Inequality Questions and Answers

Answer: D

Correct answer: Neither conclusion I or II follows

The statements are \( A > B > C \), \( D > E > F \) and \( D > C \).

There are two chains here, and the third statement is the only thing joining them - it ties D, the top of the second chain, to C, the bottom of the first. That is a weak join: it tells us D outranks C, but says nothing about where E, F, A or B stand relative to the other chain.

Conclusion I claims \( E > C \). We know \( D > E \) and \( D > C \), so E and C both sit below D with no order between them. Taking \( D = 10 \), \( E = 9, C = 1 \) works and so does \( E = 2, C = 9 \) - provided B and A are raised to suit. Does not follow.

Conclusion II claims \( F > B \). F is the bottom of the second chain and B is in the middle of the first, and no chain of statements runs between them at all. Does not follow.

Neither conclusion follows.

Answer: E

Correct answer: Both conclusions I and II follow

Chain the statements into one line: \[ K \ge M \ge N > O. \] Conclusion I claims \( O < M \). From \( M \ge N \) and \( N > O \) we get \( M > O \), so I follows. Conclusion II claims \( O < K \). Extending the same chain, \( K \ge M > O \), so \( K > O \) and II follows too. A strict inequality anywhere in a chain of \(\ge\) links makes the whole chain strict, which is what carries both conclusions. The symbol between K and M had been lost from this record and is restored above as \(\ge\); had it been the strict \(>\) instead, both conclusions would still follow, so the answer is unaffected either way.

Answer: A

Correct answer: This item cannot be worked as recorded

Translating the code: \(@\) is “greater than”, \(\#\) is “greater than or equal to”, \(\%\) is “less than or equal to”, \(*\) is “equal to” and \(\$\) is “less than”. The third statement is incomplete in this record: it reads “CP” with the symbol between C and P missing, and conclusion II has lost its symbol in the same way. That matters, because the two surviving statements give only \( R \ge D \) and \( D > C \), so \( R > C \); nothing links either letter to P without the lost symbol. Restoring it does not rescue the keyed answer either. If the missing symbol is \(@\), \(\#\) or \(*\), then \( C \ge P \), which gives \( R > P \) - conclusion I - but it also gives \( D > P \), which is conclusion III, so the answer would be “I and III are true”. If instead it is \(\$\) or \(\%\), then \( C \le P \) and neither conclusion is determined, so the answer would be “None of these”. No restoration makes I true while III is false, so “only I is true” is unreachable. The item needs its original symbols before it can be answered.

Answer: D

Correct answer: None are true

In this set the code reads: \(@\) is “neither smaller nor equal”, so \(>\); \(\#\) is “not smaller than”, so \(\ge\); \(\%\) is “not greater than”, so \(\le\); \(*\) is “neither smaller nor greater”, so \(=\); and \(\$\) is “neither greater nor equal”, so \(<\). The statements become \( B > D \), \( D \le F \) and \( F \ge H \).

The chain breaks at D. B stands above D and F stands above D, so the relation reverses direction there and nothing carries across it - and H hangs below F, on the far side of the same break.

I. \( F @ B \) asks whether \( F > B \). Both merely stand above D. With \( D = 1 \), \( B = 2 \) and \( F = 9 \) works, and so does \( B = 9 \) and \( F = 2 \). Does not follow.

II. \( H * D \) asks whether \( H = D \). H is at or below F, and D is at or below F, which permits them to be equal but does not require it. Does not follow.

III. \( H \% B \) asks whether \( H \le B \). H reaches B only through F and D, and the direction reverses at D. Does not follow.

None of the three is definite.

Answer: E

Correct answer: None of The above

In this set the code reads: \(@\) is “neither smaller nor equal”, so \(>\); \(\#\) is “not smaller than”, so \(\ge\); \(\%\) is “not greater than”, so \(\le\); \(*\) is “neither smaller nor greater”, so \(=\); and \(\$\) is “neither greater nor equal”, so \(<\). The statements become \( T > Y \), \( Y \ge G \) and \( G \le W \), and the first two join: \[ T > Y \ge G. \]

That settles T against G completely. Because the chain contains a strict link, \( T > G \).

I. \( G * T \) asks whether \( G = T \). Definitely false, since \( T > G \).

II. \( T \$ G \) asks whether \( T < G \). Definitely false for the same reason.

III. \( T @ G \) asks whether \( T > G \). True.

So exactly one conclusion holds, and no option says “only III”. The three “either…or” options are wrong for a reason worth understanding: an either-or answer applies only when neither of a complementary pair can be settled individually, so that one of them must hold without our knowing which. Here III is settled, which rules that pattern out. The answer is therefore none of the above.

Answer: C

Correct answer: Either I or II and III is true

Translating: M < R, R ≤ T, T = N, so M < R ≤ N. Since R ≤ N, either N = R or N > R, which makes conclusions I and II a complementary pair - exactly one holds, so either follows. And N ≥ R > M gives N > M, so III is true outright.

Answer: D

Correct answer: All are true

In this set the code reads: \(@\) is “not smaller than”, so \(\ge\); \(\#\) is “not greater than”, so \(\le\); \(\%\) is “neither greater nor equal”, so \(<\); \(*\) is “neither smaller nor greater”, so \(=\); and \(\$\) is “neither smaller nor equal”, so \(>\). Note that this is not the same code as the neighbouring sets use - check the direction before translating. The statements become \( J \le N \), \( K \ge N \) and \( T > K \), which join into the single chain \[ J \le N \le K < T. \]

I. \( J \% T \) asks whether \( J < T \). The chain runs from J all the way to T, and it contains the strict link \( K < T \), so the whole relation is strict. True.

II. \( T \$ N \) asks whether \( T > N \). Reading the chain backwards from T gives \( T > K \ge N \), again strict. True.

III. \( N @ J \) asks whether \( N \ge J \). That is the first statement read the other way round. True.

All three follow, so the answer is that all are true.

Answer: A

Correct answer: None is true

Translating: B = D, D ≥ H, H < F, so B = D ≥ H < F. The chain reverses direction at H, so nothing at all links B or D to F. Note also that I and II do not form a complementary pair - B = F and B > F leave B < F uncovered - so not even an either-or conclusion follows.

Answer: C

Correct answer: Either I or III is true

Translating: T > K, K ≤ R, R = M, so M = R ≥ K. That gives either M = K or M > K, which are conclusions I and III forming a complementary pair. Conclusion II asks how M compares with T, and the chain reverses at K, so nothing follows.

Answer: D

Correct answer: None is true

In this set the code reads: \(@\) is “not smaller than”, so \(\ge\); \(\#\) is “not greater than”, so \(\le\); \(\%\) is “neither greater nor equal”, so \(<\); \(*\) is “neither smaller nor greater”, so \(=\); and \(\$\) is “neither smaller nor equal”, so \(>\). Note that this is not the same code as the neighbouring sets use - check the direction before translating. The statements become \( V \ge M \), \( A > M \) and \( R \le V \).

Notice the shape of this: V and A are both anchored above M, and R hangs below V. Two elements known only to lie above the same third element are not ordered by that fact, and this is what defeats every conclusion.

I. \( R \# A \) asks whether \( R \le A \). R is tied only to V, and A is tied only to M, so no chain links them. Does not follow.

II. \( V @ A \) asks whether \( V \ge A \). Both stand above M with nothing between them: take \( M = 1 \) and either \( V = 5, A = 2 \) or \( V = 2, A = 5 \), and both satisfy the statements. Does not follow.

III. \( R \$ M \) asks whether \( R > M \). R sits at or below V, and V sits at or above M, so R could be anywhere. Does not follow.

None of the three is definite, so the answer is that none is true.