# Reasoning Practice Inequalities for SBI PO

Reasoning Practice Inequalities for SBI PO
Directions (1 – 10): In these questions, a relationship between different elements is shown in each statement. The statements are followed by two conclusions. Give answer-
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.
1. Statements : E > B ≥ A < M, N > A > P
Conclusions : I. N > B II. M > N
2. Statements : N = E ≤ P > Q, P ≥ S
Conclusions : I. N ≥ S II. S > Q
3. Statements : V < P > Q = T ≥ R, Q ≤ S
Conclusions : I. T ≥ S II. S ≥ V
4. Statements : L = I ≥ J < K, J ≥ M
Conclusions : I. K > M II. L ≥ M
5. Statements : A > B ≥ C < L, D < C ≤ E
Conclusions : I. E > L II. E ≥ B
6. Statements : S ≤ A > P, B ≥ A < C
Conclusions : I. B ≥ S II. C > P
7. Statements : M > P ≤ N < U, N > B, A ≥ U
Conclusions : I. M > B II. A > B
8. Statements : W ≥ X = V ≤ Y, U < X ≥ Z
Conclusions : I. W ≥ U II. Y ≥ U
(9 – 10): Statements : H ≤ G < V ≥ U, G ≤ S, P > V
9. Conclusions : I. H < P II. U ≤ S
10. Conclusions : I. S ≥ H II. U < P
1. D) Given:
E > B ≥ A < M ...(i)
N > A > P ...(ii)
Combining both, we get,
E > B ≥  A > P and
E > B ≥  A < N and
M > A < N
We can’t compare N and B. So, conclusion I is not true. Again, we can’t compare M and N. So, conclusion II is also not true.
2. D) Given :
N = E ≤ P > Q ...(i)
P ≥ S ...(ii)
Combining (i) and (ii) we get,
N = E ≤ P ≥ S and
Q < P ≥  S
We can’t compare N and S. So, conclusion I is not true. Again, we can’t compare S and Q. So, conclusion II is not true.
3. D) Given :
V < P > Q = T ≥ R ...(i)
Q ≤ S ...(ii)
Combining (i) and (ii) we get,
V < P > Q = T ≤ S and
S ≥ Q = T ≥  R
Hence, S ≥  T. So, conclusion I is not true.
Again, we can’t compare S and V. So, conclusion II also is not true.
4. E) Given :
L = I ≥  J < K ...(i)
J ≥ M ...(ii)
Combining (i) and (ii) we get,
L = I ≥  J ≥ M and
K > J ≥ M
Conclusion I is true.
Again, conclusion II is also true.
5. D) Given :
A > B ≥ C < L ...(i)
D < C ≤ E ...(ii)
Combining (i) and (ii) we get,
A > B ≥ C > D and
A > B ≥ C ≤ E and
D < C < L and
E ≥ C < L
We can’t compare E and L. So, conclusion I is not true.
Again, we can’t compare E and B. So, conclusion II is also not true.
6. E) S ≤ A > P ............. (i)
B ≥ A < C ........ (ii)
Combining (i), and (ii), we get
S ≤ A ≤ B and
S ≤ A < C and
B ≥ A > P and
C > A > P
(I) B ≥ S is true. So, conclusion I is true.
(II) C > P is true. So, conclusion II is true.
7. B) Given:
M > P ≤ N < U .............. (i)
N > B .......................... (ii)
A ≥ U .................................... (iii)
Combining (i), (ii) and (iii), we get
M > P ≤ N > B and
M > P ≤ N < U ≤ A and
B < N < U ≤ A
(I) We can’t compare M and B. So, conclusion I is not true.
(II) A > B is true. So, conclusion II is true.
8. D) Given:
W ≥ X = V ≤ Y ............... (i)
U < X ≥ Z ................. (ii)
Combining (i) and (ii), we get
W ≥ X > U and
U < X = V ≤ Y and
W ≥ X ≤ Z and
Z ≤ X = V ≤ Y
(I) W > U, is true. So, W U is not ture. Thus, conclusion I is not true.
(II) Y > U is true. So, Y U is not true. Thus, conclusion II is not true.
(9 – 10):
Given:
H ≤ G < V ≥ U .................. (i)
G ≤ S ............................... (ii)
P > V ............................ (iii)
Combining (i), (ii) and (iii), we get
H ≤ G ≤ S and
S ≥ G < V ≥ U and
P > V ≥ U and
H ≤ G < V < P and
P > V > G S
9. A)
(I) P > H is true. So, conclusion I is true.
(II) We can’t compare S and U. So, conclusion II is nottrue.
10. E)
(I) S ≥ H is true. So, conclusion I is true.
(II) U < P is true. So, conclusion II is also true.