Consider 3.5 kg of austenite containing 0.95 wt% c and cooled to below 727°c (1341°f). (a) what is the proeutectoid phase? (b) how many kilograms each of total ferrite and cementite form? (c) how many kilograms each of pearlite and the proeutectoid phase form?

A. The proeutectoid phase is Fe₃c because 0.95 wt/c  is greater than the eutectoid composition which is 0.76 wt/c

b.  We determine how much total territe and cementite form, we apply the lever rule expressions yields.
Wx = (fe₃c-co/cfe₃ c-cx = 6.70- 0.95/6.70- 0.022 = 0.86
The total cementite
Wfe₃C = 10-Cx/ Cfe₃c -Cx = 0.95 - 0.022/6.70 - 0.022 = 0.14
The total cementite which is formed is 
(0.14) × (3.5kg) = 0.49kg

c.  We calculate the pearule and the procutectoid phase which cementite form the equation
Ci = 0.95 wt/c
Wp = 6.70 -ci/6.70 - 0.76 = 6.70 -0.95/6.70 - 0.76 = 0.97
0.97 corresponds to mass.
W fe₃ C¹ = Ci - 0.76/5.94 = 0.03
∴ It is equivalent to 
(0.03) × (3.5) = 0.11kg of total of 3.5kg mass.