# Coded Inequality – Tips and Tricks to Solve Questions in Logical Reasoning

Wednesday, February 6th, 2019 Coding inequalities: Inequality is a phenomenon that we have known or come across for a while now. It simply means something that is not equal in all the sense whether social, economical or mathematical sense. But today we are talking about Inequalities in mathematical terms. Each one of you has seen signs such as “>”, “”, “” etc. at some point of time or other. And all those who had math’s in their high school are thoroughly acquainted with the concept and comfortable in solving problems of Inequalities. So, for all those who have little or no idea about them, I will give you a brief introduction about this concept that will be sufficient to solve problems which is the essence of the discussion of this article.

 Signs used to denote Inequality Meaning If a ≠ b, then ≠ denotes not equal to, i.e. a is not equal to b If a ≤ b, then ‘≤’ denotes less than or equal to, i.e. a is less than b or at most b If a ≥ b, then ‘>’ denotes greater than or equal to, i.e. a is greater than b or at least b Strict Inequalities If a b, then ‘>’ denotes greater than, i.e. a is greater than b

Properties of Inequalities

 Property ≥ ≤ Addition If a ≥ b, then a + c ≥ b + c If a ≤ b, then a + c ≤ b + c Subtraction If a ≥ b, then a – c ≥ b – c If a ≤ b, then a – c ≤ b – c Multiplication If a ≥ b, then ac ≥ bc, where c>0 If a ≤ b, then ac ≤ bc, where c>0 Division If a ≥ b, then a/c ≥ b/c, where c>0 If a ≤ b, then a/c ≤ b/c, where c>0 Transitivity If a ≥ b and b ≥ c then, a ≥ c If a ≤ b and b ≤ c then, a ≤ c Inverse         ·Additive If a ≥ b, then -a ≤ -b, if a>0, b>0 If a ≤ b, then -a ≥ -b, if a>0, b>0 ·Multiplicative If a ≥ b, then 1/a ≤ 1/b, if a>0, b>0 If a ≤ b, then 1/a ≥ 1/b, if a>0, b>0

The above-mentioned properties also hold for strict inequalities.

Now let’s move forward to our topic of interest. Consider the following question,

A α B means A is greater than B

A β B means A is either greater than or equal to B

A γ B means A is equal to B

A δ B means A is small than B.

A η B means A is either smaller than or equal to B

Now on the basis of the statement given below be true, find which of the following conclusion holds to be true?

Statement: E γ F, C δ D, F β G, D α E

Conclusion 1: E α G

Conclusion 2: C γ E

• If only conclusion 1 holds
• If only conclusion 2 holds
• If either 1 or 2 is true
• If neither 1 nor 2 is true
• If both 1 and 2 are true.

If you carefully examine the above question you can find out that symbols are used to represent various inequalities and with the support of these inequalities conclusions are framed which defines a relationship between two variable and you have to judge whether these relationships are true or False. In summary, you have to decode the above coded inequalities and determine their validity. These kinds of questions come in various competitive exams such as Bank PO, SSC, CAT etc. These questions can prove to be effortless and simple to solve if you now the right technique and methods to do so otherwise, they can be perplexing and time-consuming. In this blog, I will provide you straightforward and direct methods and steps on how to approach these types of problems and solve them rapidly.

Step 1: Accurately decipher: You should make sure you correctly and carefully write the meaning of the code either along with it or can make a separate table for it. To avoid mistakes, make sure that you take one code at a time and replace it with it’s original meaning everywhere then move on to the next one. For example, E γ F means E is equal to F. Thus, replace γ with = everywhere in the statement and conclusions wherever the code exists.

Step 2: Relevancy of statements for assessing the conclusion: In this step, what needs to be done is that we pick one conclusion and then judge which statements are admissible for evaluating the conclusion. And admissible statement implies that the statement is not incompetent for deriving the particular conclusion. To check that for example, if there’s a conclusion a<b, then a statement is present such as p < q is meaningless as neither x nor y has been mentioned in it. For any conclusion, the relevant statements are those that can be combined to prove or disprove the conclusion. To determine that, take the two terms of a given conclusion and see if each of them separately appears with a single common term in the given statements in the question. These statements are relevant ones. For example, in the above question take conclusion 1 i.e. E α G, now inspect the given statement in the question. You can notice that both E and G are linked with common term F in the following statements

E γ F, F β G

Hence above statements are “relevant statements” for the given conclusion.

Step 3: Combine the relevant statements and develop the conclusion from it.

Now, how to combine the statements?  There are 3 golden rules to do so.

Rule 1: There must be a common term which is present in the above example i.e. F

Rule 2: The common term must be less than (or equal) to one term and greater than (or equal to) the other term.

Note 1: You can check this instantly be following the visual rule. If the common greater than (or equal to) one term and less than (or equal to) another term; then the pointed side of the ‘>’ sign will be towards the common term in one case and away from it in the other. So, the combination is possible in such cases only. In all other cases, the pointed sign will be either towards the common term or away from it in both the cases

Rule 3: The conclusion inequality is obtained by letting the common term disappear and it has a ‘≥’ or ‘≤’ sign if and only if both inequalities in the second step had a ‘≥’ or a ‘≤’ sign. In all other cases, there will be a ‘>’ or ‘<’ sign in the conclusion.

When you are done with all these above steps and the conclusion is established and verified; well and good. If not, then follow these additional checks.

Check 1: Examine if the given conclusion directly follows from any one single statement.

Check 2:  Analyze if the conclusion inequality you reach is essentially the same as the given conclusion but written differently.

Check 3: If your derived conclusion (or given a statement) is of type A ≥ B (or A ≤ B) then check if the two conclusions are A > B or A = B (or, A < B and A = B). If yes, choice “either follows” is true.

Check 4: If neither of the conclusion has been proved correct till now, check if the given conclusions form a complementary pair. They form a complementary pair in the following 4 cases.

• A ≥ B and A <B
• A > B and A ≤ B
• A ≤ B and A > B
• A < B and A ≥ B

Note 2: Check 3 and check 4 are required only if both conclusions have the same pair of terms.

Note 3: Two terms can never be equal if in the original statements these two terms won’t appear in separate equalities. It means that P = Q may hold only if there are two statements such as P = R. Q = S present in the problem. Otherwise, if no such statement is there in the question then it is definitely false.

Therefore, with the help of above steps, we can straightaway solve the problem quoted earlier.

Step 1:  E = F, C < D, F ≥ G, D > E

Conclusion 1: E > G

Conclusion 2: C = E

Step 2: We take conclusion 1 and as stated earlier F is the common term between E and G. And their relationship is as follows: E = F, F ≥ G

Step 3: Combining the above two statements by using the three golden rules we get, E = F ≥ G i.e.  E ≥ G which doesn’t satisfy the given conclusion. Thus, conclusion 1 doesn’t hold.

Let’s perform both these two steps for conclusion 2.

Step 2: Relevant statements for conclusion 2 are C < D and D > E.

Step 3: We cannot combine inequalities here as D is greater than both C and E. Check 3 and 4 are futile here because of Note 2. Also, check 1 and check 2 cannot prove anything concrete and doesn’t bring any solution. Thus, conclusion 2 is also invalid.

Therefore, the correct answer to the question is neither of the conclusion is true. I believe the above method would make solving similar questions easily and smoothly. Though, you will get more clarity on how to use the above checks precisely with time and practice.

### CAT Questions related to Logical Reasoning

All questions from CAT Exam Logical Reasoning
Logical Reasoning – Set 1: A high security research lab requires the researchers to set a pass key sequence based on the scan of the five fingers of their left hands.
Logical Reasoning – Set 2: Eight friends: Ajit, Byomkesh, Gargi, Jayanta, Kikira, Manik, Prodosh and Tapesh are going to Delhi from Kolkata
Logical Reasoning – Set 3: In an 8 X 8 chessboard a queen placed anywhere can attack another piece if the piece is present in the same row
Logical Reasoning – Set 4: A tea taster was assigned to rate teas from six different locations – Munnar, Wayanad, Ooty, Darjeeling, Assam and Himachal.
Logical Reasoning – Set 5: Four cars need to travel from Akala (A) to Bakala (B). Two routes are available, one via Mamur (M) and the other via Nanur (N).
Logical Reasoning – Set 6: A new airlines company is planning to start operations in a country.
Logical Reasoning – Set 7: In a square layout of size 5m × 5m, 25 equal sized square platforms of different heights are built.
Logical Reasoning – Set 8: There are 21 employees working in a division, out of whom 10 are special-skilled employees (SE) and the remaining are regular skilled employees (RE).
Logical Reasoning – Set 9: Healthy Bites is a fast food joint serving three items: burgers, fries and ice cream.

You can also see: Geometry Basics for CAT – Triangle related questions and problems

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### One response to “Coded Inequality – Tips and Tricks to Solve Questions in Logical Reasoning”

1. Raje says:

Well understood

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