In each of the following questions, two statements are given, followed by four…

Concept: In syllogisms, a universal affirmative statement of the form “All A are B” can be validly converted only to the particular statement “Some B are A” — never to another universal “All B are A.” Also, two universal affirmatives chained together, “All A are B” and “All B are C,” combine into a further universal affirmative, “All A are C.”

Application:

1. Chain the two given universal statements: from “All green are blue” and “All blue are white,” it follows that “All green are white.”

2. Convert “All green are blue” (a universal affirmative) into its valid particular form: “Some blue are green” — this is guaranteed to be true.

3. Convert the derived “All green are white” the same way: “Some white are green” — this is also guaranteed to be true.

4. “Some green are not white” directly contradicts the derived “All green are white,” so it cannot follow.

5. “All white are blue” would need “All blue are white” to convert into another universal statement, which the conversion rule forbids — only “Some white are blue” follows, so this conclusion does not hold.

Cross-check:

Picture the three sets as nested circles: Green inside Blue inside White. Every member of Green is necessarily also inside Blue and inside White, which independently confirms both “Some blue are green” and “Some white are green.” Nothing in this picture forces every member of White to sit inside Blue, confirming that the reverse inclusion is not guaranteed.

So exactly two conclusions are logically guaranteed — “Some blue are green” and “Some white are green” — matching the option that states precisely these two together.