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
Answers:
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.
