SHORT TRICK FOR INEQUALITY QUESTION IN REASONING SECTION

Hello Aspirants,

in several competitive examinations generally Inequality questions in reasoning section asked. specially all IBPS, RBI, LIC, NICL, OICL CTET,SSC etc. exams 5 questions comes from this topic. in these questions some statements are given on based you have to find out the right conclusions. Here we explain SHORT TRICK FOR INEQUALITY QUESTION REASONING SECTION.

SHORT TRICK FOR INEQUALITY QUESTION REASONING:

There are Two type of inequality question asked.

  1. Indirect inequality 
  2. Direct inequality
TYPE 1 : Indirect inequalities

In these question inequality given indirect way. you need to first make inequality with help of given information then solve it.

LIKE
‘P # Q’ means ‘P is neither greater than nor equal to Q’.
‘P © Q’ means ‘P is neither equal to nor smaller than Q’.
‘P % Q’ means ‘P is neither smaller than nor greater than Q’
‘P $ Q’ means ‘P is not smaller than Q’.
‘P @ Q’ means ‘P is not greater than Q’.
When this type of information is given, first write on the paper what these symbols mean, like here
# means <
© means >
% means =
$ means
@ means

Example statement: L $ T, T % P, K © P
Now write the relation between elements in a single line by checking the above meanings of symbol as L ≥ T = P < K [Here K > P, but to write in a single line we will write as P < K]

Conclusions:         I. P @ L                    II. L © K                   III. L @ K
I – L ≥ T = P means L ≥ P which is conclusion I P ≤ L, so I is true.
II – L ≥ T < K, so we know there is no relationship between L and K, so II if false.
III – L ≥ T < K, so we know there is no relationship between L and K, so III if false.
But II and III make either or pair, so answer is – I and either II or III follow.

TYPE 2: Direct Inequality

In these questions where direct inequalities are given in the statement itself, you  need not form the relationship between elements like in above example. In these we will make relationship between elements in which conclusion is to be find.

Example Statement: A < P ≤ Q, L > Q < K, P ≥ O
Conclusions:
       I. K ≥ O               II. L > O
K > Q ≥ P ≥ O, so K > O, so I is false
L > Q ≥ P ≥ O so L > O, so II is true

SHORT TRICK FOR DIRECT INEQUALITY:

We discuss SHORT TRICK FOR INEQUALITY QUESTION REASONING. So first we start with Direct Inequality. here we discuss some short tricks to solve any kind of inequality question whether it is one statement or multiple statements.

REMEMBER: Maximum Three possible relation between two elements.

For example between A and B.  A>B,A<B,A=B  OR  A≥B,A<B OR  A>B, A≤B

PREFERENCE RULE 1: If we find these sign > ,≥ ,= between Two elements not in same order. Then we will give preference to them such as 1.> 2. ≥ 3. = . Remember not always all Three signs, may be only two sign also possible.

Examples: I. statement: A≥B>D=C     conclusion: A>C (because > has first preference)

II.statement: A≥B=C  conclusion: A≥C.

III.statement: A≥D=B≥C conclusion: A≥c

PREFERENCE RULE 2: If we find these sign < ,≤ ,= between Two elements not in same order. Then we will give preference to them such as 1.< 2. ≤ 3. = . Remember not always all Three signs, may be only two sign also possible.

Examples: I. statement: AB<D=C     conclusion: A<C (because < has first preference)

II.statement: AB=C  conclusion: AC.

III.statement: AD=BC conclusion: Ac

RULE 3: if at the same time <,> or ≥,< or >,≤ or ≤,≥ signs come together between two elements then no relation occurred between them  but either , neither case may be occur.

example. A>C=D<B. no conclusions between A and B because there is doubtful case. But conclusions may be A<B OR A>B OR A=B.(LIKE 6>4=4<7 , 6>4=4<5 , 7>4=4<7). Remeber between A and D there is conclusion A>D.

FRIENDS YOU HAVE ANY PROBLEM TO UNDERSTAND THESE SHORT TRICKS PLEASE WATCH THESE VIDEO AND LIKE . PLEASE SUBSCRIBE OUR YOUTUBE CHANNEL. WE ALSO PROVIDE OTHER TOPICS SHORT TRICKS BY THIS CHANNEL.

 

INEQUALITY EITHER OR AND NEITHER NOR CASE: 

As SHORT TRICK FOR INEQUALITY QUESTION REASONING we know there is only Three possible relation between two elements. Here we would take different examples to understands the either neither conditions.

Example I. statement: A≥C=B conclusions: 1. A>B 2. A=B 3. A<B. Solution. as we know by preference rule 1 possible relation is A≥B. So either 1 or 2 is possible. Because  A≥B means A>B or A=B.

Example II. statement: A≥C<B  conclusion: 1.A>B 2. A<B 3. A=B 4. Either 1 or 2 or 3. solution. no rule working on that but we know maximum Three possible relation can be between A and  B that are given in option 1,2,3. so either 1 or 2 or 3 is True.

Example III. statement: A>C<B conclusion: 1.A≥B 2.A<B 3.A=B 4. Either 1 or 2. solution. because 1 and 2 option are making all possible 3 relations So either 1 or 2 is correct.

Example IV: statement: A<C>B conclusions: 1. A>B 2. A<B 3. either 1 OR 2 4. neither 1 or 2. solution. Here no preference rule would work and option 1 ,2 together not making all possible Three relation, A=B is missing. So option 4 is correct neither 1 or 2.

HOW TO COMBINING TWO INEQUALITIES: 

We can combine Two or more inequality if they follow Two condition.

  1. If they have common element.
  2. If the common element is >,≥(greater than, OR greater than equal to) to one and <,(less than OR less than equal to) to other. Means simply make preference rule.

Example I. B<A , B≥C   =>   A>B≥C

II. A≥B , B<C     =>   NOT POSSIBLE

III. B≥A , C≥B    =>  A≤B≤C

IV. A≥C>B>D , P<F≥C>E  conclusions: 1. A≥E 2.F>D.  in these inequality C is common in both. Remember we need to find relation between A and E , in second option F and D. Between A and E => A≥C>E SO BY PREFERENCE RULE 1 A>E means option 1 is wrong. Now between F and D => F≥C>B>D SO by preference rule 1 F>D option 2 is correct.

V. I.P<X≤Y<Q  ,  II.S>Y<T , III.P=V>R  conclusions: 1. V<S  2. T>R  Solution. first find relation between V and S  V,S given in II and III Inequality but no common element between II and III. so by I and III R<P=V<X≤Y<Q , now combine it with II inequality so P=V<X≤Y<S. by preference rule V<S (option 1 correct) for T,R relation combine R<P=V<X≤Y<Q and S>Y<T ==> T>Y≥X>V=P>R  so T>R option 2 also correct.

VI. I. S>U>V  II. Y<U<Z  III. Z<X>W conclusions. 1. S<Z 2. X>Y

solution. for S,Z relation combine I and II S>U<Z ==> NO RELATION option 1 incorrect. for X,Y Relation III and II combine X>Z>U>Y ==>X>Y option 2 is correct.

SHORT TRICK FOR INDIRECT INEQUALITY:

In these type of inequality first convert statements and conclusions in direct inequality form. Then simply solve as Direct Inequality questions.

PRACTICE MORE QUESTIONS IN BELOW LINK.

 

 

BEST OF LUCK ASPIRANTS TRY TO PRACTICE MORE PROBLEMS.IF YOU FIND ANY DIFFICULTIES PLEASE COMMENT.

Leave a Reply

Your email address will not be published.