How to Solve Inequalities(Direct and Coded)- Preparation Tips and Tricks

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How to Solve Inequalities(Direct and Coded)- Preparation Tips and Tricks

How to Solve Inequalities(Direct and Coded)- Preparation Tips and Tricks.Welcome to the www.studywale.co online Study Portal. If you are preparing for SBI Clerk, SBI PO and NABARD, SSC CGL, SSC CHSL, MTS exams 2017-18, you will come across a How to approach Inequalities. Here we are providing you some Tips and Tricks “How to Solve Inequalities(Direct and Coded)- Preparation Tips and Tricks”.

This “How to Solve Inequalities(Direct and Coded)- Preparation Tips and Tricks” is also important for other banking exams such as SBI PO, IBPS Clerk, SBI Clerk, IBPS RRB Officer, IBPS RRB Office Assistant, IBPS SO, SBI SO, SSC CGL, SS CHSL, MTS and other competitive exams.

How to Solve Inequalities(Direct and Coded)- Preparation Tips and Tricks

Today we will discuss the Inequalities both direct and coded. This is the simplest topic of reasoning section. A good hold on it will easily help you in scoring 5 marks.How to Solve Inequalities(Direct and Coded)- Preparation Tips and Tricks

Tips to solve Inequalities (Coded & Direct)

There are two types of Inequalities
1. Direct Inequalities: In this type, signs >, <, =, ≤, ≤ and ≠ are mentioned directly in question. You just have to make it in order as per the question is asked.
2. Coded Inequalities: In this type, signs >,<,=,≤,≤ and ≠ are mentioned in the coded form. You have to find out the code of signs and solve it.
Now, let’s discuss with the help of examples.

Directions (Q. 1-5):
 In these questions, the 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.
1. Statements: A > B ≤ C = D ≤ E, C ≥ F = G > H
Conclusions: I. G ≤ E II. A > H
Solution: In conclusion I, the relation is asked between G and E. So, we will try to find the 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 the relation between A and C cannot be defined as the 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.
2. 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 change
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.

How to Solve Inequalities(Direct and Coded)- Preparation Tips and Tricks

3. 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.
4. 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.
5. 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.
Coded Inequalities
In these questions symbols @, #,$, *, and % are used with different meaning as follows:
‘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 smaller than B.
‘A * B’ means ‘A is not greater than B’.
‘A % B’ means ‘A is neither greater than nor equal to B’.
In each of the following questions assuming the given statements to be true, find out which of the two conclusions I and 1I given below them is/are definitely true.
Solution: Firstly, we will decode the given relations
1. A @ B’ means ‘A is not smaller than B, it means A is either > or = to B.
So, @ is ≥
2. ‘A # B’ means ‘A is neither smaller than nor equal to B’. it means A > B
So, # is >
3. ‘A $ B’ means ‘A is neither greater than nor smaller than B. it means A = B
So, $ is =
4. ‘A * B’ means ‘A is not greater than B’. it means A is either < or =. So, * is ≤
5. ‘A % B’ means ‘A is neither greater than nor equal to B’. it means A < B
So, % is <
1.Statements N @ W, W # H, H % T
Conclusions 
I. H % N
II. T # W
A. Only conclusion I is true.
B. Only conclusion H is true.
C. Either conclusion I or conclusion II is true.
D. Neither conclusion I nor conclusion II is true.
E. Both conclusions I and II are true.
Solution:
In statements 1 and 2, W is common.
N@W#H or N ≥ W > H
H < N i.e H % N
In statements 2 and 3 H is common.
W # H % T or W > H < T
Relation is discontinued at H. so it does not follow.

2.Statements F # R, H % R, L * H
Conclusions 
I. F # L
II. R @ L
A. Only conclusion I is true.
B. Only conclusion H is true.
C. Either conclusion I or conclusion II is true.
D. Neither conclusion I nor conclusion II is true.
E. Both conclusions I and II are true.
Solution: @ is ≥, # is >, $ is =, * is ≤ and % is <
F > R , H < R and L ≤ H
F > R > H ≥ L
So, conclusion I follows
Now, conclusion II, R > L i.e. R # L
So it does not follow.

3.Statements J @K, K % M, M # T
Conclusions 
I. K %T
II. K@T
A. Only conclusion I is true.
B. Only conclusion H is true.
C. Either conclusion I or conclusion II is true.
D. Neither conclusion I nor conclusion II is true.
E. Both conclusions I and II are true.
Solution: @ is ≥, # is >, $ is =, * is ≤ and % is <
From statement 2 and 3,
K % M # T i.e. K < M > T
Relation is discontinued at M as sign changes. So it does not follow.
The relation is asked between K and T. There is no need to check other statements.
But you can see that in both conclusions, all three (>, =, <)possible relations have been given.
So at least, one relation has to be followed.
The answer is either conclusion I or conclusion II follows.

4.Statements V * W, W $ H, H @ I
Conclusions 
I. V * I
II. I * W
A. Only conclusion I is true.
B. Only conclusion H is true.
C. Either conclusion I or conclusion II is true.
D. Neither conclusion I nor conclusion II is true.
E. Both conclusions I and II are true.
Solution: @ is ≥, # is >, $ is =, * is ≤ and % is <
V * W $ H @ I, V ≤ W = H ≥ I
Conclusion 1: Relation is already discontinued at W and H.
So, it does not follow.
Conclusion 2: W ≥ I, W @ I
You will mark that it does not follow.
But see carefully, if it is written I ≤ W, i.e. I * W.
It follows.


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