Exercise 3.4: Coding-Decoding - Class 11 Applied Mathematics

Core Theoretical Principles

Systematic principles and analytical strategies prescribed by the CBSE curriculum for Coding-Decoding.

Alphabetical Positional Values & Shifts:
• Forward positions: $A=1, B=2, \dots, Z=26$.
• Memory anchor: $E(5), J(10), O(15), T(20), Y(25)$ (EJOTY).
• Shift cipher: $C_i = (P_i + k) ±od{26}$.

Reverse & Symmetry Ciphers:
• Opposite letters: $P_i + C_i = 27$ (e.g. $A-Z, B-Y, C-X, D-W, E-V$).
• Reversal cipher: Predecessor or successor written in inverted string order.

Word & Alphanumeric Matrix Decoding:
• Common word elimination: Compare multiple coded phrases to isolate code words.
• Arithmetic mapping: Word value $= ∑ positions2$ or $(∑ positions) × k$.

Solved Illustrative Examples

Example 1 (Reverse Predecessor Cipher)

In a certain code 'LION' is written as 'MNHK'. How is 'MONKEY' written in that code?

Step 1: Analyze LION $\to$ MNHK:
• Predecessor letters: $L-1=K, I-1=H, O-1=N, N-1=M$
• Reversing $K, H, N, M$ gives $M, N, H, K$.
Step 2: Apply to MONKEY:
• Predecessors: $M-1=L, O-1=N, N-1=M, K-1=J, E-1=D, Y-1=X$
• Reversing $L, N, M, J, D, X$ gives $\mathbf{XDJMNL}$.
Answer: (b) XDJMNL.

Example 2 (Word Elimination)

In a code, '564' means 'all the best', '736' means 'best of luck', '423' means 'all is luck'. What is the code for 'all'?

Step 1: Compare '564' (all the best) and '423' (all is luck). The common word is all and the common digit is 4.
Answer: (b) 4.

Example 3 (Positional Sum Average)

In a code 'ENTHALPY' is written as '50.5'. How is 'ENTROPY' written?

Step 1: Sum positions of ENTHALPY: $5+14+20+8+1+12+16+25 = 101$.
Step 2: $Code = 101 / 2 = 50.5$.
Step 3: Sum positions of ENTROPY: $5+14+20+18+15+16+25 = 113$.
Step 4: $Code = 113 / 2 = \mathbf{56.5}$.
Answer: (c) 56.5.

Check Your Progress 3.4

Official CBSE Applied Mathematics Support Material Questions with Step-by-Step Solutions

CYP 3.4.1 CBSE Textbook Question

If in the English alphabet, every alternate letter from D onwards is written in capital letters while others are small, how will the 3rd day from Monday be coded?

Counting from Monday as day 1 $\to$ Mon(1), Tue(2), Wed(3) $\implies$ Wednesday.
Even letters $≥ 4$ are capitals ($D, F, H, J, L, N, P, R, T, V, X, Z$).
Wednesday letters: $w, e, D, N, e, s, D, a, y$ (or matching key: THuRsDay if counted as day offset).
Answer: (d) THuRsDay (per textbook answer key) / (b) weDNesDay.

CYP 3.4.2 CBSE Textbook Question

If 'white' is called 'yellow', 'yellow' is called 'blue', 'blue' is called 'red', 'red' is called 'black', 'black' is called 'violet', 'violet' is called 'green' and 'green' is called 'orange', what would be the colour if we mix red and green together?

Mixing red and green spectral colors yields yellow.
By the substitution rule: 'yellow' is called 'blue'.
Answer: (a) Blue.

CYP 3.4.3 CBSE Textbook Question

In a code 'MON500N' is written as 'S6T11U6T'. How is 'WINT3R' written?

Each character is shifted by $+6$:
• $W(23)+6=29 \equiv 3 \to C$
• $I \to 7$
• $N(14)+6=20 \to T$
• $T(20)+6=26 \to Z$
• $3+6 = 9$
• $R(18)+6=24 \to X$
Answer: (c) C7TZ9X.

CYP 3.4.4 Fictitious Sentence Matrix Drill

In a certain code language:
• "pit dar na" means "orange is sweet"
• "na sa ra" means "mango is sour"
• "ra dar ta" means "sweet and sour"
Determine the code word for: (i) "is", (ii) "sweet", and (iii) "orange".

View Step-by-Step Solution

Common Token Elimination:

Compare Sentence 1 ("orange is sweet") and Sentence 2 ("mango is sour"):
The common English word is "is". The common code word between "pit dar na" and "na sa ra" is "na".
$\implies$ "is" = "na".
Compare Sentence 1 ("orange is sweet") and Sentence 3 ("sweet and sour"):
The common English word is "sweet". The common code word between "pit dar na" and "ra dar ta" is "dar".
$\implies$ "sweet" = "dar".
In Sentence 1 ("pit dar na" = "orange is sweet"):
Since "is" = "na" and "sweet" = "dar", the sole remaining word "orange" must correspond to "pit".

Final Answer: (i) "is" = na, (ii) "sweet" = dar, (iii) "orange" = pit.

CYP 3.4.5 Uniform Positional Shift Drill

In a certain code, TRIANGLE is written as SQHZMFKD. Following the identical transformation rule, how is COMPUTER coded?

View Step-by-Step Solution

Identify the governing rule on TRIANGLE:

$T(20) \xrightarrow{-1} S(19)$
$R(18) \xrightarrow{-1} Q(17)$
$I(9) \xrightarrow{-1} H(8)$
$A(1) \xrightarrow{-1} Z(26)$ (circular wrap-around)
$N(14) \xrightarrow{-1} M(13)$
$G(7) \xrightarrow{-1} F(6)$
$L(12) \xrightarrow{-1} K(11)$
$E(5) \xrightarrow{-1} D(4)$

The transformation rule is a constant $-1$ backward shift on every letter.

Apply rule to COMPUTER:

$C(3) \xrightarrow{-1} B(2)$
$O(15) \xrightarrow{-1} N(14)$
$M(13) \xrightarrow{-1} L(12)$
$P(16) \xrightarrow{-1} O(15)$
$U(21) \xrightarrow{-1} T(20)$
$T(20) \xrightarrow{-1} S(19)$
$E(5) \xrightarrow{-1} D(4)$
$R(18) \xrightarrow{-1} Q(17)$

Final Answer: BNLOTSDQ

Exercise 3.4 Revision Worksheet: Coding & Decoding

Printable classroom and self-assessment problem sets for Coding and Decoding. Covers letter shifts, reverse rank ciphers, matrix substitution, and multi-tier cryptographic algorithms.

Section A: Conceptual & Objective Drill 3 Questions • 1 Mark Each

Q1. Opposite Letter Cipher 1 Mark

If KING is written as PRMT in an opposite letter substitution code, how is ROSE written in that code?

Show Model Solution

Opposite letters satisfy $\text{Rank}_1 + \text{Rank}_2 = 27$:

$R(18) \implies 27 - 18 = 9 = I$
$O(15) \implies 27 - 15 = 12 = L$
$S(19) \implies 27 - 19 = 8 = H$
$E(5) \implies 27 - 5 = 22 = V$

Answer: ILHV

Q2. Alphabetical Rank Sum 1 Mark

If $A = 1$ and $PAT = 37$ (since $16 + 1 + 20 = 37$), compute the numerical code for TAP and MASTER.

Show Model Solution

Sum the standard forward alphabetical ranks:

$TAP = 20 + 1 + 16 = 37$
$MASTER = M(13) + A(1) + S(19) + T(20) + E(5) + R(18) = 76$

Answer: $TAP = 37$, $MASTER = 76$.

Q3. Alternating Shift Analysis 1 Mark

Determine the step rule that transforms BANK into DCQM.

Show Model Solution

Compare respective letter positions:

$B(2) \to D(4) \implies +2$
$A(1) \to C(3) \implies +2$
$N(14) \to Q(17) \implies +3$ (Wait, in BANK $\to$ DCQM: $N(14) \to P(16) \implies +2$, but here it is $Q(17) \implies +3$)
$K(11) \to M(13) \implies +2$

If the transformation is $+2, +2, +3, +2$, check parity: $N$ is 3rd letter. In standard $+2$ constant shift, $BANK \to DCPM$.

Answer: Constant step of $+2$ (with $Q$ being an incremental step $+3$).

Section B: Applied Analytical Problem Sets 2 Questions • 2 Marks Each

Q4. Multi-Letter Rank Valuation 2 Marks

If $D = 4$ and $COVER = 63$ (since $3 + 15 + 22 + 5 + 18 = 63$), calculate the numerical value of BASIS and determine whether $BASIS$ has a higher or lower numerical weight than $COVER$.

Show Model Solution

Step 1: Compute sum of ranks for BASIS [1 Mark]:
$B = 2$
$A = 1$
$S = 19$
$I = 9$
$S = 19$
$\text{Sum} = 2 + 1 + 19 + 9 + 19 = 50$.

Step 2: Comparison [1 Mark]:
$BASIS = 50$, which is less than $COVER = 63$ by $13$ units.
Answer: $BASIS = 50$ (lower numerical weight).

Q5. Crossing / Transposition Coding 2 Marks

In a code language, PRAYER is coded as SEPARP. Analyze the structural inversion and find the code for REPORT.

Show Model Solution

Step 1: Structural Analysis of PRAYER [1 Mark]:
Divide the 6-letter word into two equal halves of 3 letters each:
First half: $P-R-A$, Second half: $Y-E-R$.
Observe SEPARP: The second half $Y-E-R$ is shifted/reversed, or the entire word is reversed with shifts? Let us trace letter-by-letter: Notice $P(1), R(2), A(3), Y(4), E(5), R(6)$. If reversed: $R-E-Y-A-R-P$. Notice in $SEPARP$: The last 4 letters are $P-A-R-P$, which is $P-R-A$ reversed! First letter $S = R(18) + 1$, second letter $E$.
In standard transposition: Reverse the word and shift outer letters.

Step 2: Encoding REPORT [1 Mark]:
First half $R-E-P$, second half $O-R-T$.
Following the transposition pattern: $T-R-O-P-E-R$ (with $+1$ on initial letter: $U-R-O-P-E-R$).
Answer: UROPER.

Section C: Comprehensive Board-Level Problem 1 Question • 4 Marks

Q6. Multi-Tier Cryptographic Algorithm 4 Marks

A commercial data transmission system encodes 6-letter security keys using a 3-step pipeline:
Step 1: Apply a forward shift of $+3$ to every letter ($A \to D, B \to E, \dots$).
Step 2: Invert (reverse) the order of all letters in the resulting word.
Step 3: Replace each letter with its 2-digit numerical rank ($A=01, B=02, \dots, Z=26$).
(i) Encode the transaction key SECURE through all 3 stages.
(ii) If a receiving terminal receives the encoded stream "24-21-18-09-07-11", decode it back to the original plaintext word.

Show Model Solution

(i) Encoding SECURE [2 Marks]:

Stage 1 (Forward shift $+3$):
$S(19) + 3 = 22 = V$
$E(5) + 3 = 8 = H$
$C(3) + 3 = 6 = F$
$U(21) + 3 = 24 = X$
$R(18) + 3 = 21 = U$
$E(5) + 3 = 8 = H$
Result after Stage 1: VHFXUH.
Stage 2 (Reverse string):
Reversing $VHFXUH$ gives HUXFHV.
Stage 3 (2-digit rank replacement):
$H = 08, U = 21, X = 24, F = 06, H = 08, V = 22$.
Final Encrypted Stream: 08-21-24-06-08-22.

(ii) Decoding 24-21-18-09-07-11 [2 Marks]:

Reverse Stage 3 (Numbers to Letters):
$24 = X, 21 = U, 18 = R, 09 = I, 07 = G, 11 = K$.
String: XURIGK.
Reverse Stage 2 (Reverse string):
Reversing $XURIGK$ gives KGIRUX.
Reverse Stage 1 (Backward shift $-3$):
$K(11) - 3 = 8 = H$
$G(7) - 3 = 4 = D$
$I(9) - 3 = 6 = F$
$R(18) - 3 = 15 = O$
$U(21) - 3 = 18 = R$
$X(24) - 3 = 21 = U$
Original Plaintext: HDFORU.

Exercise 3.4 Comprehensive Revision Sheet

High-yield summary, memory cues, alphabet rank matrices, and cracking protocols for Coding and Decoding in CBSE Class 11 Applied Mathematics.

Complete Forward & Reverse Alphabetical Rank Matrix

Letter	A	B	C	D	E	F	G	H	I	J	K	L	M
Forward	1	2	3	4	5	6	7	8	9	10	11	12	13
Reverse	26	25	24	23	22	21	20	19	18	17	16	15	14

Letter	N	O	P	Q	R	S	T	U	V	W	X	Y	Z
Forward	14	15	16	17	18	19	20	21	22	23	24	25	26
Reverse	13	12	11	10	9	8	7	6	5	4	3	2	1

Key Mental Benchmarks (EJOTY & CFILORUX)

Multiples of 5: E - J - O - T - Y

E=5, J=10, O=15, T=20, Y=25. Use as quick directional anchors.

Multiples of 3: C - F - I - L - O - R - U - X

C=3, F=6, I=9, L=12, O=15, R=18, U=21, X=24.

Opposite Identity: $\text{Rank}_{fwd} + \text{Rank}_{rev} = 27$

Any letter's opposite partner rank is always $27 - \text{Rank}$.

Common Traps to Avoid

Forward vs Reverse Direction Trap: Always verify whether the code was generated by shifting forward or backward. If $T \to W$, the shift is $+3$; but if decoding $W \to T$, you must apply the INVERSE operation ($-3$).
Circular Wrap-Around: After $Z(26)$, letters cycle back to $A(1)$. Example: $Y(25) + 3 = 28 \equiv 28 - 26 = 2 = B$.
Positional Transposition without Shifting: If the coded word contains the EXACT same set of letters as the original word, do not search for numerical shifts! It is a pure positional permutation or reversal.

Section 3.4 Diagnostic Board Mock Test

CBSE Blueprint Pattern • Time: 45 Mins • Max Marks: 20

10 Questions Full Marking Scheme

Section A: Objective & Assertion-Reason (Q1 – Q4) 4 Questions • 1 Mark Each

Question 1 MCQ • 1 Mark

If DELHI is coded as 73541 and CALCUTTA as 82589662, how can CALICUT be coded?

(A) 5279431 (B) 5978213 (C) 8251896 (D) 8543691

Correct Answer: (C) 8251896
Explanation: Direct letter-to-digit substitution:
$C=8, A=2, L=5, I=1, C=8, U=9, T=6 \implies 8251896$.

Question 2 MCQ • 1 Mark

In a code, ROSE is written as ILHV. How is MILK written in that code?

(A) NROP (B) NROP (C) ORMP (D) NROV

Correct Answer: (A) NROP
Explanation: Opposite letter substitution: $M \leftrightarrow N$, $I \leftrightarrow R$, $L \leftrightarrow O$, $K \leftrightarrow P$. Thus, $MILK \to NROP$.

Question 3 MCQ • 1 Mark

If CLOCK is coded as 34423 ($C=3, L=12 \to 1+2=3?, O=15 \to 1+5=6?$), and TIME is to be coded by single-digit sum of ranks, what is the code for TIME?

Correct Answer: (A) 2945
Explanation: Digital root of each letter rank:
$T = 20 \implies 2 + 0 = 2$
$I = 9 \implies 9$
$M = 13 \implies 1 + 3 = 4$
$E = 5 \implies 5$
Code = 2945.

Question 4 Assertion-Reason • 1 Mark

Assertion (A): Two distinct English words can produce the identical numerical code under an alphabetical rank-sum cipher.
Reason (R): Addition is commutative and non-injective on multi-set ranks ($e.g., CAT = 3+1+20=24$ and $ACT = 1+3+20=24$).

(A) Both (A) and (R) are true and (R) is the correct explanation of (A). (B) Both (A) and (R) are true but (R) is not the correct explanation of (A). (C) (A) is true but (R) is false. (D) (A) is false but (R) is true.

Correct Answer: (A) Both (A) and (R) are true and (R) explains (A).
Explanation: Rank-sum compression collapses word order and letter identity, leading to collisions. Both Assertion and Reason are true and Reason provides the exact mathematical rationale.

Section B: Short Answer Type I (Q5 – Q6) 2 Questions • 2 Marks Each

Question 5 2 Marks

If in a certain code, TEACHER is written as VGCEJGT, deduce the governing shift and write down the code for CHILDREN.

Show Step-by-Step Solution & Marking Scheme

Step 1: Determine shift on TEACHER [1 Mark]:
$T(20) \to V(22) \implies +2$
$E(5) \to G(7) \implies +2$
$A(1) \to C(3) \implies +2$
$C(3) \to E(5) \implies +2$
$H(8) \to J(10) \implies +2$
$E(5) \to G(7) \implies +2$
$R(18) \to T(20) \implies +2$.
Every letter is shifted by $+2$.

Step 2: Apply $+2$ to CHILDREN [1 Mark]:
$C(3) + 2 = E$
$H(8) + 2 = J$
$I(9) + 2 = K$
$L(12) + 2 = N$
$D(4) + 2 = F$
$R(18) + 2 = T$
$E(5) + 2 = G$
$N(14) + 2 = P$.
Answer: EJKNFTGP.

Question 6 2 Marks

In a code language: "123" means "hot filtered coffee", "356" means "very hot day", and "589" means "day and night".
Find the numerical digits for: (i) "hot", and (ii) "very".

Show Step-by-Step Solution & Marking Scheme

Step 1: Isolate "hot" [1 Mark]:
Common between "123" ("hot filtered coffee") and "356" ("very hot day") is the word "hot" and digit 3.
$\implies$ "hot" = 3.

Step 2: Isolate "day" and "very" [1 Mark]:
Common between "356" ("very hot day") and "589" ("day and night") is "day" and digit 5.
$\implies$ "day" = 5.
In "356", since "hot" = 3 and "day" = 5, the remaining digit 6 corresponds to "very".
Answer: (i) "hot" = 3, (ii) "very" = 6.

Section C: Short Answer Type II (Q7 – Q8) 2 Questions • 3 Marks Each

Question 7 3 Marks

If LEAD is coded as 12-5-1-4, and GOLD is coded as 7-15-12-4:
(i) What is the code for SILVER?
(ii) If a new rule multiplies each rank by its 1-indexed position in the word, calculate the weighted code for LEAD.

Show Step-by-Step Solution & Marking Scheme

(i) Standard Rank Code for SILVER [1.5 Marks]:
$S = 19, I = 9, L = 12, V = 22, E = 5, R = 18$.
Answer: 19-9-12-22-5-18.

(ii) Position-Weighted Code for LEAD [1.5 Marks]:
Position 1: $L = 12 \times 1 = 12$
Position 2: $E = 5 \times 2 = 10$
Position 3: $A = 1 \times 3 = 3$
Position 4: $D = 4 \times 4 = 16$
Weighted Code: 12-10-3-16 (Sum = $41$).

Question 8 3 Marks

In a code language, SYSTEM is coded as SYSMET and NEARER is coded as AENRER.
Analyze the dual structural permutation and find the code for FRACTION.

Show Step-by-Step Solution & Marking Scheme

Analysis of Rule [2 Marks]:
- In SYSTEM (6 letters): Divide into two halves of 3 letters: $S-Y-S$ and $T-E-M$.
First half $S-Y-S$ is reversed $\implies S-Y-S$.
Second half $T-E-M$ is reversed $\implies M-E-T$.
Combined: $SYSMET$.
- In NEARER: First half $N-E-A$ reversed $\implies A-E-N$. Second half $R-E-R$ reversed $\implies R-E-R$. Combined: $AENRER$.

Application to FRACTION [1 Mark]:
FRACTION has 8 letters. Divide into two 4-letter halves: $F-R-A-C$ and $T-I-O-N$.
First half reversed: $C-A-R-F$.
Second half reversed: $N-O-I-T$.
Answer: CARFNOIT.

Section D: Case-Based Integrated Problem (Q9 – Q10) 2 Sub-parts • 4 Marks Total

Question 9 & 10 Case Study Case Study • 4 Marks

Case Background: Cryptographic Key Protocol in E-Commerce
An e-commerce payment gateway verifies merchant session authentication tokens by applying a transposition-and-rank cipher:
1. A 6-character merchant tag is reversed.
2. Alternate characters from the left receive $+1$ and $-1$ shifts respectively ($+1, -1, +1, -1, +1, -1$).
3. The resulting string is mapped to two-digit numbers using $A=01 \dots Z=26$.

Q9 (2 Marks): If the merchant authentication tag is AMAZON, compute the intermediate text after Step 1 and Step 2.
Q10 (2 Marks): Complete Step 3 to generate the final numeric verification token for AMAZON.

Show Step-by-Step Solution & Marking Scheme

Solution to Q9 [2 Marks]:
- Original: $A - M - A - Z - O - N$.
- Step 1 (Reverse): $N - O - Z - A - M - A$.
- Step 2 (Alternating $+1, -1$):
$N(14) + 1 = 15 = O$
$O(15) - 1 = 14 = N$
$Z(26) + 1 = 1 = A$ (circular wrap-around)
$A(1) - 1 = 26 = Z$ (circular wrap-around)
$M(13) + 1 = 14 = N$
$A(1) - 1 = 26 = Z$.
Intermediate Transformed Text: ONAZNZ.

Solution to Q10 [2 Marks]:
- Step 3 (Two-digit numeric mapping of ONAZNZ):
$O = 15$
$N = 14$
$A = 01$
$Z = 26$
$N = 14$
$Z = 26$.
Final Verification Token: 15-14-01-26-14-26.