Amit travelled from A to B at an average speed of 80 km/h. He travelled the first 75% of the distance in two-third of the time and the rest at a constant speed of x km/h. The value of is:

Let the distance between AB be S and time taken to cover AB be T 
Now we get S/T =80 
So S=80T 
Now 75% (S) =60T
He covered  in 2/3T
Now in remaining T/3 he covered 20T
so speed = $$\frac{20T}{\left(\frac{T}{\left(3\right)}\right)}=60km/hr$$

Alternate Method ; 

Let any distance multiple of 80 , we take 160 km

Time taken to cover 160km = $$\frac{160}{80}=\ 2\ hr$$

75% of the distance = $$160\times\ \frac{3}{4}=120km$$

Time taken is 2/3 of original time = $$2hr\times\ \frac{2}{3}=\ 80\ \min$$

Remaining distance = 160 - 120 = 40 km

time taken = 2 hour - 80 min= 40 min

Now , According to question, 

$$x=\frac{\left(remaining\ dis\tan ce\right)}{\left(remaining\ time\right)}$$

$$\frac{40}{40\times\ \frac{1}{60}}=\ 60\ \frac{km}{hr}$$