27) C

Let speed of courier = S

Let speed of convoy = P

Tine taken to go back = 30 seconds = 1/2 x 1/60 = (1 / 120) hours

Time taken to come to the front = 3 minutes = (1 / 20) hours

__Going back__

we have to add the speed of convoy and courier to get the net speed.

1 km = (1 / 120) x (S + P)

S + P = 120

__Return to front__

Obviously, the speed of thr courier must be more than speed o convoy.

We need to subtract the speed of convoy from speed of courier to get the net speed.

1 km = (1 / 20) (S – P)

S – P = 20

P = S – 20

Using simultaneous equations,

S + P = 120

Substituting value of P,

S + (S – 20) = 120

2S – 20 = 120

2S = 140

S = 70