A large bin contains forty colored balls numbered \(161\) to \(200.\) If each ball is colored either green or red, how

What we see: a Data Sufficiency question about fractions of groups. 40 balls numbered 161 to 200, each red or green. We want the number of red balls.

Step 1. Get the hidden totals from the numbering.
What we see: the balls are numbered 161, 162, ..., 200, which is \(200 - 161 + 1 = 40\) balls.
What it means: odd and even numbers take turns, so there are exactly 20 even-numbered balls (162 to 200) and 20 odd-numbered balls (161 to 199).
What we do and why: the question stem already gives us the size of each group. Any fraction of a group now becomes a real count.

Step 2. Statement (1): \(\frac{3}{5}\) of the even balls are red, \(\frac{1}{4}\) of the odd balls are green.
What we see: one fraction for each group.
What it means: even red \(= \frac{3}{5} \times 20 = 12\). Odd green \(= \frac{1}{4} \times 20 = 5\), so odd red \(= 20 - 5 = 15\).
What we do and why: red in all \(= 12 + 15 = 27\). One value, so it is sufficient.

Step 3. Statement (2): there are 3 fewer even red balls than odd red balls.
What we see: even red \(=\) odd red \(- 3\).
What it means: red in all \(= 2 \times\) (odd red) \(- 3\). If odd red is 5, there are 7 red balls; if odd red is 10, there are 17.
What we do and why: two different answers, so it is not sufficient.

Step 4. Decide.
Statement (1) alone gives one count. Statement (2) alone does not.

The trap: C, from thinking that fractions in statement (1) are useless without group sizes. The sizes were in the question all along: 20 even and 20 odd.
Takeaway: before judging a statement, squeeze every number out of the question stem. A list of numbered items tells you how many are odd and how many are even.

Answer: A