Q4:
Part (e)
At 600°C, the alloy is within the α + δ region. This means both the α phase and the δ phase are present.
The tie line at 600°C extends from about 25% Cu (α-phase side) to roughly 85% Cu (δ-phase side).
Fraction of α phase (fα)=Cδ-C0/Cδ- Cα
Fraction of δ phase (fδ)=C0−Cα/ Cδ−Cα
fα=85−50/85−25
=35/60
fα=0.5833 or 58.33
fδ=50−25/85−25
fδ=25/60=0.4167 or 41.67
Answer: At 600°C, the alloy is:

58.33% in the α phase

41.67% in the δ phase.