# Reasoning Notes: Inequalities

Reasoning Notes: Inequalities
INTRODUCTION
In these questions, you will be provided with a statement consisting of a group of elements. These elements will be having a certain coded relationship among them which is denoted by different inequality symbols, “=, >, <, ≥, ≤”.
You should know the meaning of different symbols which will help you in finding the relationship between the different elements of a statement.
For the sake of convenience, the relationship between certain statements and their meaning is given below in a tabular form. Now we are clear about the meaning of different symbols. Check the below table for the relationship gives Conclusion for Statements. Types of Questions asked in Inequality
Now a days, inequality based questions are provided in two types
a) Direct Inequality in which direct symbols will be given in the statement.
b) Coded Inequality in which coded symbols (like @, %, \$ etc) will be given and they signify will be provided separately.
Both kinds of questions can be solved easily once you have gone through the above tables.
Inequalities Golden Rules
Rule 1: The combination between two inequalities can be established, if they have a common term.
For e.g.
(i) A > B, B > C – combination can be easily established as: A > B > C. Here we can make conclusion – A > C or C < A
(ii) A < B, B < Q - combination can be easily established as: A < B < Q. Here we can make conclusion – A < Q or Q > A
(iii) > B, B > C combination can be easily established as: A > B > C. Here we can make conclusion – A > C or C < A
Rule 2: The combination between two inequalities cannot be established, if they don’t have a common term.
For e.g.
(i) A > B, B < C – combination cannot be established. (Here relationship between A & C cannot be established.)
(ii.) A <B, B > D - combination cannot be established. (Here relationship between A & C cannot be established.)
(iii.) > B, B < C combination cannot be established. (Here relationship between A & C cannot be established.)
Rule 3: The combination between two inequalities can be established, if and only if the common term is greater than (or ‘greater than or equal to’) one and less than (or ‘less than or equal to’) the other.
For e.g.
(i) > B, C < B. (Here common term B is less than or equal to one term A, and greater than other term C. So here combination between the elements can be easily established.)
> B, C < B
Possible inequality - A > B > C or C < B < A
Note: Here we can make conclusion as: A > C or C < A
(ii) > Q, Q < C – (Here common term Q is less than both the term, so combination between the elements cannot be established.)
(iii) > M, L > N (Here common term N is greater than or equal to one term M, and less than other term L. So here combination between the elements can be easily established.)
> M, L > N
Possible inequality - L > N > M or M < N < L
Note: Here we can make conclusion as: L > M or M < L
(iv) > M, N > L (Here common term N is greater than both the term, so combination between the elements cannot be established.)
If we combined the inequality – L ≤ N > M; so here we cannot make combined inequality.)
Rule 4: Complementary Pairs: (Either & or) – Either and or cases only takes place in complementary pairs. We cannot combine two elements with common elements in which no relation is established.
For e.g.
> B, B < C
Solved Examples:
1. Statement: P ≥ R > Q = T ≥ S
Conclusions:
I. P ≥ Q → False
II. P > Q → True
III. Q ≥ S → True
2. Statements: A > B > F > C; D > E > C
Conclusions:
I.    C < A → True
II.   C > A → False
Directions (3 – 5): In the following questions, the symbols \$,@,%, © and # are used with the following meanings as illustrated below:
'P©Q' means 'P' is greater than 'Q'.
'P%Q' means 'P' is smaller than 'Q'.
'P@Q' means 'P' is either greater than or equal 'Q'.
'P\$Q' means 'P' is either smaller than or equal to 'Q'.
'P#Q' means 'P' is equal to 'Q'.
A) If only conclusion I is true.
B) If only conclusion II is true.
C) If either conclusion I or II is true.
D) If neither conclusion I nor II is true.
E) If both conclusions I and II are true
3. Statements: K@B,  B#J,  J©T
Conclusions: I. K#T  II. B@T
Solution:
K ≥ B = J >T, K > T. Hence, Conclusion I is not true. B >T. So, conclusion II is not true.
4. Statements: F\$M,  M@L,  L#W
Conclusions: I. W\$M  II. F@L
Solution:
F ≤M ≥ L = W, M ≥ W. By conversion W ≤ M. Hence, conclusion I is true. We can't compare F and L. Hence, conclusion II is not true.
5. Statements: R #Q,  Q@F,  F%A
Conclusions: I. R ©A  II. R#F
Solution:
R = Q ≥ F < A We can't compare R and A. Hence, conclusion I is not true. R ≥ F. Hence, conclusion II is not true.
Directions (6 – 10): In the following questions, the symbols \$,@,%, © and # are used with the following meanings as illustrated below:
'P©Q' means 'P' is greater than 'Q'.
'P%Q' means 'P' is smaller than 'Q'.
'P@Q' means 'P' is either greater than or equal 'Q'.
'P\$Q' means 'P' is either smaller than or equal to 'Q'.
'P#Q' means 'P' is equal to 'Q'.
A) If only conclusion I is true.
B) If only conclusion II is true.
C) If either conclusion I or II is true.
D) If neither conclusion I nor II is true.
E) If both conclusions I and II are true
6) Statements: M @ R, R ©F, F#L
Conclusions: I. R@L II.M@L
Solution:
(D) M ≥ R>F=L. So, R>L. Hence, conclusion I is not true. Even, the Conclusion II is not true.
7) Statements: T % J, J @ V, V # W
Conclusions: I. T©W II. W@T
Solution:
(C) T < J ≥ V= W Either I or II follows.
8) Statements: J @ D, D\$ L, L#N
Conclusions: I. J # L II. J \$ L
Solution:
(D) J ≥ D ≤ L = N Both the conclusions are not true
9) Statements: R \$ M, M%H,H\$F
Conclusions: I. R % F II. M\$F
Solution:
(A) R ≤ M < H ≤ F. Hence, R< F. Conclusion I is true. As M< F conclusion II is not true.
10) Statements: K \$ H, H % I, I © F
Conclusions: I. K \$ I II.H % F
Solution:
(D) K ≤ H < I > F. As K< I, conclusion I is not true. H and F can't be compared. Hence, conclusion II is not true
Directions (11 – 15): In these questions, relationship between different elements is shown in the statements. These statements are followed by two conclusions. Mark answer
a) If only conclusion I follows.
b) If only conclusion II follows.
c) If either conclusion I or II follows.
d) If neither conclusion I nor II follows.
e) If both conclusions I and II follow.
11. Statements: A > B ≤ C = D ≤ E, C ≥ F = G > H
Conclusions: I. G ≤ E II. A > H
Solution: In conclusion I relation is asked between G and E. So, we will try to find relation between it and we can see that G and E are in different statements. So, first we will identify the element which is common in both statements given i.e. C.
C = D ≤ E and C ≥ F = G
Let’s combine these both relation
G = F ≤ C = D ≤ E
Hence, conclusion I follows.
In conclusion II, relation is asked between A and H, C is also between two statements.
A > B ≤ C and C ≥ F = G > H
Combining these,
A > B ≤ C ≥ F = G > H. In this, C > H but relation between A and C cannot be defined as relation is changed at B.
So, Conclusion II does not follow.
How to approach this in Exam without taking more time?
You should check that inequality sign should follow in same direction like in statement 1 from C to E signs are in same direction and In statement 2 from G to C are also following in same direction.
So this follows.
But in case of A and H, direction of sign between A and C changes at B. It discontinued the relation between A and C, from C to H relation are in same direction. So conclusion does not follows.
12. Statements: H ≥ T > S ≤ Q, T ≥ U = V > B
Conclusions: I. V > S II. B ≤ H
Solution: In statements, T element is common.
For relation between S and V, sign changes at T itself. So this does not follow.
For relation between B and H, sign does not changes
i.e. H ≥ T ≥ U = V > B but you can see that sign between V and B is ‘>’. In conclusion H ≥ B
So, it also does not follow.
13. Statements: F < K ≤ L, H ≥ R > K
Conclusions: I. H > L II. R > F
Solution: The element common between given statements is K
For conclusion I: K ≤ L and H ≥ R > K
combining these, H ≥ R > K ≤ L
For H to L, relation is discontinued at K. So it does not follow.
For conclusion II: R > K and F < K or K > F
combining these, R > K > F
Hence this follows.
14. Statements: N ≥ P > K = L, P ≤ Q < Z, T > K
Conclusions: I. N < Q II. Z > T
Solution: There are three statements in statement 1 and 2 P is common, statement 1 and 3 K is common.
Conclusion I: Relation is asked between N and Q
N ≥ P and P ≤ Q, i.e. N ≥ P ≤ Q
We can see that, directions of signs are changed at P.
So it does not follow.
Conclusion II: Now, we have to connect all three statements.
P > K, P ≤ Q < Z  and T > K
K < P ≤ Q < Z and  T > K
In this also direction of sign changes at K.
It does not follow.
15. Statements: P < H = O ≥ N, E ≥ H < S
Conclusions: I. N ≤ E II. S > P
Solution: In statement 1 and 2, element H is common.
Conclusion I: In statement 1 and 2, H = O ≥ N , E ≥ H
Combining these, E ≥ H = O ≥ N
Hence E ≥ N follows.
Conclusion II: In statement 1 and 2, P < H and H < S
Combining these, P < H < S
Hence S > P.