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 TÂ
Now we get S/T =80Â
So S=80TÂ
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 ;Â
Let any distance multiple of 80 , we take 160 km
Time taken to cover 160km =Â $$\frac{160}{80}=\ 2\ hr$$
75% of the distance = $$160\times\ \frac{3}{4}=120km$$
Time taken is 2/3 of original time =Â $$2hr\times\ \frac{2}{3}=\ 80\ \min$$
Remaining distance = 160 - 120 = 40 km
time taken = 2 hour - 80 min= 40 min
Now , According to question,Â
$$x=\frac{\left(remaining\ dis\tan ce\right)}{\left(remaining\ time\right)}$$
$$\frac{40}{40\times\ \frac{1}{60}}=\ 60\ \frac{km}{hr}$$
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