Step 1: Think in terms of carbon content instead of phase names:
Every steel type on the Fe-C diagram is really just defined by how much carbon it has compared to the eutectoid point at about 0.76 percent carbon.
So instead of memorizing definitions, we can work out which side of 0.76 percent this alloy must sit on.
Step 2: Use the phase that forms first as a clue:
If the steel had exactly 0.76 percent carbon, cooling through the eutectoid line would give a single-step change straight from austenite to 100 percent pearlite, with nothing forming early.
Since this alloy forms a solid phase (ferrite) before it even reaches the eutectoid line, its carbon content has to be below 0.76 percent, because ferrite is the low-carbon phase (about 0.02 percent carbon or less) and the material rejects carbon into the remaining austenite as ferrite grows.
If the alloy had carbon above 0.76 percent, the excess carbon would instead push out pro-eutectoid cementite (a high-carbon phase), not ferrite.
\[ \%C_{alloy} < 0.76\%, \quad \text{ferrite (low-C phase) precipitates first} \]
Step 3: Confirm the classification:
Carbon content below the eutectoid value, with ferrite forming first and pearlite forming at the eutectoid line, is the standard definition of hypo-eutectoid steel.
This also rules out hyper-eutectoid (needs cementite first), eutectoid (needs no pro-eutectoid phase), and hyper-eutectic (that term belongs to cast iron, above 4.3 percent carbon, not steel at all).
Final Answer:
Working backward from which phase forms first confirms low carbon content and a hypo-eutectoid classification.
\[ \boxed{\text{Hypo-eutectoid steel}} \]