In the question, symbols $, #, % are used for different meanings as follows. $…

2025

In the question, symbols $, #, % are used for different meanings as follows.

$ means ‘neither greater nor equal to’ (i.e., less than).

# means ‘neither greater nor smaller than’ (i.e., equal to).

% means ‘neither smaller nor equal to’ (i.e., greater than).

In the following question, 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: T % I, I # L, L % U

Conclusions:

I. T $ L

II. U $ T

  1. A.

    Only I is true

  2. B.

    Only II is true

  3. C.

    Either I or II is True

  4. D.

    Both I and II are true

Attempted by 25 students.

Show answer & explanation

Correct answer: B

Concept: In coded-inequality reasoning, each custom symbol stands for one of the standard relations (<, >, =). Once every symbol is decoded, chain the statements using transitivity — if A > B and B = C, then A > C — and a conclusion is ‘definitely true’ only when it holds in every case consistent with that chain.

Application:

  1. Decode the symbols: $ = ‘<’, # = ‘=’, % = ‘>’.

  2. Translate the statements: T % I → T > I; I # L → I = L; L % U → L > U.

  3. Chain them in order: T > I = L > U, which collapses to T > L and L > U, so also T > U.

  4. Check Conclusion I (T $ L, i.e., T < L): the chain gives T > L, the opposite relation, so Conclusion I is false.

  5. Check Conclusion II (U $ T, i.e., U < T): the chain gives T > U, i.e., U < T, so Conclusion II holds in every case.

Cross-check: Assign consistent sample values, e.g., I = L = 3, T = 4, U = 2. This satisfies T > I (4 > 3), I = L (3 = 3) and L > U (3 > 2). Here T is not less than L (4 is not < 3), confirming Conclusion I fails, while U is less than T (2 < 4) in this and every other valid assignment, confirming Conclusion II always holds.

Result: Only Conclusion II is definitely true.

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