A statement is given followed by two conclusions. Find which conclusion is TRUE based on the given statement. Statement: C < A ≤ T = D ≥ O; G ≥ D < S; T ≥ A Conclusions: I: A ≤ O II: C < G

Correct — D.
Rule: a conclusion holds only if one chain joins its two letters with every sign facing the same way.

II: C < A ≤ T = D, and G ≥ D means D ≤ G.
So C < A ≤ T = D ≤ G, and the strict < makes C < G certain. II is true.

I: A ≤ T = D, but D ≥ O.
The signs meet at D from opposite sides, so A and O are unrelated. I is not true.

Only II is true → option (d).

Why the others are wrong

• (a)
  Either I or II — Either I or II is for two uncertain conclusions on the same pair of letters that between them cover every case. Here II is certain, and I and II name different letters.
• (b)
  Neither I nor II — II is certain: C < A ≤ T = D ≤ G joins C to G with every sign facing one way, and the strict < at the start makes C < G definite.
• (c)
  Only I — I breaks at D: A ≤ D and O ≤ D put both letters at or below D, so O could be above, equal to or below A.

A coded inequality is solved by building one chain between the two letters a conclusion names. Every sign on the way must face the same direction, and one strict sign makes the result strict.

Where two signs point at a shared letter from opposite sides, as in A ≤ D ≥ O, the chain breaks and the outer letters are unrelated. That break is what sinks conclusion I here.

Key facts

• A chain holds only when every sign in it faces the same way.
• One strict sign (< or >) in a valid chain makes the whole relation strict.
• Signs pointing at a shared letter from opposite sides, as in A ≤ D ≥ O, leave the outer letters unrelated.

Common traps

• Reading A ≤ T = D ≥ O as A ≤ O by ignoring the direction of the last sign
• Leaving G ≥ D unread, so C is never linked to G

No directly related past PYQ was found.