Coded Inequality problems are the advanced version of Mathematical Inequality questions. To solve Coded Inequality in the Reasoning Sections of exams, we must know how to solve Mathematical Inequality where direct inequality operators are used (>, >, ≤, ≤ and =). These types of questions are important for competitive exams like IBPS Clerk, SBI Clerk, SSC CGL, Placement Aptitude, IBPS PO, SBI PO, NICL AO, LIC AAO, SBI Associate Clerk, SBI Associate PO, CAT and others.
In problems on mathematical equalities, operator ‘>’ and ‘<’ has the highest priority. This is followed by ‘≥’ and ‘≤’. On the other hand, ‘=’ has the least priority. So if a statement is A > B ≥ C = D then A > C, D because ‘>’ has highest priority. B ≥ D because ≥ has priority more than ‘=’.
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Now we know how to get the relation between two elements. So, we can go ahead and solve Coded Inequality in Reasoning.
In problems on coded inequalities, operators are coded with some symbols. So we first need to decode it and then check the conclusions.
Example of Coded Inequality in Reasoning
Directions: In the following questions, the symbols δ, @, ©, % and ⋆ are used with the following meaning as illustrated below.
‘A © B’ means ‘A is not smaller than B’.
‘A % B’ means ‘A is neither smaller than nor equal to B ’.
‘A ⋆ B’ means ‘A is neither greater than nor equal to B’.
‘A δ B’ means ‘A is not greater than B’.
‘A @ B’ means ‘A is neither greater than nor smaller than B’.
Now in each of the following questions assuming the given statements to be true, find which of the four conclusions I, II, II and IV given below them is / are definitely true and give your answer accordingly.
Statements:
P δ T, T @ R, R © O, O % K
Conclusions:
I. R @ P
II. R % P
III. K ⋆ T
IV. O δ T
1) Only either I or II is true
2) Only III and IV are true
3) Only either I or II and III are true
4) Only either I or II and IV are true
5) Only either I or II and III and IV are true
Follow the steps given below to simplify the process.
Steps Involved in Solving Coded Inequality in Reasoning
Step 1: Make Decoding Table.
The easiest method is to first make a table as shown below.
NOTE: Elements used in question are A and B so we have added A and B in table.
TIP: Sometimes, to make questions more complicated, reverse relations may be given as:
‘A * B’ means ‘B is not smaller than A’.
So here we will write B in the first row and A in the last row.
Step 2: Add Symbols to Table
Step 3: Start decoding symbols one by one. Then add decoded operator into the table.
Here symbols are:
© → not smaller than → means greater than or equal to → ‘≥’
% → neither smaller than nor equal to → means greater than → ‘>’
⋆ → neither greater than nor equal to → means smaller than → ‘<’
δ → not greater than → means smaller than or equal to → ‘≤’
@ → neither greater than nor smaller than → means equal to → ‘=’
So our decoding table becomes:
We will now use this decoding table to solve the actual questions.
Step 4: Decode Statements using Decoding Table.
Statements: P δ T, T @ R, R © O, O % K
Decoded statements: P ≤ T, T = R, R ≥ O, O > K
Step 5: Combine Decoded Statements
Combined statement will be: P ≤ T = R ≥ O > K
Step 6: Conclude Individually
Look at conclusions one by one, decode each conclusion using the Decoding Table. Then check whether the conclusion follows or not.
Conclusion I: R @ P → R = P
Now from the combined statement we get, P ≤ T = R.
According to priority level we get, P ≤ R.
Thus R = P is false.
Conclusion II: R % P → R > P
From the combined statement we get, P ≤ T = R.
Thus again we get P ≤ R.
So R > P is false.
But we know from the combined statement that P ≤ R. Hence either conclusion I or II has to be true as they form complementary pair.
TIP: Whenever there are 2 conclusions having the same 2 elements and both conclusions are false, always check for complementary pair.
Complementary means combining in such a way as to enhance or emphasize the qualities of each other or another.
Like, ‘<’, ‘>’ and ‘=’ form complementary pairs for any case. This is because there cannot be any relation other than these if these three cases are included.
Conclusion III: K ⋆ T → K < T
Now from the combined statement we get T = R ≥ O > K.
This can be shortened as T ≥ O > K.
Now according to priority, > has more priority than ≥. So the final relation between T and K will be T > K.
Thus the conclusion is true.
Conclusion IV: O δ T → O ≤ T
Now from the combined statement we get, T = R ≥ O.
According to priority, ≥ has more priority than =. So the final relation between T and O will be T ≥ O.
Thus the conclusion is true.
So our final answer will be, conclusion III, IV and either conclusion I or conclusion II follows.
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