Two towns A and B are 120 km apart. A cyclist going from A to B covers half of the distance at twice his usual speed and the rest of the distance at half of his usual speed. If he has taken 75 minutes longer than his usual time to complete the journey, then find the usual speed of the cyclist.
A. 24 km/hr
B. 25 km/hr
C. 30 km/hr
D. 40
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Answer:
A. 24 km/h
Step-by-step explanation:
Let usual speed=v
" " time=t
S=120 km.
let new speed=vn
let usual time=tn
vt=120
vn x tn=120
vn(avg speed)= (2v1 x v2)/(v1 + v2)
v1 = 2v (given)
v2 = v/2 (given)
On substituting the values
vn = 4v/5
since...
vt=vn x tn
therefore...
vt = 4v/5 x tn
tn= 5t/4
since...tn= t+75 min.
there fore...
5t/4 = t+5/4 hrs.
On solving...
t= 5 hrs.
Since...vt=120 km
v= 120/t
= 120 km/5hrs.
= 24 km/hr.
Hence, his usual speed is 24km/hr.
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