 4 and B=2 ★ (3 ★ 4), with exactly these brackets. Find A−B.

A. −8 B. 8 C. 0 D. 40

17. Move five English-alphabet positions forward each time, returning to A after Z. In ?, D, I, N, S, which letter precedes D?

A. Z B. Y C. X D. W

18. From the third term onward, each term equals the sum of the previous two: 3, 8, 11, ?, 30, 49. Find the missing term.

A. 19 B. 16 C. 22 D. 30

19. Every distinct word has one distinct code token used consistently; token order within each line may change. “tall trees sway”→“be ru fo”, “trees sway gently”→“fo da ru”, and “tall walls stand”→“be zi ka”. What is justified about the code for trees?

A. ru B. fo C. be D. Cannot be determined uniquely

20. Each term is two more than the square of consecutive increasing positive integers: ?, 27, 38, 51, 66. Find the first term.

A. 16 B. 14 C. 18 D. 27

Explained key

1. B — S

O(15),K(11),G(7) establish a −4 step. Moving back one step from W(23) gives S(19), and S→O is also −4. The full positions are 23,19,15,11,7. T gives −3 then −5 around the blank; R gives −5 then −3; Q gives −6 then −2. Each alternative breaks the fixed step.

2. D — −10

The given gaps are 4−(−3)=7, 11−4=7 and 18−11=7. The preceding term is −3−7=−10. Then −10+7=−3 checks the target. Positive 10 reverses the required sign and does not connect to −3 by +7. −4 reaches −3 by only +1. −17 reaches −3 by +14 and skips a step.

3. A — IXCC

Both examples consistently use +3: W→Z, A→D, V→Y, E→H; Z→C, O→R, N→Q, E→H. FUZZ becomes F→I, U→X, Z→C, Z→C, so IXCC is correct. HWBB uses +2. IXCF assigns different results to the two identical Z inputs. IWCC shifts U by only +2 while the other positions use +3.

4. C — 18

The translation is (4+5)×2=9×2=18. The value 14 comes from dropping the bracket and doing 4+5×2. The value 11 treats both operators as addition, giving 4+5+2. The value 40 treats both as multiplication, giving 4×5×2. Those expressions do not preserve the supplied legend and grouping.

5. C — K

Positions 2,4,7 start with gaps +2,+3. Continue +4,+5,+6 to obtain 11,16,22, so the blank is K(11). This also checks K→P and P→V. J(10) gives gaps +3,+6 around the blank; L(12) gives +5,+4; I(9) gives +2,+7. None follows the required increasing-gap row.

6. A — 192

48÷96=24÷48=12÷24=1/2. The predecessor must therefore be 96÷(1/2)=192; checking forward, 192×1/2=96. The option 96 repeats the second term and would make the first multiplier 1. The option 48 moves forward rather than backward from 96. The option 384 goes back two halving steps, skipping 192.

7. D — BOOK

Every supplied pair shifts +2: Y→A, A→C, R→T, D→F; D→F, O→Q, V→X, E→G. Undo by −2: D→B, Q→O, Q→O, M→K, giving BOOK. Encoding BOOK returns DQQM. DQQM makes no change. FSSO applies +2 again. KOOB reverses the decoded word without permission.

8. B — 18 ÷ 3 + 4 − 2 = 8

The original signs +,−,÷ become ÷,+,− in one pass. Thus 18÷3+4−2=6+4−2=8. The rewrite with 17 changes the first inserted ÷ again into −, which is forbidden cascading. The rewrite with 4 uses the wrong meanings for the last two original signs. The rewrite with 19 keeps the original signs unchanged. Each displayed arithmetic total is correct for its own expression, but only the expression giving 8 is the authorized translation.

9. A — W

Odd positions contain ?,S(19),O(15),K(11),G(7), so the step is −4 and the first letter is W(23). Even positions B(2),F(6),J(10),N(14),R(18) all use +4. V→S would be −3, X→S would be −5 and U→S would be −2; none matches the odd-row step. The target belongs to the odd row because its term position is 1.

10. C — 30

The first gaps are 6 and 9, so the constant increase is 3. The complete gaps must be 6,9,12,15,18, giving 3,9,18,30,45,63. The option 27 repeats the gap 9 and then jumps by 18. The option 33 uses 15 before 12, reversing the needed two gaps. The option 36 jumps by 18 then 9. Only 30 checks both sides and the final known gap.

11. B — DEHS

The requested stage is reversal only: SHED→DEHS, with D,E,H,S under positions 1,2,3,4. The later shift would produce EDIR, but that is not the requested intermediate stage. SHED omits reversal. DEIR shifts only the last two reversed positions, mixing stages. The examples check as GLOW→WOLG→XNMF and TURN→NRUT→OQVS.

12. D — Line 2

The correct translation is line 1: 16−(3×2)=16−6=10. Line 2 changes the grouping to (16−3)×2=26, so it is the first invalid line. Line 3 is a correct simplification of the already wrong line 2, and line 4 continues that same earlier error; neither is the first failure. Line 1 itself correctly follows the legend and original brackets.

13. D — PL

The first column is V(22),T(20),R(18),P(16),N(14), with −2. The second is C(3),F(6),I(9),L(12),O(15), with +3. Thus PL fits both neighbours. OL gives a −3 then −1 first-column gap. PM gives +4 then +2 in the second column. LP reverses the required two positions.

14. B — 63

The odd row is 11,17,23,29 with +6. In the even row 70,?,56,49, the final step is −7; restoring the same step gives 70,63,56,49. The option 64 subtracts 6 from 70 and then needs −8 to reach 56. The option 62 needs −8 then −6. The option 35 extends the odd row past 29 and answers a different position.

15. C — Cannot be determined uniquely

The examples fix R=6, O=2, A=4 and D=8, so OARX begins 246. X is unobserved and may consistently take any unused digit: 0,1,3,5,7 or 9. For example, X=0 yields 2460 and X=1 yields 2461; both preserve all clues. Neither 2460 nor 2461 is forced. The option 2468 is invalid because it would give X and D the same digit, violating the stated distinct-code condition.

16. A — −8

For A, 2★3=2+6=8 and 8★4=8+8=16. For B, 3★4=3+8=11 and 2★11=2+22=24. Hence A−B=16−24=−8. Positive 8 reverses the subtraction. Zero assumes the two bracketings have the same value, which the calculations disprove. The option 40 adds 16 and 24 instead of subtracting.

17. B — Y

Going five moves forward from Y gives Z,A,B,C,D. Equivalently, move backward five from D: C,B,A,Z,Y. The full positions are 25,4,9,14,19 under the stated cycle. Z reaches D in four forward moves, X in six and W in seven; they do not meet the fixed five-step rule.

18. A — 19

The first check is 3+8=11. Then 8+11=19, 11+19=30 and 19+30=49, so both later clues confirm 19. The option 16 doubles 8 instead of adding 8 and 11. The option 22 doubles 11. The option 30 copies the following term and would make the next sum 11+30=41, not 30.

19. D — Cannot be determined uniquely

tall=be follows from lines 1 and 3. The remaining shared words trees and sway can match ru and fo in either order. One consistent full map is tall=be, trees=ru, sway=fo, gently=da, walls=zi, stand=ka. Swapping only trees and sway gives another valid map. Thus ru and fo are both possible, neither is forced. be is already tall and cannot also be trees under the one-to-one condition.

20. C — 18

Subtract 2 from the known terms: 25,36,49,64 are 5²,6²,7²,8². The preceding positive base is 4, so the missing term is 4²+2=18. This restores 18,27,38,51,66 under the stated rule. The option 16 omits +2. The option 14 subtracts 2 instead. The option 27 repeats the next term instead of using the preceding base.

Where to revisit

• Alphabet positions and letter series: questions 1,5,9,13,17. Rebuild the relevant position row and check every column/subsequence
• Number series and missing terms: questions 2,6,10,14,18,20. Recheck direction, term position and both sides of a blank
• Coding and decoding: questions 3,7,11,15,19. Record all supplied mappings and distinguish a fixed target from several valid possibilities
• Mathematical operations with new symbols: questions 4,8,12,16. Separate once-only translation from arithmetic, and keep brackets and input order

Source and scope

RRB CEN 09/2025, §14.1, printed p.28, names alphabetical and number series, coding and decoding, and mathematical operations. The topic list is illustrative, not exhaustive. NCERT Ganita Prakash, Chapter 1, §§1.2–1.4, supports the basic pattern ideas. University of Cambridge NRICH supports the fixed-shift and explicit-cycle coding idea. We consistently use A=1 through Z=26. OpenStax Prealgebra 2e, §2.1 states the grouping and equal-priority arithmetic convention used here. All explanations, examples and questions here are original teaching material. The selected subtypes and question counts are learning choices, not official topic weightage. These are not previous-year questions or a full reasoning course.

Analogy

Use this review like checking four different instruction cards: where to move in an alphabet, how to generate a numerical term, how to transform a code, and how to interpret an operation. The first job is choosing the card actually named by the question. A correct result should leave a short trace that another learner can follow.

Quick reference

• 20 original, untimed, free learning questions; no pass cutoff
• Explicit rule family or definition controls every item
• Check every term/example and preserve raw code-bearing data
• An underdetermined target needs two consistent alternatives
• Translate original operators once; then calculate
• Preserve all brackets and left/right inputs
• Review each error by skill; do not infer full-exam readiness from this score