Coding-Decoding is one of those topics where the question looks unfamiliar the first time but becomes completely mechanical once you recognise the pattern type. The exam is not testing creativity — it is testing whether you can identify the rule quickly and apply it without error.
Here is the core idea: a "code" is nothing but a systematic transformation applied uniformly to every element of a word, number, or sentence. Your job is to reverse-engineer that transformation from the examples given, then apply it to the target word.
Think of it like a lock-and-key cipher. Every question gives you at least two "known locks and keys" (coded examples). From those, you extract the pattern, then use it to open the unknown lock (the question word).
There are four major types you will encounter in SBI Clerk:
Type 1 — Letter Shifting: Each letter in a word moves a fixed number of positions forward or backward in the alphabet. GAME → HBNF means every letter shifted +1. This is by far the most common type.
Type 2 — Positional or Mixed Shifting: The shift value is not constant — it alternates, or follows a pattern like +1, −1, +1, −1, or depends on the letter's position in the word. These appear in higher-difficulty questions.
Type 3 — Number Coding: Letters are replaced by numbers, or numbers carry encoded values. WATER = 12345 type questions. The logic is almost always one-to-one substitution.
Type 4 — Sentence/Word Coding: Individual words in sentences are replaced by code words. Three statements are given, and you have to isolate what each word maps to by finding common elements.
The analogy that works best: imagine each type as a different kind of key cut. Type 1 is a master key (uniform shift). Type 2 is a multi-cut key (different shifts per position). Types 3 and 4 are codebooks (look-up tables). In the exam hall, your first job is to identify which key type you are holding.
This is the Caesar cipher variant. Every letter in the original word is shifted by the same number of positions in the English alphabet (A=1, B=2, ... Z=26, and it wraps around cyclically).
How to find the shift: Pick the first letter of the given word and its coded counterpart. Count the gap. Verify with the second letter. If it matches, you have your shift.
Example: BOARD → ERDUG
For the target word, add 3 to each letter's position. If the result exceeds 26, subtract 26 (wrap-around).
Speed note: Do not count on your fingers. Memorise the alphabet positions for the 10 most-shifted letters. More practically, for a +3 shift: just go three letters ahead in the alphabet. T → W, A → D, B → E, L → O, E → H. That is exactly how TABLE → WDEOH falls out instantly.
Here the shift is not uniform. A question like FRIEND → GSJFOE looks like +1 across the board, but check position 3: I(9) → H(8) = −1. The pattern is +1, +1, −1, +1, +1, +1 (alternating at position 3).
How to handle it: Write down each letter-pair shift explicitly for the first two example words before touching the question word. This takes 30 seconds but saves you from misapplying the rule.
WATER = 12345, POWER = 67845. Build a mapping table from the given codes:
Then substitute directly for the target word. The mapping is always one-to-one — a letter always gets the same number.
Watch out: The question always gives you enough information to derive the code for every letter in the target word. If it seems like you are missing a letter's code, re-read the question — you have missed a mapping somewhere.
This is solved entirely by set intersection. Here is the method step by step:
Step 1 — Pair two sentences that share exactly one word. The code word that appears in both sentences corresponds to that shared word.
Step 2 — Repeat for other pairs until you have identified all words you need.
Step 3 — Assign by elimination for any remaining word.
Example from the PYQs:
pen pencil eraser = na sa tabook pencil pen = ta ma naeraser book paper = sa ma kaS1 ∩ S2 share: pencil, pen → codes na, ta are shared. But we cannot assign individually yet.
S2 ∩ S3 share: book → code ma is shared. So book = ma.
S1 ∩ S3 share: eraser → code sa is shared. So eraser = sa.
Back in S1: pen pencil eraser = na sa ta, eraser=sa, so pen and pencil use na and ta.
Back in S2: book pencil pen = ta ma na, book=ma, so pencil and pen use ta and na.
Now in S3: eraser book paper = sa ma ka, eraser=sa, book=ma, so paper = ka.
This method always works. Do not guess — execute the intersection systematically.
Memorise the position of every 5th letter and interpolate:
This alone cuts your letter-shift calculation time by half.
Memorise just five letters and their positions: E=5, J=10, O=15, T=20, Y=25. For any letter, find the nearest anchor and count from there. For +3 shifts: instead of counting from A each time, shift the anchor itself (E→H, J→M, O→R, T→W, Y→B). Then interpolate for any letter near the anchor. This reduces a 26-letter lookup to a 5-anchor lookup. Standard counting method: 8-10 seconds per letter. EJOTY method: 2-3 seconds per letter. For a 5-letter word, that is a 25-35 second saving per question.
In any letter-shift question, calculate the shift from only the first two letters of the example word. If both give the same shift value, apply it to the target word immediately — do not check all letters. If the first two letters give different shifts, you are looking at a mixed-shift question and must map all positions before answering. This decision point takes 5 seconds and prevents you from misclassifying the question type. Standard approach (checking all letters): 30-40 seconds. This rule: 5 seconds to classify, then proceed accordingly.
In sentence coding, always start with the sentence pair that shares exactly one common word (not two or three). That pair gives you a 1-to-1 mapping immediately without any ambiguity. If no pair shares exactly one word, look for the word that appears in all three statements — its code appears in all three coded statements. Applying this selection rule before writing anything saves 20-30 seconds of confused trial-and-error, reducing the average solve time from 90 seconds to 55 seconds.
For decoding (going from code to original), simply apply the opposite shift. If the encoding rule is +3, the decoding rule is −3. For wrap-around: if subtracting takes you below A(1), add 26. Practise this on the answer options — if an option decodes back to the question word under your identified rule, that is your answer without needing to independently encode the question word. This substitution approach takes 15 seconds versus 25 seconds for full encoding, when the answer options are close to each other.
In letter-shift questions, the first letter of the coded word is deterministic. Calculate only the first letter of your answer and eliminate all options that do not start with that letter. In most SBI Clerk questions, this alone narrows 4 options to 1 or 2. If it narrows to 1, mark and move. If 2 remain, calculate only the letter that differs between them. Full encoding: 5 letters × 3 seconds = 15 seconds. First-letter elimination: 3 seconds, often sufficient to pick the answer.
When you see a Coding-Decoding question, run this decision tree:
Step 1 — Classify the type in under 5 seconds.
Step 2 — For letter shift (Type 1/2): Find the shift from letter 1 and letter 2. Same shift? Apply uniformly. Different shifts? Map all positions first.
Step 3 — For number coding (Type 3): Build a letter-to-number table from all given examples. Substitute directly. Never assume — always verify each letter is covered.
Step 4 — For sentence coding (Type 4): Find the sentence pair with exactly one common word. Assign its code. Repeat. Use elimination for the final word.
Step 5 — Apply first-letter elimination on answer options before committing to full calculation.
Target time: 45-60 seconds per question. If you are past 75 seconds, mark your best guess and move on — Coding-Decoding questions rarely yield to brute force faster than a fresh attempt later.
Why this question: The most common question format in SBI Clerk — uniform +3 shift. This is the baseline you must get right every time.
Solving path: Find shift: B→E = +3. Verify: O→R = +3. Confirmed. Apply to TABLE: T(20)+3=W(23), A(2)+3=D(5), B(2)+3=E(5), L(12)+3=O(15), E(5)+3=H(8). Answer: WDEOH. Time: 25 seconds using EJOTY anchors.
Why this question: The +1 shift variant — simpler, but the most error-prone because aspirants often second-guess themselves and look for a more complex pattern when the answer is straightforward.
Solving path: G→H = +1. A→B = +1. Pattern confirmed as uniform +1. CODE: C→D, O→P, D→E, E→F. Answer: DPEF. First-letter check: C+1=D. Only option starting with D is DPEF. Mark immediately. Time: 15 seconds.
Why this question: Tests whether you can recognise the +1 shift even with a 6-letter word, and whether you handle all letters cleanly.
Solving path: L→M = +1. A→B = +1. Pattern: uniform +1. TABLET: T→U, A→B, B→C, L→M, E→F, T→U. Answer: UBCMFU. Time: 20 seconds.
Why this question: The +3 shift applied to a 6-letter word — tests whether you can maintain accuracy across more letters without slowing down.
Solving path: P→S = +3. R→U = +3. SCHOOL: S(19)+3=V(22), C(3)+3=F(6), H(8)+3=K(11), O(15)+3=R(18), O(15)+3=R(18), L(12)+3=O(15). Answer: VFKRRO. Use EJOTY: O is anchor 15, O+3=R is anchor 15+3=18=R. Clean. Time: 30 seconds.
Why this question: Sentence coding — the type that trips up the most aspirants. Systematic intersection method is non-negotiable here.
Solving path: S2 ∩ S3: book is the only common word; ma is the only common code → book=ma. S1 ∩ S3: eraser only common word; sa only common code → eraser=sa. S3: eraser book paper = sa ma ka. eraser=sa, book=ma, so paper=ka. Answer: ka. Time: 55 seconds.
Miscounting the shift for letters near the wrap-around. When applying a +3 shift to X(24), Y(25), Z(26): X→A(1), Y→B(2), Z→C(3). Aspirants often stop at Z and do not wrap. Always check if the target letter exceeds Z and subtract 26.
Applying the encoding shift when asked for decoding. If the question gives you a coded word and asks for the original, you must apply the inverse shift (−n, not +n). This is the single most common careless error in this topic.
Assuming uniform shift without verifying the second letter. In mixed-shift questions (like the FRIEND→GSJFOE type), the first letter always matches the pattern, but position 3 diverges. Always check at least letters 1 and 2 before concluding the shift is uniform.
In sentence coding, pairing sentences that share two common words. If S1 and S2 share both pencil and pen, you know two codes belong to those two words but you cannot assign individually. Aspirants incorrectly assign arbitrarily. Only pair sentences that share exactly one word for definitive assignment.
In number coding, assuming a letter not given in examples can be inferred. Number coding questions are always self-contained — the target word only uses letters whose codes are already given. If you think a letter is missing, re-read all given examples carefully before assuming a pattern.
Spending more than 75 seconds on a single Coding-Decoding question. This topic rewards quick pattern recognition. If you are stuck past 75 seconds, it almost always means you have misclassified the type. Reset — reclassify from scratch — rather than pushing further with a wrong approach.