Reasoning Equality And InequalityPage 2
Section-2
Direction:In these questions relationship between different elements is shown in statements. The statements are followed by conclusions. Study the conclusions based on the given statement and select the appropriate answer.
1. | Statements:
M > U > L ≤ N
L ≥ Y > A
Conclusions:
Y < N
M > N
N = Y
M > A | | (a)Only 1 , 2 and 3 are true | | (b)Only 1 and 2 are true | | (c)All 1, 2 3 and 4 are true | | (d)Only 4 and Either 1 or 3 are true |
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Answer is: DM > U > L ≤ N ……..(1)
L ≥ Y > A ………..(2)
Combining 1 and 2, we get
M > U > L ≥ Y > A and A < Y ≤ L ≤ N
Thus, Y ≤ N is true.
Hence, Y < N may be true.
And Y = N may be true.
So, conclusion 1 and 3 make a complementary pair.
Thus, either 1 or 3 is true.
Again, we can't compare M and N. Hence 2 is not true.
But M > A is true. Hence, 4 is true.
Hence only 4 and either 1 or 3 are true.
2. | Statements:
M > U > L ≤ N
L ≥ Y > A
Conclusions:
Y < N
M > N
N = Y
M > A | | (a)Only 1 , 2 and 3 are true | | (b)Only 2 is true | | (c)Only 4 and Either 1 or 3 are true | | (d)All 1, 2 3 and 4 are true |
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Answer is: CM > U > L ≤ N ……..(1)
L ≥ Y > A ………..(2)
Combining 1 and 2, we get
M > U > L ≥ Y > A and A < Y ≤ L ≤ N
Thus, Y ≤ N is true.
Hence, Y < N may be true.
And Y = N may be true.
So, conclusion 1 and 3 make a complementary pair.
Thus, either 1 or 3 is true.
Again, we can't compare M and N. Hence 2 is not true.
But M > A is true. Hence, 4 is true.
Hence only 4 and either 1 or 3 are true.
3. | Statements:
J ≥ A > D = E
L < A < M
Conclusions:
M < J
J > L
D > L
E < M | | (a)Only 1 , 2 and 3 are true | | (b)Only 2 and 4 are true | | (c)Only 1 and Either 2 or 4 are true | | (d)All 1, 2 3 and 4 are true |
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Answer is: BJ ≥ A > D = E …….(1)
L < A < M ……….(2)
Combining equation (1) and (2), we get
M > A > D = E …….(3)
J ≥ A < M ……….(4)
J ≥ A > L ……….(5)
L < A > D …….(6)
Thus, from equation 4, we can't compare J and M. Hence, 1 is not true.
From 5, J > L is true. Hence 2 is true.
Again, from 6, we can't compare D and L. Hence 3 is not true.
Now, from (3), M > E or E < M is true. Hence (4) is true.
Thus, only 2 and 4 are true.
4. | Statements:
Y > F ≤ O ≤ P
F ≥ U < T
Conclusions:
Y > P
T < F
O > T
P < U | | (a)Only 1 , 2 and 3 are true | | (b)Only 1 and 2 are true | | (c)All 1, 2 3 and 4 are true | | (d)none is true |
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Answer is: DY > F ≤ O ≤ P …………(1)
F ≥ U < T ………….(2)
Thus, from (1) we can't compare Y and P. Hence 1 is not true. Again, from (2) we can't compare F and T. So, 2 is not true.
T > U ≤ F ≤ O ≤ P
Thus, we can't compare O and T. Hence (3) is not true.
Again, U ≤ P is true. So, (4) (P < U) is not true. Hence none is true.
5. | Statements:
M > H ≤ Y ≤ R < U = Z ≥ E
Conclusions:
M > R
Z ≤ R
R > E
Z > H | | (a)Only 1 , 2 and 3 are true | | (b)Only 4 is true | | (c)Only 2 is true | | (d)All 1, 2 3 and 4 are true |
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Answer is: BM > H ≤ Y ≤ R < U = Z ≥ E
Thus, we can't compare M and R. Hence (1) is not true.
Again, H < Z or Z > H is true. Hence 4 is true.
And R < Z or Z > R is true. Hence 2 (Z ≤ R) is not true. We can't compare R and E. Hence 3 is not true.
6. | Statements:
P > Q ≤ C ≤ B = M > D
Conclusions:
M > Q
D ≤ Q
M = Q
C > D | | (a)Only 1 or 3 are true | | (b)Only 1 , 2 and 3 are true | | (c)Only 1 and Either 2 or 4 are true | | (d)All 1, 2 3 and 4 are true |
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Answer is: AP > Q ≤ C ≤ B = M > D
Thus, Q ≤ M is true. So, M > Q may be true.
And M = Q may be true. So, conclusion 1 and 3 make a complimentary pair.
Again, we can't compare D and Q, or C and D. So, 2 and 4 are not true.
7. | Statements:
A ≥ B = C, B < D ≤ E
Conclusions:
D > A
E > C | | (a)If only conclusion 1 follows | | (b)If only Conclusion 2 follows | | (c)If either conclusion 1 or 2 follows | | (d)If neither conclusion 1 nor 2 follows |
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Answer is: BA ≥ B = C < D ≤ E
D > A is not true.
E > C is true.
8. | Statements:
L > U ≥ K, Z < U < R
Conclusions:
L > Z
K < R | | (a)If only Conclusion 1 follows | | (b)If only Conclusion 2 follows | | (c)If either Conclusion 1 or 2 follow | | (d)If both conclusions 1 and 2 follow |
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Answer is: DGiven that-
L > U ……..(1)
U ≥ K . …..(2)
Z < U …….(3) and
U < R …….(4).
Combining 1 and 3, we get
L > U > Z or L > Z. hence 1 follows.
Combining 2 and 4, we get
K ≤ U < R or K < R. Hence 2 is follows.
9. | Statements:
L > U ≥ K, Z < U < R
Conclusions:
L > Z
K < R | | (a)If only Conclusion 1 follows | | (b)If only Conclusion 2 follows | | (c)If either Conclusion 1 or 2 follow | | (d)If both conclusions 1 and 2 follow |
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Answer is: DGiven that-
L > U ……..(1)
U ≥ K . …..(2)
Z < U …….(3) and
U < R …….(4).
Combining 1 and 3, we get
L > U > Z or L > Z. hence 1 follows.
Combining 2 and 4, we get
K ≤ U < R or K < R. Hence 2 is follows.
10. | Statements:
Y < J = P ≥ R > I
Conclusions:
J > I
Y < R | | (a)If only Conclusion 1 follows | | (b)If only Conclusion 2 follows | | (c)If either Conclusion 1 or 2 follow | | (d)If neither Conclusion 1 nor 2 follow |
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Answer is: AY < J = P ≥ R > I
J > I is true.
Y < R is not true
11. | Statements:
V ≥ K > M = N, M > S, T < K
Conclusions:
T < N
V = S | | (a)If only Conclusion 1 follows | | (b)If only Conclusion 2 follows | | (c)If either Conclusion 1 or 2 follow | | (d)If neither conclusion 1 nor 2 follows |
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Answer is: CV ≥ K > M = N > S
K > T
T < N is not true.
V = S is not true.
12. | Statements:
F ≤ X < A, R < X ≤ E
Conclusions:
F ≤ E
R < F | | (a)If only Conclusion 1 follows | | (b)If only Conclusion 2 follows | | (c)If either Conclusion 1 or 2 follow | | (d)If neither conclusion 1 nor 2 follows |
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Answer is: AGiven that –
F ≤ X ……..(1)
X < A ……..(2)
R < X ………(3) and
X ≤ E …….(4)
Combining 1 and 4, we get
F ≤ X ≤ E or F ≤ E.
Hence 1 follows
From 1 and 3, R and F can't compared. Hence 2 does not follow.
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