Year-on-year exam counts by state: In 1998, A:B:C = 3:5:6. In 1999, A and C both increase by 20%, B by 10%, and A:C becomes 1:2. What is the number of students in state A in 1999?

Introduction / Context:
This ratio problem spans two years with percentage increases. However, without at least one absolute number, pure ratios do not determine unique counts, even if a ratio condition holds in the next year.

Given Data / Assumptions:

• 1998 counts: A:B:C = 3x : 5x : 6x.
• 1999 increases: A by 20%, B by 10%, C by 20%.
• 1999 ratio A:C = 1:2.

Concept / Approach:
Compute 1999 A and C as 1.2 * 3x = 3.6x and 1.2 * 6x = 7.2x. The ratio 3.6x : 7.2x simplifies to 1 : 2 automatically for any x. Thus the A:C condition provides no new information to fix x, so absolute values remain undetermined.

Step-by-Step Solution:
1999 A = 3x * 1.2 = 3.6x1999 C = 6x * 1.2 = 7.2xA:C = 3.6x : 7.2x = 1 : 2, true for all xNo absolute count (x) is given ⇒ cannot compute numeric A for 1999

Verification / Alternative check:
Any x (e.g., x = 1,000) will satisfy the given ratio in 1999, producing different absolute values for A. Hence a unique solution is impossible.

Why Other Options Are Wrong:
7200, 6000, 7500 are arbitrary choices that cannot be derived from the provided data alone.

Common Pitfalls:
Assuming x = 1000 or some default base; without an absolute reference, ratios do not yield unique totals.

Final Answer:
Data inadequate