Ferrite and pearlite in high strength steel

The question gives no carbon content, so the common textbook case is taken: high strength steel is high carbon steel of eutectoid composition, 0.80% C. Pearlite holds 0.80% C and cementite holds 6.67% C (carbon dissolved in ferrite is neglected).

$$\text{Pearlite} = \frac{0.80}{0.80} \times 100 = 100\%$$ $$\text{Cementite} = \frac{0.80}{6.67} \times 100 = 11.99\%, \qquad \text{Ferrite} = 100 - 11.99 = 88.01\%$$

So the structure is 100% pearlite, made of about 88% ferrite and 12% cementite layers. Free ferrite is zero.

General method for any hypo-eutectoid steel with carbon $C$ (%): pearlite $= \dfrac{C}{0.80} \times 100$ and free ferrite $= 100 -$ pearlite. For example, for 0.60% C: pearlite $= 75\%$, free ferrite $= 25\%$.

Answer: Pearlite = 100%, with 88.01% ferrite and 11.99% cementite inside it (0.80% C steel).