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Breakup Type Inequalities MBA CET 2022 Actual Questions

Inequalities

Q1. Statement:  R < O ≤ L ≤ E; G = E ≥ S; P ≤ S

Conclusions:

I.  R > P                     

II. P ≤ E

(A) Both conclusion I and II are true.

(B) Either conclusion I or II is true.

(C) Only conclusion I is true.

(D) Neither conclusion I nor II is true.

(E) Only conclusion II is true.

EXPLANATION

Correct Option: E

Given statements:    R < O ≤ l ≤ E       …(i)

                                     G = E ≥ S         …(ii)

                                     P ≤ S                 …(iii) 

Combining (i), (ii) and (iii) , we get

       R  <  O  ≤  L  ≤  E  =  G  ≥  S  ≥  P 

Hence, only conclusion II is true.

Q2. Statements:  A > B ≤ C = D ≤ E,  C ≥ F = G > H

Conclusions: 

I. C < H                              

II. A > H

(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 conclusion I and II are true

EXPLANATION

Correct Option: D

Given Statements: 

A > B ≤ C = D ≤ E    ….(i) 

C ≥ F = G > H        …(ii) 

Combining (i) and (ii), we get

A > B ≤ C ≥ F = G > H 

As the relation between C & H clearly confirms that C is greater than H while the given conclusion states that C is less than H.

Hence C1 doesn’t follow.

As it’s clear that inequality signs are opposite, we therefore can’t find a definite relationship between A & H.

Q3. Statements:  H  ≥  T  >  S  ≤  Q, T  ≥  U  =  V  >  B 

Conclusions:

I. V  >  S                          

II. B  ≤  H

(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

EXPLANATION

Correct Option: D

Given Statements: 

H  ≥  T  >  S  ≤  Q            …(i) 

T  ≥  U  =  V  >  B           …(ii) 

Combining (i) and (ii), we get

H  ≥  T  ≥  U  =  V  >  B 

As we can see that the relationship between B & H clearly concludes that B is only less than H (B < H). Therefore C2 (B ≤ H) is not true.

Again, from (i) and (ii), we get

S  <  T  ≥  U  =  V 

As it’s clear that inequality signs are opposite, we therefore can’t find a definite relationship between V & S. Hence CI (V > S) is not true.

Q4. Statements:   L ≥ A ≥ C ,  K = Y ≤ C ,  H > D ≤ K ,   A > E < Y

Conclusions: 

I. D < A 

II. A = D

III. L > Y

(A) All the conclusions follow

(B) Either conclusion I or II follows

(C) Only conclusion III follows

(D) Only conclusions II and III follow

(E) None of the conclusions follows

EXPLANATION

Correct Option: B

Statements:   L ≥ A ≥ C ,  K = Y ≤ C ,  H > D ≤ K

Conclusions:  D < A  ,  A = D ,  L > Y

For conclusion I: D < A

From statements I, II and III, we get:

D ≤ K = Y ≤ C ≤ A

Here, the common sign between D and A is ‘≤’, hence D ≤ A.

Thus conclusion I does not follow individually.

For conclusion II: A = D

From statements I, II and III, we get:

D ≤ K = Y ≤ C ≤ A

Here, the common sign between D and A is ‘≤’, hence D ≤ A.

Thus conclusion II does not follow individually.

On combining conclusion I and II we get D ≤ A.

Therefore either conclusion I or II follows.

For conclusion III: L > Y

From statements I and II, we get:

Y ≤ C ≤ A ≤ L

Thus the common sign between Y and L is ‘≤’, Therefore Y ≤ L is the true relationship

Hence conclusion III does not follow.

Therefore either conclusion I or II follows.

Hence option B is correct.

Category :

Inequalities

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