5.2 Coded & Mathematical Inequalities

Coded & Mathematical Inequalities

Mathematical Inequalities and Coded Inequalities form one of the highest-yield scoring topics in the SBI PO examination. In Prelims, candidates routinely encounter 5 direct or coded inequality questions that must be solved in under 2 minutes. In Mains, inequalities appear as conditional statements, blank-filling expressions, or reverse inequalities. Success depends on mastering the operator priority hierarchy and using a symbol priority box to avoid rewriting full expressions.

---

Operator Priority Hierarchy

Mathematical inequalities are governed by five basic symbols: Greater than ($>$), Less than ($<$), Greater than or equal to ($\ge$), Less than or equal to ($\le$), and Equal to ($=$). These symbols are categorized into three distinct priority levels:

Chain Continuity Rule

For a conclusion between two variables $X$ and $Y$ to be valid, all operators along the path from $X$ to $Y$ must point in the same direction (either all left-facing $> , \ge, =$ or all right-facing $<, \le, =$).

• Unbroken Chain Example: $A \ge B = C > D \ge E \implies A > E$ (Valid, Priority 1 operator $>$ dictates the relationship).
• Broken Chain Example: $A \ge B > C \le D = E \implies$ Relation between $A$ and $D$ is Undetermined due to opposite operators ($>$ and $\le$).

---

The Symbol Priority Box Technique

In Coded Inequalities, standard operators are replaced by arbitrary symbols (e.g., $@, #, $, %, &$). Rather than wasting time converting every symbol in every option back into standard math notation, construct a Symbol Priority Box.

   [ Top Left: > ]     --- (Priority 1) ---     [ Top Right: < ]
          |
   [ Mid Left: >= ]    --- (Priority 2) ---     [ Mid Right: <= ]
          |
                     [ Center: = ] (Priority 3)

Constructing the Box from Statements

Suppose the question gives the following code key:

• '$P @ Q$' means '$P$ is not smaller than $Q$' $\implies P \ge Q$ (Mid Left)
• '$P # Q$' means '$P$ is neither smaller than nor equal to $Q$' $\implies P > Q$ (Top Left)
• '$P $ Q$' means '$P$ is neither greater than nor smaller than $Q$' $\implies P = Q$ (Center)
• '$P % Q$' means '$P$ is not greater than $Q$' $\implies P \le Q$ (Mid Right)
• '$P & Q$' means '$P$ is neither greater than nor equal to $Q$' $\implies P < Q$ (Top Right)

Rules for Box Navigation

1. Forward Movement (Left to Right): If reading variables in original left-to-right order (e.g., $A \to B \to C$), retain the operator's exact position in the box.
2. Backward Movement (Right to Left): If reading variables backwards (e.g., $C \to B \to A$), flip the operator across the box diagonal (Top Left $# [>]$ flips to Top Right $& [<]$; Mid Left $@ [\ge]$ flips to Mid Right $% [\le]$).
3. Combining Symbols: The resultant symbol between two variables is the highest priority symbol along the path. If symbols lie on opposite sides of the box (e.g., Top Left and Top Right), the chain is broken.

---

Definite vs. Either-Or Cases

An inequality conclusion pair can result in an 'Either-Or' determination under two distinct structural conditions:

Type 1: Definite Relation Split

Occurs when the combined unbroken chain yields a Priority 2 relation ($\ge$ or $\le$), and the two conclusions split this relation into its strict components.

• Given Chain Result: $A \ge B$
• Conclusion I: $A > B$
• Conclusion II: $A = B$
• Evaluation: Since $A$ must be either strictly greater than $B$ OR equal to $B$, the pair forms a Type 1 Either-Or.

Type 2: Broken Chain Full Coverage (No-Relation Either-Or)

Occurs when the chain between two variables is broken (opposite operators), making all relations undetermined, BUT the two conclusions collectively cover all three possible mathematical states ($> , <, =$):

• Given Chain Result: $A$ and $B$ have a broken chain ($A$ ? $B$).
• Conclusion I: $A > B$
• Conclusion II: $A \le B$ (which incorporates both $<$ and $=$).
• Evaluation: Because $A$ must be either greater than, smaller than, or equal to $B$, and these two statements cover all three possibilities without overlap, the pair forms a Type 2 Either-Or.

---

Advanced Mains Patterns: Blank-Filling Inequalities

In Mains exams, you are presented with an incomplete expression containing blanks, such as: $M \text{ \_\_\_ } N \text{ \_\_\_ } O \text{ \_\_\_ } P \text{ \_\_\_ } Q$

You are asked to find the sequence of operators from the options that simultaneously satisfies two target conditions, e.g., Condition 1: $M > O$ is Definitely True, and Condition 2: $N \le O$ is Definitely False.

Step-by-Step Solution Algorithm

1. Isolate First Condition ($M > O$): Examine the first two blanks between $M, N,$ and $O$. To make $M > O$ definitely true, the sequence of the first two operators must contain at least one Priority 1 left-facing operator ($>$) and no right-facing operators ($<, \le$). Eliminate options violating this.
2. Isolate Second Condition ($N \le Q$ Definitely False): Examine the remaining blanks between $N, O, P,$ and $Q$. For $N \le Q$ to be definitely false, either $N > Q$ must be definitely true OR the relationship must be completely broken such that $N \le Q$ cannot be proven.
3. Test Remaining Candidate Sequences: Plug candidate sequences into the full chain to verify both conditions.

---

Step-by-Step Worked Example

Decoded Symbol Key:

• $# \implies >$
• $@ \implies \ge$
• $$ \implies =$
• $% \implies \le$
• $& \implies <$

Given Statements: $H @ I \$ J \# K \% L$

Combined Decoded Expression: $H \ge I = J > K \le L$

Conclusions to Evaluate:

• Conclusion I: $H # J$ ($H > J$)
• Conclusion II: $H $ J$ ($H = J$)
• Conclusion III: $I # K$ ($I > K$)
• Conclusion IV: $H # L$ ($H > L$)

Step-by-Step Evaluation

1. Evaluate $H$ vs $J$: The path from $H$ to $J$ is $H \ge I = J$. Combining $\ge$ and $=$ yields $H \ge J$. This is Priority 2. Conclusion I ($H > J$) and Conclusion II ($H = J$) split this Priority 2 relation. Therefore, Conclusion I or II follows (Type 1 Either-Or).
2. Evaluate $I$ vs $K$: The path from $I$ to $K$ is $I = J > K$. Combining $=$ and $>$ yields $I > K$ (Priority 1). Conclusion III states $I # K$ ($I > K$). Therefore, Conclusion III is Definitely True.
3. Evaluate $H$ vs $L$: The path from $H$ to $L$ is $H \ge I = J > K \le L$. The chain breaks between $J$ and $L$ due to $>$ and $\le$. Conclusion IV ($H > L$) cannot be determined. Therefore, Conclusion IV is False.

Final Result: Either Conclusion I or II follows, and Conclusion III follows definitely.