The Wheatstone bridge shown in Fig. here, gets balanced when the carbon resistor used as `R_(1)` has the colour code (Orange, Red, Brown). The resistors `R_(2) and R_(4)` are `80 Omega` and `40 Omega`, respectively.
Assuming that the colour code for the carbon resistor, used as `R_(3)`, would be:

Options

A. Brown, Blue, Black
B. Grey, Black, Brown
C. Red, Green, Brown
D. Brown, Blue, Brown

Explanation

`R_1=32 times 10^1=320 Omega, R_2=80 Omega,R_4=40 Omega`
`R_1R_4=R_2=R_3 implies R_3=(R_1R_14)/R_2=(320 times 40)/80=160 Omega`