Step by Step Strategy to Solve Inequality Problems

Ques 1. 

W ≥ D < M < P < A = F

Conclusions:

I. F > D
II. P < W

Solution:

To solve Conclusion
I. F>D,
Consider the statement from D to F
D < M < P < A = F
Symbols between D and F are uniform i.e. < which implies D will be definitely lesser than F.
Therefore conclusion I follows

To solve conclusion
II. P < W
Consider the statement from P to W i.e. W ≥ D < M < P
Symbols between P and W are not uniform, mixer of lesser than and greater than symbol implies P can’t be definitely lesser than W
Therefore Conclusion II doesn’t follow

Ques 2.

H  ≥  M > F <A = B> S

Conclusion:

I. H > B
II. F < S

Solution:

To solve Conclusion
I. H > B
Consider the statement from H to B
H ≥ M > F <A=B
Symbols between H and B are mixer of both lesser than and greater than the symbol which implies H cannot be greater than B.
Therefore conclusion I doesn’t follow

To solve conclusion
II. F<S
Consider the statement from F to S i.e. F <A=B> S
Symbols between F and S are mixer of lesser than and greater than symbol implies F can’t be definitely lesser than S
Therefore Conclusion 2 doesn’t follow

Ques 3. 

B > T > Q > R = F

Conclusion:

I. Q ≥ F
II. T> F

Solution:

To solve Conclusion
I. Q ≥ F
Consider the statement from Q to F
Q > R = F
Above statement implies that Q is definitely greater than F
Therefore conclusion I doesn’t follow

To solve conclusion
II. T>F
Consider the statement from T to F i.e.
T>Q>R=F
Symbols between T and Fare uniform which consist of only > symbol which implies Twill be definitely greater than F
Therefore Conclusion 2 follows

Ques 4.

 H < J, F < H, I  ≤  J = K

Conclusion:

I. H > I
II. I ≥ F

Solution:

To solve Conclusion
I. H>I
Since it is split statements combine statement with H and I variable
H < J  ≥  I
Symbol between H and I are mixer of both > and < which implies H can’t be definitely greater than I.
Therefore conclusion 1 doesn’t follow

To solve conclusion
II. I ≥ F
Combine the statement of I and F
F < H < J ≥ I
Symbols between T and Fare not uniform which consist of both < and > symbol which implies I can’t be ≥ F
Therefore Conclusion 2 doesn’t follow

Direction (Q5-Q7): Read the information given below and solve the questions that follow. 
% means not greater than (% ➔ ≤)
& means not smaller than (& ➔ ≥)
# means neither equal to nor smaller than (# ➔ >)
$ means neither equal to nor greater than ($ ➔ <)
@ means neither smaller than nor greater than (@ ➔ =)
From the above statements, we can conclude,

Ques5.

Statements

M $ K,K & T,
T $ J

Conclusions:

I. J # K
II. T # M
III. M # J

Solution:

Convert the statement and conclusion from symbols to mathematical operation.
Statement: M < K ≥ T < J
Conclusion:
I. J > K
II. T > M
III. M > J

To solve Conclusion I
J>K, consider statement from J to K
K ≥ T < J
Since it is the mixer of both < and > symbol
Conclusion 1 is false

To solve Conclusion II
T > M, consider statement from T to M
M < K ≥ T
Since it is the mixer of both < and > symbol
Conclusion 2 is false

To solve Conclusion III
M > J, consider statement from M to J
M < K ≥ T < J
Since it is the mixer of both < and > symbol
Conclusion 3 is false
Therefore none of the three conclusions is true

Ques 6. 

Statements: 

F @ T
T % M
M # R

Conclusion:

R $ T
F @ M
F $ M

Solution:

Convert the statement and conclusion from symbols to mathematical operation.
Statement: F = T ≤ M > R
Conclusion:
I. R < T
II. F = M
III. F < M

To solve Conclusion I
R < T, consider statement from R to T
T ≤ M > R
Since it is the mixer of both < and > symbol
Conclusion 1 is false

To solve Conclusion 2
F = M, consider statement from F to M
F = T ≤ M
From the above statement, It is possible that F can be equal to M
Conclusion II may be True

To solve Conclusion 3
F < M, consider statement from F to M
F = T ≤ M
From the above statement, it is possible that F can be lesser than M
Conclusion III may be True
Therefore Either Conclusion 2 or 3 can be true

Ques 7. 

Statements: 

J & H,
H @ B,
B % N

Conclusion:

I. N & H
II. N @ J
III. J & B

Solution:

Convert the statement and conclusion from symbols to mathematical operation.
Statement: J ≥ H = B ≤ N
Conclusion:
I. N ≥ H
II. N = J
III. J ≥ B

To solve Conclusion I
N ≥ H, consider statement from H to N
H = B ≤ N
It clearly implies N ≥ H
Conclusion I is True

To solve Conclusion II
N=J, consider statement from N to J
J ≥ H = B ≤ N
Since it is combination of <and > symbol we can’t predict N=J
Conclusion II is False

To solve Conclusion III
J ≥ B, consider statement from J to B
J ≥ H = B
It clearly implies J ≥ B
Conclusion III is True
Therefore Conclusion I and III are True

Must Read 
Reasoning Ability: Concepts of Inequality with Examples: Part 1
Reasoning Ability: Concepts of Inequality with Examples: Part 2

Smart Prep Kit for Banking Exams by Ramandeep Singh - Download here


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