Plotting the biker's motion in a graph:

The portion from O to A shows uniform acceleration. The portion from A to B shows uniform velocity; and the portion from B to C shows uniform retardation.
In this figure, we know the following:
t1 + t2 + t3 = 25
t1 = t2
Also, t3 = t1/2 because the deceleration is twice that of the acceleration.
This gives t1 = t2 = 10 min; and t3 = 5 mins
v = 0 + 3*10*60 = 1800 m/s.
The distance covered is the area under the curve.
Distance = \frac{1}{2}\cdot v\cdot t1+v\cdot t2+\frac{1}{2}\cdot v\cdot t3=v\left(\frac{t1}{2}+t2+\frac{t3}{2}\right)