# How to solve Inequalities (Coded and Direct)?

# How to solve Inequalities (Coded and Direct)?

# How In the banking examinations, inequality is the easiest and marks scoring topic of the Reasoning section. to help you out in understanding all the basic facts of it, we are here with the tips to solve coded and direct inequality questions. If you are having a good hold on it, you easily fetch 5 marks. After understanding the concepts, do practice a lot of questions so that you get a good command over these questions.

## 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.

### Examples of Direct Inequality

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 follow.

(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. 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, the conclusion I follow.

In conclusion II, the 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.

Now, a question often came into student's mind that How to approach inequality questions in Exam without taking more time?

You should check that inequality sign should follow in the same direction like in statement 1 from C to E signs are in the same direction and In statement 2 from G to C are also following in the same direction. So conclusion 1 follows.

But in case of A and H, the 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 the same direction. So conclusion 2 does not follow.

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 the relation between S and V, the sign changes at T itself. So Conclusion 1 does not follow.

For the relation between B and H, the 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, Conclusion 2 does not follow.

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, the 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 the 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.

These were examples of Direct Inequality. Now we will move on to the examples of Coded Inequalities.

### Coded Inequalities

In these questions symbols @, #,$, *, and % are used with the 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

The 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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