In these questions, the relationship between different elements is shown in…
2025
In these questions, the relationship between different elements is shown in the statements. The statements are followed by two conclusions. Study the conclusions based on the given statements and select the appropriate 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 both conclusions I and II are true
(e) If neither conclusion I nor II is true
Statements: Q ≤ M > D > E = T; F = G ≥ H < C < T ≥ A
Conclusions:
I. A ≤ M
II. F > Q
Answer: A. a — Concept: In a coded-inequality chain, combining relations along one consistent overall direction keeps the strongest available link — when a chain running the…
- A.
a
- B.
b
- C.
c
- D.
d
- E.
e
Attempted by 38 students.
Show answer & explanation
Correct answer: A
Concept:
In a coded-inequality chain, combining relations along one consistent overall direction keeps the strongest available link — when a chain running the same way includes even one strict step (< or >), the combined result between its endpoints is strict, even when the rest of the chain is ≤, ≥, or =. (A chain that reverses direction partway, such as X > Y < Z, does not combine at all — no relation between the endpoints can be fixed.) Once the relation between two letters is fixed, a conclusion phrased with ≤ (or ≥) is true whenever the fixed relation is any of <, =, or ≤ (respectively >, =, or ≥) — a strict result also satisfies the weaker, non-strict conclusion, because it never contradicts it. A conclusion phrased with < (or >) needs the fixed relation to be exactly that strict direction. An 'either/or' answer is valid only when both conclusions compare the very same pair of letters in complementary directions.
Given relations:
Q ≤ M > D > E = T; F = G ≥ H < C < T ≥ A.
Conclusion I — is A ≤ M definitely true?
Trace M downward: M > D, D > E, E = T, so D > T (since D > E and E = T), and hence M > T — every step in this stretch is strict, so M > T is a strict result.
Trace A: the statement gives T ≥ A directly.
Combine: M > T ≥ A. Because the M > T link is strict and the T ≥ A link continues in the same direction, the combined relation is strict: M > A, i.e., A < M.
Conclusion I asks whether A ≤ M. Since A < M is fixed, and a strict ‘<’ satisfies the weaker ‘≤’ claim (A is never established greater than M), Conclusion I is true.
Conclusion II — is F > Q definitely true?
Trace F: F = G, and G ≥ H, while H < C < T. At H the direction reverses — G ≥ H comes down to H, then H < C goes back up — so F (via G) cannot be linked past H toward C or T in a single direction.
Trace Q: Statement 1 only fixes Q ≤ M; M is linked to T (M > T, from Conclusion I above), but that link is between M and T, not between Q and T, C, or F.
With F blocked at the H reversal and Q never linked to F’s side of the puzzle, no chain connects F and Q in either direction.
Therefore F > Q is not fixed by the statements — Conclusion II is not definitely true.
Cross-check (numeric substitution):
Let T = E = 5, D = 6, M = 7, Q = 7, so Q ≤ M > D > E = T holds. With A = 5 (T = A) or A = 2 (T > A), A ≤ M (5 ≤ 7, or 2 ≤ 7) holds either way — Conclusion I holds in every valid case. For Conclusion II, keep these values and vary the second statement: F = G = 10, H = 1, C = 2 satisfies F = G ≥ H < C < T ≥ A, and here F > Q (10 > 7); but F = G = 1, H = 0, C = 0.5 also satisfies every statement, and here F > Q fails (1 > 7 is false). Since valid assignments exist on both sides, F > Q is never forced.
Result:
Only Conclusion I is true.