Option 2 : 43

Let the runs scored by the player in 1^{st}, 2^{nd} and 3^{rd} matches are ‘x’, ‘y’ and ‘z’ respectively.

Given that,

(x + y + z)/3 = 42

x + y + z = 126 ---- (1)

Also given,

(y + z)/2 = 31

y + z = 62 ---- (2)

And,

z = y × 55/100 = 11y/20 ---- (3)

From equations (1) and (2):

x + 62 = 126

x = 64

From equations (2) and (3):

y + 11y/20 = 62

31y/20 = 62

y = 40 and z = 22

∴ Required average = (x + z)/2 = (64 + 22)/2 = 43