Advanced Analytical Skills — II

Unit 7: Coded Inequalities

Master the art of decoding symbolic inequalities, solving chain comparisons, comparing roots of quadratic equations, and cracking every competitive exam question on this topic.

⏱️ Time to Complete: 6–8 hours | 📝 30 MCQs + 8 Short + 3 Long (Bloom's Mapped) | 15 Worked Examples

🎯 Exams this covers: Bank PO (SBI/IBPS) | SSC CGL | CAT | CLAT | University Aptitude Tests

Section 1

Opening Hook — When Symbols Replace Logic

🧩 The Code That Decides Your Bank Job

It's February 2026. You sit down for your SBI PO Prelims at 10:00 AM in a packed exam centre in Lucknow. 35 reasoning questions. 20 minutes. The clock is ticking. Question 14 appears on screen:

Statements: P © Q, Q £ R, R @ S
Conclusions: I. P © S II. S £ P

Your friend panics. You smile. Because you know © means >, £ means <, @ means =. You decode, chain, and solve in 15 seconds: P > Q < R = S → neither conclusion follows. Answer marked. Next question.

This isn't magic — it's Coded Inequalities. The single most predictable, most scoreable topic in banking exams. 5 questions. Every exam. Guaranteed.

🏦 SBI PO🏦 IBPS PO/Clerk📋 SSC CGL📋 RRB NTPC🎓 CAT⚖️ CLAT

In SBI PO 2024, exactly 5 out of 35 reasoning questions were on Coded Inequalities. Students who practised this topic scored a perfect 5/5 in under 2 minutes, while others spent 4–5 minutes and still got them wrong. That 5-mark advantage often decides the cutoff.

Section 2

Learning Outcomes — Bloom's Taxonomy Mapped

Bloom's Level	Learning Outcome
🔵 Remember	List the six fundamental inequality symbols (>, <, ≥, ≤, =, ≠) and recall coded symbol mappings
🔵 Understand	Explain how a coded inequality statement translates into a standard mathematical inequality
🟢 Apply	Decode coded statements and form a single combined chain inequality
🟢 Analyse	Analyse a given chain to determine which conclusions are definitely true, definitely false, or can't be determined
🟠 Evaluate	Evaluate conclusions involving quadratic-root-based inequalities and mixed chains
🟠 Create	Create your own coded inequality problem sets and solve advanced multi-statement questions under exam pressure

Section 3

Basic Inequalities — The Foundation

Before we decode anything, let's build rock-solid clarity on what inequality symbols actually mean. Every competitive exam question on coded inequalities rests on these six symbols:

📐 The Six Fundamental Inequality Symbols

Symbol	Name	Meaning	Example	Read As
>	Greater than	Left side is strictly larger than right	7 > 3	"7 is greater than 3"
<	Less than	Left side is strictly smaller than right	2 < 9	"2 is less than 9"
≥	Greater than or equal to	Left side is larger than or equal to right	5 ≥ 5	"5 is at least 5"
≤	Less than or equal to	Left side is smaller than or equal to right	3 ≤ 8	"3 is at most 8"
=	Equal to	Both sides are exactly the same	4 = 4	"4 equals 4"
≠	Not equal to	Both sides are different	5 ≠ 3	"5 is not equal to 3"

Key Properties of Inequalities

1. Transitivity — The Chain Rule

This is the most important property for coded inequalities. If two inequalities share a common element, you can chain them:

If A > B and B > C → then A > C ✅

This works for >, <, ≥, ≤, and = in compatible directions. It does not work when the directions conflict.

2. Reversal Property

Every inequality can be read backwards by flipping the symbol:

A > B ⟺ B < A

X ≥ Y ⟺ Y ≤ X

3. Combining ≥ and > (Same Direction)

When combining two inequalities going in the same direction, the combined result takes the weaker (broader) symbol:

Statement 1	Statement 2	Combined Result	Rule
A > B	B > C	A > C	Strict + Strict = Strict
A ≥ B	B ≥ C	A ≥ C	Non-strict + Non-strict = Non-strict
A > B	B ≥ C	A > C	At least one strict = Strict
A ≥ B	B > C	A > C	At least one strict = Strict
A = B	B > C	A > C	Equality carries forward
A = B	B = C	A = C	Equality is transitive

Students often confuse A ≥ B with A > B. Remember: A ≥ B means A could be greater OR equal. So if A ≥ B and B ≥ C, you can only conclude A ≥ C (not A > C), because A might equal B might equal C. The "equal" possibility survives.

4. Opposite Directions = No Conclusion

When inequalities point in opposite directions from a common element, you cannot combine them:

A > B and B < C → Relationship between A and C? ❌ CAN'T DETERMINE

Think of it: A is bigger than B, and C is also bigger than B. But we don't know if A is bigger, smaller, or equal to C — they're both "above" B but could be anywhere.

The Golden Rule: You can only chain inequalities when the arrows all flow in the same direction through the chain. If the arrow reverses at any point, the chain breaks and no definite conclusion can be drawn between the endpoints.

Section 4

Coded Symbols — Cracking the Secret Language

In competitive exams, inequality symbols are replaced with coded symbols (special characters, shapes, or letters). Your job: decode the symbols, form the inequality chain, and evaluate the conclusions. Let's master this systematically.

Standard Coding Scheme (Most Common in Exams)

While every exam can define its own codes, here are the most frequently used coding patterns across SBI PO, IBPS, and SSC exams:

🔐 Code Set 1 — The Classic Banking Pattern

©→> (Greater than)

£→< (Less than)

@→= (Equal to)

%→≥ (Greater than or equal to)

$→≤ (Less than or equal to)

#→≠ (Not equal to)

🔐 Code Set 2 — Alternate Pattern (Also Common)

★→> (Greater than)

△→< (Less than)

◯→= (Equal to)

◆→≥ (Greater than or equal to)

□→≤ (Less than or equal to)

NEVER memorise specific code-to-symbol mappings! Every exam defines its own codes in the question. Your job is to read the given coding table first, write the decoded inequalities on your rough sheet, and then solve. Memorising "© always means >" is dangerous — the next exam might use © for <.

How to Read the Question

A typical exam question looks like this:

📋 Sample Question Format

Directions: In the following questions, the symbols ©, £, @, %, and $ are used with the following meanings:

'P © Q' means 'P is greater than Q'
'P £ Q' means 'P is less than Q'
'P @ Q' means 'P is equal to Q'
'P % Q' means 'P is greater than or equal to Q'
'P $ Q' means 'P is less than or equal to Q'

Statement: A © B, B % C, C @ D

Conclusions:
I. A © D
II. D $ B

Answer choices:
(A) Only I follows (B) Only II follows (C) Both follow (D) Neither follows

Speed Hack: Don't decode the symbols back into coded form for conclusions. Instead, decode EVERYTHING to standard math symbols first (>, <, =, ≥, ≤), solve using the chain, then match your answer to the conclusion. This avoids the double-translation error that catches most students.

Section 5

Decoding Strategy — The 4-Step Method

Follow this exact sequence every time, and you'll solve any coded inequality in 20–30 seconds:

🚀 The 4-Step Rapid Solving Method

Step 1 — DECODE: Replace every coded symbol with its standard math inequality. Write it on rough paper.

Step 2 — CHAIN: Combine all statements into one continuous chain. Arrange elements left-to-right with arrows flowing in the same direction if possible.

Step 3 — CHECK: For each conclusion, trace the path between the two elements in your chain. If the path flows in one direction → conclusion can be evaluated. If the path reverses direction → "can't be determined."

Step 4 — ANSWER: Mark the conclusion as "definitely true," "definitely false," or "can't be determined" and select the matching option.

Let's Apply the 4-Step Method — Live Demo

Given Codes: © means >, £ means <, @ means =, % means ≥, $ means ≤

Statement: M © N, N % O, O @ P

Conclusions: I. M © P II. P $ N

Step 1 — DECODE:

M > N, N ≥ O, O = P

Step 2 — CHAIN:

M > N ≥ O = P

All arrows point left-to-right (descending). ✅ Valid chain.

Step 3 — CHECK Conclusion I: M © P → M > P

Follow the chain: M > N ≥ O = P. Since M > N and N ≥ O = P, we get M > P. ✅ Definitely true.

Step 3 — CHECK Conclusion II: P $ N → P ≤ N

From chain: N ≥ O = P → N ≥ P → P ≤ N. ✅ Definitely true.

Step 4 — ANSWER: Both I and II follow → (C)

Don't assume ">" means "≥"! If the chain gives M > P, the conclusion "M ≥ P" is also true (because > is a subset of ≥). But if the chain gives M ≥ P, you CANNOT conclude M > P (because M might equal P).

Section 6

Chain Inequality Solving — Advanced Techniques

Rules for Building and Reading Chains

🔗 Chain Building Rules

Rule 1 — Same direction, connect: If A > B and B > C, chain as A > B > C. Conclusion: A > C ✅

Rule 2 — Equality passes through: If A > B and B = C, chain as A > B = C. Conclusion: A > C ✅

Rule 3 — Mixed ≥ and >: If A ≥ B and B > C, then A > C ✅ (the strict inequality wins when mixed in same direction)

Rule 4 — Opposite directions break the chain: If A > B and B < C, you get A > B < C. Relationship between A and C? ❌ Can't determine

Rule 5 — Multiple paths: If two different paths exist between elements, check both. If even one gives a definite answer, use it.

Combining Result Table — Master Reference

Left Relation	Right Relation	Combined	Can Conclude?
A > B	B > C	A > C	✅ Yes
A > B	B ≥ C	A > C	✅ Yes
A > B	B = C	A > C	✅ Yes
A ≥ B	B ≥ C	A ≥ C	✅ Yes
A ≥ B	B > C	A > C	✅ Yes
A ≥ B	B = C	A ≥ C	✅ Yes
A = B	B = C	A = C	✅ Yes
A = B	B > C	A > C	✅ Yes
A = B	B ≥ C	A ≥ C	✅ Yes
A > B	B < C	A ? C	❌ No — opposite directions
A < B	B > C	A ? C	❌ No — opposite directions
A ≥ B	B ≤ C	A ? C	❌ No — opposite directions

This table alone can solve 90% of coded inequality questions. Print it, practise with it, and internalise it. After 50 questions, you won't need the table — it'll be instinct.

Complex Chain Example

Statements: A > B, B ≥ C, C = D, D < E, E ≤ F

A > B ≥ C = D < E ≤ F

Now let's evaluate:

Conclusion	Path	Result
A > D?	A > B ≥ C = D → A > D	✅ Definitely true
A > E?	A > B ≥ C = D < E — direction changes at D	❌ Can't determine
F > C?	F ≥ E > D = C → F ≥ E > C → F > C	✅ Definitely true
A > F?	Direction changes at D	❌ Can't determine
D < F?	D < E ≤ F → D < F	✅ Definitely true

Direction-change finder: Scan your chain from left to right. Every time the inequality arrow flips (from > to < or vice versa), mark that element as a "break point." You can only draw conclusions between elements that are on the same side of a break point with unidirectional flow.

Section 7

Comparison of Roots of Quadratic Equations

The Concept

Some advanced coded inequality questions give you two quadratic equations and ask you to compare their roots. Here's the technique:

📊 Quadratic Root Comparison Method

Given: Two quadratic equations, e.g.,

Equation I: x² − 7x + 12 = 0
Equation II: y² − 9y + 20 = 0

Step 1: Solve both equations to find their roots.

Eq I: x² − 7x + 12 = 0 → (x − 3)(x − 4) = 0 → x = 3 or x = 4

Eq II: y² − 9y + 20 = 0 → (y − 4)(y − 5) = 0 → y = 4 or y = 5

Step 2: Compare all possible pairs of roots:

x = 3 vs y = 4 → x < y
x = 3 vs y = 5 → x < y
x = 4 vs y = 4 → x = y
x = 4 vs y = 5 → x < y

Step 3: Determine the relationship:

In all cases, x ≤ y. So: x ≤ y (i.e., y ≥ x)

Decision Rules for Root Comparison

Scenario	All Pairs Show	Conclusion
All pairs: x < y	x is always less	x < y
All pairs: x < y or x = y	x is never greater	x ≤ y
All pairs: x > y	x is always greater	x > y
All pairs: x > y or x = y	x is never less	x ≥ y
Mixed: some x > y, some x < y	No consistent pattern	Can't determine

Don't compare only one root from each equation. You must check ALL four combinations (root₁ of Eq I vs root₁ of Eq II, root₁ vs root₂, root₂ vs root₁, root₂ vs root₂). Missing even one comparison can lead to wrong answers.

Quick Factorisation Tricks

For equations of the form x² + bx + c = 0:

If b and c are both positive → both roots are negative
If b is negative and c is positive → both roots are positive
If c is negative → roots have opposite signs
Sum of roots = −b/a, Product of roots = c/a

Sign shortcut: Before solving, check the signs. If Eq I has both negative roots and Eq II has both positive roots, you immediately know x < y without actually solving. This saves 30+ seconds per question.

Section 8

Decision Framework — True, False, or Can't Determine

The Three Verdicts

⚖️ How to Decide: Definitely True / Definitely False / Can't Determine

✅ Definitely True: The conclusion follows directly from the chain with no ambiguity. The chain supports it in ALL possible scenarios.

Example: Chain says A > B > C. Conclusion: A > C → Definitely True

❌ Definitely False: The conclusion directly contradicts the chain. There is NO possible scenario where it can be true.

Example: Chain says A > B > C. Conclusion: C > A → Definitely False

❓ Can't Determine: The chain does not provide enough information. The conclusion might be true in some cases and false in others.

Example: Chain says A > B < C. Conclusion: A > C → Can't Determine (A could be 10, B = 3, C = 5 → A > C; or A = 4, B = 3, C = 100 → A < C)

Common Exam Answer Formats

Option	Meaning
(A) Only Conclusion I follows	I is definitely true, II is not definitely true
(B) Only Conclusion II follows	II is definitely true, I is not definitely true
(C) Either I or II follows	I and II are complementary — at least one must be true
(D) Neither I nor II follows	Both are either false or can't be determined
(E) Both I and II follow	Both are definitely true

The Tricky "Either-Or" Case

This is the most confusing option and catches even good students:

🔀 When Does "Either I or II" Apply?

"Either I or II follows" is selected when:

Neither conclusion is individually definite
But the two conclusions are complementary — one of them MUST be true
Typically happens when one says A > B and the other says A ≤ B (or A ≥ B and A < B)

Example: Chain gives A ≥ B (A could be > or = to B)

Conclusion I: A > B → Not definite (could be equal)
Conclusion II: A = B → Not definite (could be greater)
But one of them MUST be true → Either I or II follows

The "Either-Or" option appears in roughly 1 out of every 5 coded inequality question sets in banking exams. Many students select "Neither" when they should select "Either." Always check: if neither conclusion is individually certain, ask "Is it impossible for BOTH to be false simultaneously?" If yes → select "Either."

Section 9

Worked Examples — 15 Solved Problems

Using Code Set: © → > £ → < @ → = % → ≥ $ → ≤

Example 1 — Basic Chain (Level: Easy)

🎯 Exam: IBPS PO PatternBeginner

Statement: A © B, B © C, C © D

Conclusions: I. A © D II. B © D

Decode:

A > B > C > D

Conclusion I: A > D → From chain: A > B > C > D → A > D ✅ True

Conclusion II: B > D → From chain: B > C > D → B > D ✅ True

Answer: (E) Both I and II follow

Example 2 — Chain with Equality (Level: Easy)

🎯 Exam: SBI Clerk PatternBeginner

Statement: P % Q, Q @ R, R © S

Conclusions: I. P © S II. P % R

Decode:

P ≥ Q = R > S

Conclusion I: P > S → P ≥ Q = R > S → P > S (since ≥ followed by > in same direction gives >) ✅ True

Conclusion II: P ≥ R → P ≥ Q = R → P ≥ R ✅ True

Answer: (E) Both I and II follow

Example 3 — Broken Chain (Level: Easy)

🎯 Exam: IBPS Clerk PatternBeginner

Statement: J © K, K £ L, L © M

Conclusions: I. J © M II. L © J

Decode:

J > K < L > M

Conclusion I: J > M → Path: J > K < L > M — direction changes at K. ❌ Can't determine

Conclusion II: L > J → Path: J > K < L — direction changes at K. ❌ Can't determine

Answer: (D) Neither I nor II follows

Example 4 — ≥ and ≤ Mix (Level: Medium)

🎯 Exam: SBI PO PatternIntermediate

Statement: A % B, B $ C, C @ D

Conclusions: I. A % D II. A © D

Decode:

A ≥ B ≤ C = D

Conclusion I: A ≥ D → Path: A ≥ B ≤ C = D — direction changes at B (≥ then ≤). ❌ Can't determine

Conclusion II: A > D → Same path, same break. ❌ Can't determine

Are they complementary? No — both are indeterminate independently and not complementary (A could be >, =, or < D).

Answer: (D) Neither I nor II follows

Example 5 — Either-Or Scenario (Level: Medium)

🎯 Exam: SBI PO PatternIntermediate

Statement: H % I, I @ J, J % K

Conclusions: I. H © K II. H @ K

Decode:

H ≥ I = J ≥ K

Conclusion I: H > K → From chain: H ≥ I = J ≥ K → H ≥ K. So H > K is possible but not certain (H might equal K). ❌ Not definite.

Conclusion II: H = K → H ≥ K means H might equal K, but might not. ❌ Not definite.

But: H ≥ K means H is either > K or = K. One of these MUST be true. Conclusions I and II are complementary!

Answer: (C) Either I or II follows

Example 6 — Five-Element Chain (Level: Medium)

🎯 Exam: IBPS PO PatternIntermediate

Statement: T $ U, U @ V, V % W, W © X

Conclusions: I. T £ W II. U © X

Decode:

T ≤ U = V ≥ W > X

Conclusion I: T < W → Path: T ≤ U = V ≥ W — direction changes at V (ascending then descending). ❌ Can't determine

Conclusion II: U > X → Path: U = V ≥ W > X → U ≥ W > X → U > X ✅ True

Answer: (B) Only Conclusion II follows

Example 7 — Reversed Reading (Level: Medium)

🎯 Exam: SSC CGL PatternIntermediate

Statement: D © E, F £ E, F @ G

Conclusions: I. D © G II. G £ E

Decode:

D > E, F < E, F = G

Rearrange by linking common elements: D > E > F = G (since F < E means E > F)

D > E > F = G

Conclusion I: D > G → D > E > F = G → D > G ✅ True

Conclusion II: G < E → G = F < E → G < E ✅ True

Answer: (E) Both I and II follow

Example 8 — Only One Follows (Level: Medium)

🎯 Exam: Bank PO PatternIntermediate

Statement: M © N, N % O, P £ O

Conclusions: I. M © P II. O @ M

Decode:

M > N, N ≥ O, P < O → Rearrange: M > N ≥ O > P

M > N ≥ O > P

Conclusion I: M > P → M > N ≥ O > P → M > P ✅ True

Conclusion II: O = M → From chain M > N ≥ O → M > O or M ≥ O but not M = O necessarily. Since M > N ≥ O, if N > O then M > O (strict). If N = O then M > N = O → M > O. Either way M > O. So O = M is ❌ False.

Answer: (A) Only Conclusion I follows

Example 9 — Quadratic Root Comparison (Level: Hard)

🎯 Exam: SBI PO Mains PatternAdvanced

Equation I: x² − 5x + 6 = 0

Equation II: y² − 7y + 10 = 0

Find the relationship between x and y.

Solve Eq I: x² − 5x + 6 = 0 → (x − 2)(x − 3) = 0 → x = 2 or x = 3

Solve Eq II: y² − 7y + 10 = 0 → (y − 2)(y − 5) = 0 → y = 2 or y = 5

Compare all pairs:

x	y	Comparison
2	2	x = y
2	5	x < y
3	2	x > y
3	5	x < y

Results are mixed (some <, some >, some =). Answer: Relationship cannot be established.

Example 10 — Quadratic with Clear Result (Level: Hard)

🎯 Exam: IBPS PO Mains PatternAdvanced

Equation I: x² + 7x + 12 = 0

Equation II: y² − 3y + 2 = 0

Solve Eq I: x² + 7x + 12 = 0 → (x + 3)(x + 4) = 0 → x = −3 or x = −4

Solve Eq II: y² − 3y + 2 = 0 → (y − 1)(y − 2) = 0 → y = 1 or y = 2

Compare all pairs:

x	y	Comparison
−3	1	x < y
−3	2	x < y
−4	1	x < y
−4	2	x < y

All pairs: x < y. Answer: x < y (i.e., y > x)

Sign shortcut worked here: Eq I has x² + 7x + 12 = 0 → b is positive, c is positive → both roots are negative. Eq II has y² − 3y + 2 = 0 → b is negative, c is positive → both roots are positive. Negative < Positive, so x < y without solving!

Example 11 — Six-Element Chain (Level: Hard)

🎯 Exam: SBI PO Mains PatternAdvanced

Statement: A © B, B % C, C @ D, D £ E, E $ F

Conclusions: I. A © F II. F % C III. A © D

Decode:

A > B ≥ C = D < E ≤ F

Conclusion I: A > F → Path from A to F passes through D where direction changes (> then <). ❌ Can't determine

Conclusion II: F ≥ C → F ≥ E > D = C → F ≥ E > C → F > C. Since F > C, then F ≥ C is also true. ✅ True

Conclusion III: A > D → A > B ≥ C = D → A > D ✅ True

Answer: Only II and III follow

Example 12 — Multiple Breaks (Level: Hard)

🎯 Exam: RRB PO PatternAdvanced

Statement: P £ Q, Q © R, R £ S, S © T

Conclusions: I. T © R II. Q © T

Decode:

P < Q > R < S > T

Two break points: at Q (ascending→descending) and at S (ascending→descending).

Conclusion I: T > R → T is on the right side of S (S > T), R is on the left side of S (R < S). So R < S > T — direction change at S. ❌ Can't determine

Conclusion II: Q > T → Q is left of R, T is right of S. Path: Q > R < S > T — multiple breaks. ❌ Can't determine

Answer: (D) Neither I nor II follows

Example 13 — Quadratic with Negative Coefficient (Level: Hard)

🎯 Exam: IBPS PO Mains PatternAdvanced

Equation I: 2x² + 5x − 3 = 0

Equation II: 2y² − 7y + 3 = 0

Solve Eq I: 2x² + 5x − 3 = 0 → 2x² + 6x − x − 3 = 0 → 2x(x + 3) − 1(x + 3) = 0 → (2x − 1)(x + 3) = 0 → x = ½ or x = −3

Solve Eq II: 2y² − 7y + 3 = 0 → 2y² − 6y − y + 3 = 0 → 2y(y − 3) − 1(y − 3) = 0 → (2y − 1)(y − 3) = 0 → y = ½ or y = 3

Compare all pairs:

x	y	Comparison
½	½	x = y
½	3	x < y
−3	½	x < y
−3	3	x < y

All pairs: x ≤ y (x is less than or equal to y in every case). Answer: x ≤ y

Example 14 — Complementary Conclusions (Level: Medium)

🎯 Exam: SBI Clerk PatternIntermediate

Statement: W $ X, X @ Y, Y $ Z

Conclusions: I. W £ Z II. Z @ W

Decode:

W ≤ X = Y ≤ Z

Conclusion I: W < Z → From chain: W ≤ X = Y ≤ Z → W ≤ Z. So W < Z is possible but not certain (W might equal Z). ❌ Not definite.

Conclusion II: Z = W → Also possible but not certain. ❌ Not definite.

Are they complementary? W ≤ Z means W < Z or W = Z. Conclusions I (W < Z) and II (Z = W) cover both cases. One MUST be true.

Answer: (C) Either I or II follows

Example 15 — Mixed Codes with Alternate Symbols (Level: Hard)

🎯 Exam: SBI PO Mains PatternAdvanced

Given Codes: ★ means >, △ means <, ◯ means =, ◆ means ≥, □ means ≤

Statement: A ◆ B, B ◯ C, C ★ D, D □ E, E △ F

Conclusions: I. A ★ D II. F ★ C III. A ◆ E

Decode:

A ≥ B = C > D ≤ E < F

Conclusion I: A > D → A ≥ B = C > D → A > D (since ≥ then = then > in same direction → strict >) ✅ True

Conclusion II: F > C → C > D ≤ E < F. Path from C to F: C > D ≤ E < F — direction changes. ❌ Can't determine

Conclusion III: A ≥ E → A ≥ B = C > D ≤ E — direction change at D. ❌ Can't determine

Answer: Only Conclusion I follows

Self-Test: Go back through Examples 1–15 and try solving each one BEFORE reading the solution. Time yourself — your target is 20–30 seconds per question. If you can do all 15 in under 8 minutes, you're exam-ready!

Section 10

MCQ Assessment Bank — 30 Questions (Bloom's Mapped)

Codes for Q1–Q24: © → > £ → < @ → = % → ≥ $ → ≤

Remember / Recall (Q1–Q5)

If 'P © Q' means 'P is greater than Q', what does 'P £ Q' mean?

P is greater than Q
P is less than Q
P is equal to Q
P is not equal to Q

Remember

✅ Answer: (B) P is less than Q — £ is defined as the "less than" symbol in this code set.

The symbol '≥' means:

Strictly greater than
Strictly less than
Greater than or equal to
Less than or equal to

Remember

✅ Answer: (C) Greater than or equal to — The '≥' symbol includes both the possibility of being greater and being equal.

If A > B and B > C, which property allows us to conclude A > C?

Commutative property
Transitive property
Associative property
Distributive property

Remember

✅ Answer: (B) Transitive property — Transitivity states that if A > B and B > C, then A > C. This is the foundation of chain inequality solving.

In a quadratic equation ax² + bx + c = 0, the sum of roots is:

b/a
−b/a
c/a
−c/a

Remember

✅ Answer: (B) −b/a — By Vieta's formulas, for ax² + bx + c = 0, sum of roots = −b/a and product of roots = c/a.

If A > B, then which of the following is the same statement written in reverse?

A < B
B > A
B < A
B = A

Remember

✅ Answer: (C) B < A — Reading A > B from right to left gives B < A. The inequality flips when you reverse the order.

Understand / Explain (Q6–Q10)

Statement: A © B, B @ C. Which conclusion is true?

A @ C (A = C)
A © C (A > C)
C © A (C > A)
A £ C (A < C)

Understand

✅ Answer: (B) A > C — Decode: A > B, B = C → A > B = C → A > C.

Why can't we determine the relationship between A and C if A > B and B < C?

Because A and C are not in the same equation
Because the inequality directions reverse at B, creating ambiguity
Because B is an unknown variable
Because we need a fourth variable

Understand

✅ Answer: (B) — When A > B and B < C, both A and C are greater than B, but we don't know their relationship to each other. The direction reversal at B creates ambiguity.

If X ≥ Y and Y > Z, what can we conclude about X and Z?

X = Z
X ≥ Z
X > Z
Cannot determine

Understand

✅ Answer: (C) X > Z — X ≥ Y means X is at least Y. Y > Z is strict. Combined: X ≥ Y > Z → X > Z (the strict inequality from Y > Z ensures X > Z, since X ≥ Y).

What does "Either Conclusion I or II follows" mean in coded inequality questions?

Both conclusions are true simultaneously
Exactly one conclusion is true but we can't tell which, and they are complementary
Both conclusions are false
We need more data to decide

Understand

✅ Answer: (B) — "Either-or" applies when neither conclusion is individually certain, but they are complementary (e.g., one says A > B, the other says A ≤ B), so exactly one must be true.

Q10

In quadratic root comparison, if Equation I gives x = −2, −5 and Equation II gives y = 1, 3, what is the relationship?

x > y
x < y
x = y
Cannot determine

Understand

✅ Answer: (B) x < y — All roots of x are negative (−2, −5) and all roots of y are positive (1, 3). Every negative number is less than every positive number, so x < y in all pairs.

Apply / Solve (Q11–Q15)

Q11

Statement: P © Q, Q % R, R @ S. Conclusions: I. P © S II. S $ Q

Only I follows
Only II follows
Both follow
Neither follows

Apply

✅ Answer: (C) Both follow — Decode: P > Q ≥ R = S. I: P > S? P > Q ≥ R = S → P > S ✅. II: S ≤ Q? S = R ≤ Q → S ≤ Q ✅.

Q12

Statement: A $ B, B @ C, C % D. Conclusions: I. A % D II. D $ A

Only I follows
Only II follows
Both follow
Neither follows

Apply

✅ Answer: (D) Neither follows — Decode: A ≤ B = C ≥ D. I: A ≥ D? A ≤ B = C ≥ D → direction change at B/C. Can't determine. II: D ≤ A? Same issue. Neither follows.

Q13

Statement: M © N, O £ N, O @ P. Conclusions: I. M © P II. N % P

Only I follows
Only II follows
Both follow
Neither follows

Apply

✅ Answer: (C) Both follow — Decode: M > N, O < N (i.e., N > O), O = P. Chain: M > N > O = P. I: M > P ✅. II: N ≥ P? N > O = P → N > P, and > implies ≥ ✅.

Q14

Statement: E % F, F © G, G $ H. Conclusions: I. E © H II. E © G

Only I follows
Only II follows
Both follow
Neither follows

Apply

✅ Answer: (C) Both follow — Decode: E ≥ F > G ≤ H. Wait — direction change at G for Conclusion I. Reassess: E ≥ F > G. I: E > H? E ≥ F > G ≤ H — break at G. Can't determine E vs H. ❌. II: E > G? E ≥ F > G → E > G ✅. Answer: (B) Only II follows.

Q15

Statement: J $ K, K @ L, L £ M, M © N. Conclusions: I. N £ L II. J £ M

Only I follows
Only II follows
Both follow
Neither follows

Apply

✅ Answer: (B) Only II follows — Decode: J ≤ K = L < M > N. I: N < L? N is right of M (M > N), L is left of M (L < M). L < M > N → can't determine L vs N. ❌. II: J < M? J ≤ K = L < M → J < M ✅.

Analyse / Evaluate (Q16–Q20)

Q16

Only I follows
Only II follows
Both follow
Neither follows

Analyse

✅ Answer: (C) Both follow — Decode: A > B ≥ C ≤ D < E. I: A > D? A > B ≥ C ≤ D — break at C. ❌ Can't determine. II: E > C? E > D ≥ C → E > C ✅. Corrected: Only II follows → (B).

Q17

Only I follows
Only II follows
Either I or II follows
Neither follows

Analyse

✅ Answer: (C) Either I or II follows — Decode: P ≥ Q = R ≥ S = T → P ≥ T. So P > T or P = T. Neither is individually certain, but one must be true. They are complementary → Either-Or.

Q18

Only I follows
Only II follows
Both follow
Neither follows

Analyse

✅ Answer: (D) Neither follows — Decode: W > X < Y > Z. Two direction changes. W vs Z: path goes W > X < Y > Z — multiple breaks. Can't determine. Y vs W: Y > X < W — break at X. Can't determine.

Q19

Only I follows
Only II follows
Both follow
Neither follows

Analyse

✅ Answer: (B) Only II follows — Decode: A ≥ B ≤ C = D > E. I: A ≥ E? Path: A ≥ B ≤ C — break at B. ❌. II: C > E? C = D > E → C > E ✅.

Q20

Only I follows
Only II follows
Either I or II follows
Neither follows

Analyse

✅ Answer: (D) Neither follows — Decode: L = M > N ≤ O = P. L vs P: L = M > N ≤ O = P — direction change at N. Can't determine. Not complementary either (L could be >, <, or =).

Evaluate / Apply to Complex Scenarios (Q21–Q25)

Q21

Equation I: x² − 8x + 15 = 0. Equation II: y² − 5y + 6 = 0. What is the relationship between x and y?

x > y
x < y
x ≥ y
Cannot be determined

Evaluate

✅ Answer: (C) x ≥ y — Eq I: (x−3)(x−5)=0 → x=3,5. Eq II: (y−2)(y−3)=0 → y=2,3. Pairs: 3≥2✅, 3≥3✅, 5≥2✅, 5≥3✅. All pairs: x ≥ y.

Q22

Equation I: x² + x − 6 = 0. Equation II: y² − y − 6 = 0. What is the relationship between x and y?

x > y
x < y
x = y
Cannot be determined

Evaluate

✅ Answer: (D) Cannot be determined — Eq I: (x+3)(x−2)=0 → x=−3,2. Eq II: (y−3)(y+2)=0 → y=3,−2. Pairs: −3 vs 3 (xy). Mixed results → can't determine.

Q23

Only I follows
Only II follows
Both follow
Neither follows

Evaluate

✅ Answer: (B) Only II follows — Decode: A > B, C ≤ B, C ≥ D, E < D. Rearrange: A > B ≥ C ≥ D > E. I: A > D? A > B ≥ C ≥ D → A > D ✅. II: E < B? E < D ≤ C ≤ B → E < B ✅. Both follow → (C).

Q24

Only I follows
Only II follows
Both follow
Neither follows

Analyse

✅ Answer: (D) Neither follows — Decode: M ≤ N > O ≥ P = Q. I: M < P? M ≤ N > O — break at N. ❌. II: Q ≤ N? Q = P ≤ O < N → Q < N, which means Q ≤ N ✅. Corrected: Only II follows → (B).

Create / Advanced Problem Solving (Q25–Q30)

Q25

Equation I: 3x² + 14x + 8 = 0. Equation II: 3y² + 10y + 8 = 0. What is the relationship?

x > y
x < y
x ≤ y
Cannot be determined

Create

✅ Answer: (C) x ≤ y — Eq I: 3x²+14x+8=0 → (3x+2)(x+4)=0 → x=−2/3, −4. Eq II: 3y²+10y+8=0 → (3y+4)(y+2)=0 → y=−4/3, −2. Pairs: −2/3 vs −4/3 (x>y), −2/3 vs −2 (x>y), −4 vs −4/3 (x

Q26

Only I follows
Only II follows
Both follow
Neither follows

Analyse

✅ Answer: (B) Only II follows — Decode: R > S ≤ T > U ≥ V. I: R > V? R > S ≤ T — break at S. ❌. II: T > V? T > U ≥ V → T > V ✅.

Q27

Only I follows
Only II follows
Either I or II follows
Both follow

Evaluate

✅ Answer: (C) Either I or II follows — Decode: A ≥ B = C ≥ D = E ≥ F → A ≥ F. So A > F or A = F. Neither individually definite, but complementary → Either-Or.

Q28

Equation I: x² − 12x + 35 = 0. Equation II: y² − 8y + 15 = 0. What is the relationship?

x > y
x ≥ y
x < y
Cannot be determined

Evaluate

✅ Answer: (B) x ≥ y — Eq I: (x−5)(x−7)=0 → x=5,7. Eq II: (y−3)(y−5)=0 → y=3,5. Pairs: 5≥3✅, 5≥5✅, 7≥3✅, 7≥5✅. All x ≥ y.

Q29

Only I follows
Only II follows
Both follow
Neither follows

Evaluate

✅ Answer: (D) Neither follows — Decode: G < H ≥ I > J ≥ K. I: G < J? G < H ≥ I > J — direction changes at H and I. ❌. II: K < H? K ≤ J < I ≤ H → K < H ✅. Corrected: Only II → (B).

Q30

Only I follows
Only II follows
Both follow
Neither follows

Create

✅ Answer: (B) Only II follows — Decode: P > Q ≤ R > S ≥ T = U. I: P > U? P > Q ≤ R — break at Q. ❌. II: R > U? R > S ≥ T = U → R > U ✅.

Section 11

Short Answer Questions — 8 Questions

SA-1

Define the transitive property of inequalities. Give one example.

Answer: The transitive property states that if A > B and B > C, then A > C. The relationship "carries through" the common element. Example: If Rahul is taller than Priya, and Priya is taller than Amit, then Rahul is definitely taller than Amit. Mathematically: If 8 > 5 and 5 > 3, then 8 > 3.

SA-2

Explain the difference between '>' and '≥' with an example relevant to coded inequalities.

Answer: '>' (strictly greater than) means the left value is always larger — no possibility of equality. '≥' (greater than or equal to) means the left value is larger OR could be equal. In coded inequalities, if a chain gives A > B, then the conclusion "A ≥ B" is also true (since > is a subset of ≥). But if a chain gives A ≥ B, we cannot conclude A > B because A might equal B. Example: A ≥ B, B > C. We can say A > C (definite). But A > B? Not certain.

SA-3

What happens when inequality directions reverse in a chain? Explain with an example.

Answer: When directions reverse, we cannot determine the relationship between elements on opposite sides of the reversal point. Example: A > B < C. Here, A is greater than B, and C is also greater than B. But we don't know if A is greater, less, or equal to C. Consider: If A=10, B=3, C=5 → A > C. But if A=4, B=3, C=100 → A < C. The reversal at B makes the A-to-C relationship indeterminate.

SA-4

Describe the 4-step method for solving coded inequality questions.

Answer: Step 1 — DECODE: Replace each coded symbol with its standard inequality (>, <, =, ≥, ≤). Step 2 — CHAIN: Combine all decoded statements into a single continuous chain, linking common elements. Step 3 — CHECK: For each conclusion, trace the path between the two elements. If the path flows in one direction → evaluate; if direction reverses → can't determine. Step 4 — ANSWER: Mark each conclusion as definitely true, definitely false, or can't determine, then select the appropriate answer option.

SA-5

When does the "Either Conclusion I or II follows" option apply? Give a specific example.

Answer: The "Either-Or" option applies when (1) neither conclusion is individually definite, but (2) the two conclusions are complementary — at least one MUST be true. Example: If the chain gives P ≥ Q, and Conclusion I says "P > Q" while Conclusion II says "P = Q", neither is individually certain. But since P ≥ Q means P is either greater than or equal to Q, one of the two conclusions must be true. Hence, "Either I or II follows."

SA-6

How do you compare roots of two quadratic equations? List the steps.

Answer: Step 1: Solve both quadratic equations to find their individual roots (using factorisation, formula, or sign analysis). Step 2: List all roots. Eq I gives x₁, x₂ and Eq II gives y₁, y₂. Step 3: Compare ALL four pairs: (x₁,y₁), (x₁,y₂), (x₂,y₁), (x₂,y₂). Step 4: If all pairs show the same direction (e.g., all x < y), that's the conclusion. If pairs are mixed (some x > y, some x < y), the relationship cannot be determined.

SA-7

If a chain shows A ≥ B ≥ C, can we conclude A > C? Justify.

Answer: No, we cannot conclude A > C. From A ≥ B ≥ C, we can only conclude A ≥ C. The reason: it's possible that A = B = C (say A = B = C = 5), in which case A is NOT strictly greater than C. The ≥ symbol preserves the possibility of equality through the chain. We need at least one strict inequality (>) in the chain to guarantee a strict result at the endpoints.

SA-8

In the quadratic equation x² + bx + c = 0, how can you determine the signs of the roots without solving?

Answer: Using the relationships: Sum of roots = −b and Product of roots = c (for a=1). Case 1: If b > 0, c > 0 → sum is negative, product is positive → both roots are negative. Case 2: If b < 0, c > 0 → sum is positive, product is positive → both roots are positive. Case 3: If c < 0 → product is negative → roots have opposite signs (one positive, one negative). This sign analysis is a powerful shortcut in root comparison questions.

Section 12

Long Answer Questions — 3 Questions

LA-1 (10 Marks)

Given the following coded inequalities, decode, build the chain, and evaluate all three conclusions. Show complete working.

Step 1 — Decode each statement:

Step 2 — Build the chain:

A > B ≥ C = D < E ≤ F

The chain has a direction change at D (descending from A to D, then ascending from D to F).

Step 3 — Evaluate each conclusion:

Conclusion I: A © D → A > D
Path: A > B ≥ C = D. All arrows point the same direction (left-to-right descending). A > B means A is greater than B. B ≥ C means B is at least C. C = D. Combining: A > B ≥ C = D → A > D. ✅ Conclusion I is DEFINITELY TRUE.

Conclusion II: F % C → F ≥ C
Path from F to C: F ≥ E > D = C (reading the chain from right to left: F ≥ E, E > D, D = C). So F ≥ E > D = C → F > D = C → F > C. Since F > C, then F ≥ C is also true (> is a subset of ≥). ✅ Conclusion II is DEFINITELY TRUE.

Conclusion III: A © F → A > F
Path: A > B ≥ C = D < E ≤ F. The direction changes at D (from descending to ascending). We know A > D and F > D, but we cannot determine the relationship between A and F. A could be greater, less, or equal to F. ❌ Conclusion III CANNOT BE DETERMINED.

Final Answer: Only Conclusions I and II follow.

LA-2 (10 Marks)

Solve the following quadratic equations, compare the roots of both, and determine the relationship between x and y. Verify using the sign-based shortcut method.

Equation I: x² + 9x + 20 = 0 Equation II: y² − 11y + 28 = 0

Solving Equation I: x² + 9x + 20 = 0

Find two numbers that multiply to 20 and add to 9: 4 × 5 = 20, 4 + 5 = 9.
Factorisation: (x + 4)(x + 5) = 0
Roots: x = −4 or x = −5

Solving Equation II: y² − 11y + 28 = 0

Find two numbers that multiply to 28 and add to 11: 4 × 7 = 28, 4 + 7 = 11.
Factorisation: (y − 4)(y − 7) = 0
Roots: y = 4 or y = 7

Comparing all root pairs:

• x = −4 vs y = 4 → −4 < 4 → x < y
• x = −4 vs y = 7 → −4 < 7 → x < y
• x = −5 vs y = 4 → −5 < 4 → x < y
• x = −5 vs y = 7 → −5 < 7 → x < y

All four pairs show x < y. Conclusion: x < y (i.e., y > x).

Verification using sign-based shortcut:

Eq I: x² + 9x + 20 = 0. Here b = +9, c = +20. Since b > 0 and c > 0, the sum of roots is negative (−b = −9) and product is positive (c = 20). Both roots must be negative. ✓ (We got −4 and −5.)

Eq II: y² − 11y + 28 = 0. Here b = −11, c = +28. Since b < 0 and c > 0, the sum of roots is positive (−b = 11) and product is positive (c = 28). Both roots must be positive. ✓ (We got 4 and 7.)

Shortcut conclusion: All roots of Eq I are negative. All roots of Eq II are positive. Every negative number is less than every positive number. Therefore x < y — confirmed without individual pair comparison!

LA-3 (10 Marks)

Explain the complete decision framework for coded inequality questions: when does a conclusion "definitely follow," when is it "definitely false," when "can't be determined," and when does the "either-or" case apply? Illustrate each case with a unique example.

The decision framework for coded inequality questions involves four possible verdicts for any conclusion:

Case 1 — Definitely Follows (True):
A conclusion definitely follows when the chain inequality directly supports it with no ambiguity. The path between the two elements in the conclusion must flow in one consistent direction.

Example: Chain: A > B > C > D. Conclusion: A > D.
The path A → B → C → D flows consistently downward. A > D in every possible scenario. ✅ Definitely follows.

Case 2 — Definitely False:
A conclusion is definitely false when it directly contradicts the chain. The chain proves the opposite with certainty.

Example: Chain: A > B > C. Conclusion: C > A.
The chain proves A > C. The opposite (C > A) is impossible. ❌ Definitely false.

Case 3 — Can't Be Determined:
When the path between two elements involves a direction reversal, the chain doesn't provide enough information. The conclusion might be true in some scenarios and false in others.

Example: Chain: A > B < C. Conclusion: A > C.
Scenario 1: A=10, B=3, C=5 → A > C (true). Scenario 2: A=4, B=3, C=100 → A < C (false). The direction reversal at B makes the A-C relationship indeterminate. ❓ Can't determine.

Case 4 — Either-Or:
This special case applies when TWO conclusions are individually indeterminate, but they are logically complementary — one of them MUST be true. This typically occurs when the chain gives a non-strict inequality (≥ or ≤), and the two conclusions split it into its strict and equality components.

Example: Chain: P ≥ Q (from P ≥ R = Q). Conclusion I: P > Q. Conclusion II: P = Q.
Neither is individually certain (P could be greater or equal). But P ≥ Q means exactly one of {P > Q, P = Q} must be true. They are exhaustive and mutually exclusive. → Either I or II follows.

Important notes for exam application:

1. Always check for "either-or" last — after confirming neither conclusion individually follows.
2. Two conclusions are complementary only if they cover all possibilities without overlap (e.g., > and =, or < and ≥).
3. "A > B" and "A < B" are NOT complementary because "A = B" is missing — use either-or only for complete partitions like > vs ≤, or < vs ≥, or > vs = when the chain gives ≥.
4. In exam marking: if you select "Neither" when the answer is "Either," you lose marks. This is the most commonly lost mark in banking exams on this topic.

Section 13

Chapter Summary — Quick Revision Card

🎯 Unit 7 — Coded Inequalities: Key Takeaways

1. Six Symbols: > (greater), < (less), ≥ (greater or equal), ≤ (less or equal), = (equal), ≠ (not equal).

2. Coded Symbols: Exams replace standard symbols with codes (©, £, @, %, $, ★, △, etc.). Always read the code table FIRST.

3. The 4-Step Method: DECODE → CHAIN → CHECK → ANSWER. Follow this for every question.

4. Chain Rule: Same direction = can combine. Direction reversal = chain breaks, can't determine.

5. Combining Rules: Strict + Strict = Strict. Non-strict + Non-strict = Non-strict. Strict + Non-strict (same dir) = Strict. Equality passes through.

6. Quadratic Roots: Solve both equations, compare ALL four root pairs. Consistent result = definite answer. Mixed results = can't determine. Use sign shortcut for speed.

7. Three Verdicts: Definitely True (chain supports), Definitely False (chain contradicts), Can't Determine (chain has direction break).

8. Either-Or: When neither conclusion alone is certain, but they are complementary (one MUST be true). Common with ≥ split into > and =.

9. Exam Strategy: 5 questions guaranteed in banking exams. Target: all 5 correct in under 2 minutes. This is the easiest scoring section.

Concept	Difficulty	Exam Frequency	Mastery Check
Basic inequality symbols	Easy	Every exam (foundation)	✅ Can list all 6 symbols with meanings
Symbol decoding	Easy	Every question	✅ Can decode any code table in 10 seconds
Chain building	Medium	Every question	✅ Can build chains with 4–6 elements
Direction-break detection	Medium	70% of questions	✅ Can spot breaks instantly
Either-Or identification	Hard	20% of questions	✅ Can identify complementary conclusions
Quadratic root comparison	Hard	Bank Mains exams	✅ Can solve and compare in 45 seconds
Sign-based shortcuts	Advanced	Mains level	✅ Can determine root signs without solving

Your exam action plan: Practise 10 coded inequality sets daily for 7 days. By day 7, you should be solving each set (5 questions) in under 90 seconds. This topic is a guaranteed 5/5 with practice — don't leave free marks on the table.

✅ Unit 7 complete. You are now ready to tackle Coded Inequalities in any competitive exam!

[QR: Link to EduArtha practice set — Coded Inequalities]