Work and Rate Problems

What we see: a work-rate Data Sufficiency question. Stark bakes alone, then Roggers joins, and 100 pies are done \(x\) hours later. We want how many pies Stark made before Roggers arrived.

Step 1. Shrink the question to what we really need.
What we see: after Roggers joins, they bake together for \(x\) hours at \(M + J\) pies per hour.
What it means: pies made together \(= (M + J) \times x\). Everything else of the 100 was baked by Stark alone, so Stark's early pies \(= 100 - (M + J)x\).
What we do and why: we do not need \(M\) or \(J\) on its own. We only need the product \((M + J) \times x\). Knowing that is enough.

Step 2. Statement (1): together they bake 25 pies per hour.
What we see: \(M + J = 25\).
What it means: we still do not know \(x\). If \(x = 1\), Stark made \(100 - 25 = 75\) early. If \(x = 2\), Stark made \(100 - 50 = 50\).
What we do and why: two different answers, so it is not sufficient.

Step 3. Statement (2): they finish together in 2 hours.
What we see: \(x = 2\).
What it means: we still do not know their combined speed. At 10 pies per hour, Stark made 80 early; at 30 pies per hour, Stark made 40.
What we do and why: two different answers, so it is not sufficient.

Step 4. Use both statements together.
What we see: \(M + J = 25\) and \(x = 2\).
What it means: they baked \(25 \times 2 = 50\) pies together, so Stark made \(100 - 50 = 50\) pies before Roggers came.
What we do and why: one value, so together they are sufficient.

The trap: E, because we never learn \(M\) or \(J\) alone. We do not need them: after Roggers joins, only the combined speed matters.
Takeaway: in Data Sufficiency, rewrite the question as the smallest thing you need (here, \((M + J) \times x\)). A sum or a product can be enough even when its parts stay unknown.

Answer: C