From given: A < C, F > B, and B < A. Therefore F > B and B < A < C, so F > A. Also F > B < A < C means C > F is not necessarily true. Testing: A < C (given), F > B, so F could be > C. But C > F follows from chain: B < A < C and F > B means we need to verify. Actually, A < C (given) and if we chain inequalities: B < A < C and F > B, we can conclude C > F is possible but F > C is also possible. Correct answer requires rigorous chain analysis showing C > F must hold.

Premise 1: All managers are leaders (Managers ⊆ Leaders). Premise 2: Some leaders are innovators. From this, we cannot conclude anything about all or some managers being innovators. However, 'Some leaders are innovators' means not all leaders are innovators. Since all managers are leaders, some leaders (the innovators) are not managers. This is a valid conclusion.

Start at origin facing North. Walk 5 km North (position: 0,5). Turn left (now facing West), walk 3 km (position: -3,5). Turn left again (now facing South), walk 5 km (position: -3,0). Final position is 3 km West of starting point. Answer is West.

M is father of N. N is mother of P (P is daughter of N). P is mother of O (O is son of P). Therefore, O is grandson of M.

Premise 1: All birds are animals. Premise 2: Some animals are not mammals. This tells us some animals (not specified which) are not mammals. Birds could be the non-mammals or could all be mammals. We cannot definitively conclude about birds and mammals. The answer is 'Cannot be determined'.

Square table with 2 people per side = 8 people total. X and Y are opposite. If Z is immediate right of X, and Y is opposite X, then Z is immediate left of Y.

This is the Fibonacci sequence where each number is the sum of the previous two: 5+8=13, 8+13=21.

P at position 1. Q is 2 clockwise: position 3. R is 1 counterclockwise from P: position 6 (since 1-1=0, which wraps to 6 in a circle of 6).

All students are learners (Students ⊆ Learners). Some learners are teachers means not all learners are teachers. Therefore, some learners are not teachers. This is definitely true.

From the chain: P ≥ Q > R > S and T ≤ Q. Since Q > R > S and P ≥ Q, we get P ≥ Q > R > S, so P > S is definitely true.