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.
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
