Maths

speed for the rest of the journey = v
speed for the first 30 km of the journey = v - 48
30/(v - 48) + 10/v = 1/2
60/(v - 48) + 20/v = 1
60v + 20(v - 48) = (v - 48)v
60v + 20v - 960 = v^2 - 48v
v^2 - 128v + 960 = 0
(v - 120)(v - 8) = 0
v = 120 or v = 8 (rejected)
driving speed for the rest of the journey is 120 km/hr

Let the driving speed for the rest of the journey be s km/h.

Then, 30 / (s - 48) + (40 - 30) / s = 1/2
=> 30 / (s - 48) + 10 /s = 1/2
=> 60 / (s - 48) + 20 / s = 1
=> [60s + 20(s-48)]/[s(s-48)] = 1
=> 80s - 960 = s(s-48)
=> 80s - 960 = s^2 - 48s
=> s^2 - 128s + 960 = 0
=> (s - 120)(s - 8) = 0
=> s = 120 or s = 8 (rejected since the driving speed for the 1st 30 km = 8 - 48 = -40 < 0)
Therefore the driving speed for the rest of the journey = 120.

Let the driving speed for the first 30 km of the journey be x km/h and
the driving speed for the rest of the journey be y km/h

x = y - 48 ... (1)
(30/x) + (10/y) = 1/2 ... (2)

Substitute (1) into (2),
[30/(y - 48)] + (10/y) = 1/2
30y + 10(y - 48) = (1/2)(y)(y - 48)
60y + 20y - 960 = y^2 - 48y
y^2 - 128y + 960 = 0
(y - 120) (y - 8) = 0
y = 120 or y = 8 (rejected as y > = 48)

Substitute (y = 120) into (1),
x = 120 - 48
x = 72

Therefore, the driving speed for the rest of the journey is 120 km/h.

If there is a mistake, please inform me.