This is a great median problem that tests whether you can recognize when you have enough information about a set of numbers to determine their median — a classic Data Sufficiency concept!
The question asks: If Lee, Dimitra, and Jan each paid a fine and the total was $900, what was the median fine?
Since we have 3 values, the median will be the middle value when the three fines are arranged in order.
Statement (1): Jan was fined $300
From this, we know:
- Jan = $300
- Lee + Dimitra = $900 - $300 = $600
Now here's the key insight: No matter how we split $600 between Lee and Dimitra, when we arrange all three values in order, $300 will ALWAYS be the median.
Let me show you why with a few examples:
- If Lee = $200 and Dimitra = $400: Ordered = {$200, $300, $400} → Median = $300
- If Lee = $100 and Dimitra = $500: Ordered = {$100, $300, $500} → Median = $300
- If Lee = $300 and Dimitra = $300: Ordered = {$300, $300, $300} → Median = $300
- If Lee = $450 and Dimitra = $150: Ordered = {$150, $300, $450} → Median = $300
Since $300 is exactly one-third of the total ($900), and the other two values must sum to twice that amount ($600), $300 will always fall in the middle position.
Statement (1) is SUFFICIENT.
Statement (2): Lee and Dimitra were fined a total of $600
This tells us:
- Lee + Dimitra = $600
- Jan = $900 - $600 = $300
Wait a minute — this gives us exactly the same information as Statement 1! We know Jan = $300, and the analysis is identical. The median must be $300.
Statement (2) is SUFFICIENT.
Answer: D
Common traps to avoid:
A common mistake is thinking that Statement 1 is insufficient because "we don't know how $600 is split between Lee and Dimitra." But in Data Sufficiency, we don't need to find all individual values — we only need to determine if we can find the median. The beautiful thing about this problem is that the median doesn't depend on how the remaining money is distributed. Another trap is not recognizing that Statements 1 and 2 are actually giving the same information in different forms — if you know Jan's fine, you know the sum of the other two, and vice versa.
Key takeaway: For median problems with an odd number of values, you don't always need to know every individual value. If you know enough about the structure of the data set, you can sometimes determine the median even with some unknowns. In this case, knowing one value that equals exactly one-third of the total locks in the median position. This is a powerful concept that applies whenever you have constraints that force certain values into fixed positions in the ordered list!