In the following questions, the symbols #, %,@,$ and ⦾ are used with the following meanings as illustrated below
‘P # Q’ means ‘P is not smaller than Q’
‘P % Q’ means ‘P is not greater than Q’
‘P @ Q’ means ‘P is neither smaller than nor equal to Q’
‘P $ Q’ means ‘P is neither greater than nor equal to Q’
‘P ⦾ Q’ means ‘P is neither smaller than nor greater than Q’
In each of the following questions assuming the given statements to be true, find out which of the two conclusions I and II given below them is/are definitely true.
Statements :
H ⦾ W, W%R,R @F
Conclusions:
I R ⦾ H
II R @ H
Answer: C
Correct answer: either conclusion I or II is true
Translating: H = W, W ≤ R, so R ≥ H. That leaves exactly two possibilities, R = H or R > H, which are conclusions I and II. Neither can be established alone, but between them they exhaust the cases - which is what an either-or conclusion means.
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Statements:
M$T, T@K, K ⦾ D
Conclusions:
I D$ T
II K$M
Answer: A
Correct answer: if only conclusion I is true
Translating: M < T, T > K, K = D. Conclusion I asks whether D < T, and D = K < T gives it directly. Conclusion II asks how K compares with M, but both lie below T with no link between them, so nothing follows.
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Statements:
R%N, N#F, F@B
Conclusions:
I F⦾ R
II B$N
Answer: B
Correct answer: only conclusion II is true
Translating: R ≤ N, N ≥ F, F > B. Conclusion II asks whether B < N, and B < F ≤ N gives it. Conclusion I asks whether F = R, but R and F both lie below N with no relation to one another, so it cannot be established.
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Statements:
H@W, W$M, M#K
Conclusions:
I K$W
II H@M
Answer: D
Correct answer: neither conclusion I or II is true
Translating: H > W, W < M, M ≥ K. The chain reverses at W, which is the smallest element named, so H and M are each above it with no link between them, and K sits below M with no link to W. Neither conclusion can be reached.
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Statements:
R#T, T⦾ M, M@D
Conclusions:
I D $ T
II R#M
Answer: E
Correct answer: both conclusion I and II is true
Translating: R ≥ T, T = M, M > D, giving one unbroken chain R ≥ T = M > D. Conclusion I follows since D < M = T, and conclusion II since R ≥ T = M. An unbroken chain with no reversal is what allows both to be established.
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Statements:
F#K, K$B, B%M
Conclusions:
I M@F
II B@F
Answer: D
Correct answer: neither conclusion I or II is true
Translating: F ≥ K, K < B, B ≤ M. The chain reverses at K, so F sits above it on one side and B and M above it on the other, with nothing joining the two branches. Neither M nor B can therefore be compared with F.
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In the following questions, the symbols * , $, #, δ and % are used with the following meaning as illustrated below:
‘P $ Q’ means ‘P is neither greater than nor smaller than Q’.
‘P δ Q’ means ‘P is neither greater than nor equal to Q’.
‘P % Q’ means ‘P is neither smaller than nor equal to Q’.
‘P*Q’ means ‘P is not smaller than Q’.
‘P # Q’ means ‘P is not greater than Q’.
Now in each of the following question assuming the given statements to be true, find which of the conclusions I, II, III and IV given below them is/are definitely true and give your answer accordingly.
Statements: A * P, P % E, E δ F, F # S
Conclusions:
I. S % E
II. A % E
III.F % P
Answer: C
Correct answer: Only I and II are true
Translating this set’s code: \(\$\) is “neither greater nor smaller”, so \(=\); \(\delta\) is “neither greater than nor equal”, so \(<\); \(\%\) is “neither smaller than nor equal”, so \(>\); \(*\) is “not smaller than”, so \(\ge\); and \(\#\) is “not greater than”, so \(\le\). The statements become \[ A \ge P > E < F \le S. \] E is a low point: the chain falls to it from the left and rises from it to the right, so nothing carries across E.
I. \( S > E \): both lie on the right of the pivot, joined by \( E < F \le S \), which is strict. True.
II. \( A > E \): both lie on the left, joined by \( A \ge P > E \), again strict. True.
III. \( F > P \): P sits left of the pivot and F sits right of it, so no chain connects them. With \( E = 1 \), \( P = 2, F = 9 \) works and so does \( P = 9, F = 2 \). Not determined.
I and II follow, III does not.
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Statements: P δ W, W $ Q, Q % S, S * A
Conclusions:
I. A δ Q
II. Q % P
III. W % A
Answer: D
Correct answer: All I, II and III are true
Translating this set’s code: \(\$\) is “neither greater nor smaller”, so \(=\); \(\delta\) is “neither greater than nor equal”, so \(<\); \(\%\) is “neither smaller than nor equal”, so \(>\); \(*\) is “not smaller than”, so \(\ge\); and \(\#\) is “not greater than”, so \(\le\). The statements become \[ P < W = Q > S \ge A. \] Q is a high point here, but every conclusion compares an element with Q or W rather than reaching across the peak, so each has an unbroken run to use.
I. \( A < Q \): reading down the right branch, \( Q > S \ge A \), strict at the first link. True.
II. \( Q > P \): the left branch, \( P < W = Q \). True.
III. \( W > A \): W equals Q, so this is conclusion I again with the peak named by its other letter: \( W = Q > S \ge A \). True.
All three follow. The comparison the chain would not settle is P against A - the two ends, one on each side of the peak - and the question does not ask for it.
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Statements: L % N, N # T, T $ D, D δ A
Conclusions:
I. L % A
II. L # A
III. A % N
Answer: A
Correct answer: Only either I or II and III are true
Translating this set’s code: \(\$\) is “neither greater nor smaller”, so \(=\); \(\delta\) is “neither greater than nor equal”, so \(<\); \(\%\) is “neither smaller than nor equal”, so \(>\); \(*\) is “not smaller than”, so \(\ge\); and \(\#\) is “not greater than”, so \(\le\). The statements become \[ L > N \le T = D < A. \] N is a low point, so L on its left and A on its right are not connected by any chain.
III. \( A > N \): both sit on the right of the pivot, joined by \( N \le T = D < A \), which is strict at the last link. True.
Now take I and II together, since that is what the answer turns on. I claims \( L > A \) and II claims \( L \le A \). Neither can be established, because L and A stand on opposite sides of the pivot: with \( N = 1 \) we may have \( L = 9, A = 2 \) or \( L = 2, A = 9 \). But the two claims are exhaustive - L is either greater than A or it is not, and “not greater” is exactly \( \le \) - so one of them must hold. That is the “either…or” case, and it arises here from a complementary pair rather than from the statements settling anything.
So III is definitely true and either I or II is true, which is the keyed answer.
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Statements: M # Q, Q $ K, A % K, V * A
Conclusions:
I. K * M
II. A % Q
III. A % M
Answer: E
Correct answer: All I, II and III are true
Translating this set’s code: \(\$\) is “neither greater nor smaller”, so \(=\); \(\delta\) is “neither greater than nor equal”, so \(<\); \(\%\) is “neither smaller than nor equal”, so \(>\); \(*\) is “not smaller than”, so \(\ge\); and \(\#\) is “not greater than”, so \(\le\). The statements become \( M \le Q \), \( Q = K \), \( A > K \) and \( V \ge A \). Since Q and K are equal they can be treated as one element, which lines the rest up as \[ V \ge A > K = Q \ge M. \]
That single descending chain settles all three conclusions.
I. \( K \ge M \): from \( M \le Q = K \). True.
II. \( A > Q \): from \( A > K = Q \). True.
III. \( A > M \): the run \( A > K = Q \ge M \) points one way and contains a strict link. True.
All three follow. Spotting the equality early is what makes this quick: an \(=\) link lets two letters be merged, and a chain with nothing but one reversal-free direction settles every comparison along it.
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