How to tackle Inequalities in Banking exams

How to tackle Inequalities in Banking exams

5 marks in 1 min

There is no need of pen and paper in solving inequality. You need to only give some time to read this article and brush up your inequality concept.

       If you are already familiar with inequality topic, but not forget there is some special cases in inequality viz. Either or . So connect with us and continue read this article and enjoy your confirm 5 marks in exam.

Firstly, I introduce the concept of open/ close gate to all of you by which we solve inequality questions.

Open gate ->   >
close gate ->    <
both side open ->    =

If we need to go one place to another place than we need to all Gate should be open in path of both houses.

Secondly, Family members of Inequalities
i) <, > are oldest members
ii) ≤, ≥ are middle members
iii) = is smallest member

Importance,
i) = (Less important)
ii) ≥, ≤ (More important)
iii) <, > (Most important)

Take examples :-
statement :
A > B > C ≥ D > E = F > G < H ≤ I < J
conclusion :
i) A > D
ii) F < I
iii) C ≥ G
iv) D = F

Note :- when we solve this questions , we always move from open Gate side.

Take first conclusion,
                          We go from A to D
                          And A > (oldest member) D
Note :- If there is oldest member between two element, we don’t care about all members of family. But oldest member is also present in statement.
so, we start to go from A in statement
=> A open Gate& oldest member B (reached) open Gate& oldest member C (reached) open Gate & middle member D (Reached)
so, first conclusion is right.

Take second conclusion,
                          We go from I to F
                          And F < (oldest member) I
so, we start to go from I in statement
=> I open Gate& middle member H (reached) open Gate& oldest member G (reached) close Gate & oldest member F (Not reached)
so, Second conclusion is wrong.

Take Third conclusion,
                          We go from C to G
                          And C < (Middle member) G
Note :- If there is middle member between two elements, we don’t care about smallest member but we care about oldest members of family. If there is middle member between two elements, oldest member can’t present in statement for true condition.
So, we start to go from C in statement
=> C open Gate& middle member D (reached) open Gate& oldest member E (reached) Both side open & smallest member G (reached)
so, Third conclusion is wrong.

Note :- Here we reached G but conclusion is wrong, Because I said earlier If there is middle member is present than oldest member can’t present in statement and oldest member is present in statement.

Take Fourth conclusion,
                          We go from D to F or We go from F to D
                          And D = (Smallest member) F
Note :- If there is smallest member between two elements, we care about middle and oldest members of family. If there is smallest member between two elements, oldest member and middle member can’t present in statement for true condition.
So, we start to go from D in statement
=> D open Gate& oldest member E (reached) Both side open & smallest member F (reached)
so, Fourth conclusion is wrong.

Note :- Here we reached F but conclusion is wrong, Because I said earlier If there is smallest member is present than oldest member and middle member can’t present in statement and oldest member is present in statement.

Special cases viz. Either or discusses in detail in next post. please connect with us. Thankyou very much.

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