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2008 Section 1 Question 27

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

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