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Asked by Jane Last Modified
Answer 
et the initial time taken be "t" hours.
Let the initial speed be V km/hr.
Case 1: when speed increases by 20 %, it takes 1hour less to cover the same distance. Let the distance be "x" km.
Equation 1: t*v= x
Equation 2: 1.2*(t-1)= x
Equating the L.H.S. on both sides, we get
t= 1.2t-1.2
So,t= 6 hours.
Case 2: As the car travels at a uniform speed of vkm/hr for the first 120km and 25% higher speed for the rest of the distance saving 48 minutes of journey time on the whole,
Equation 3 is
(120/v) +((x-120)/(1.25*v))= t-0.8
Às t*v=x and we have already computed t= 6 hours, 6*v=x.
Substituting the value of (6*v) instead of x and 6 hours as "t" value in equation 3, we get
V= 60km/hr.
Hence distance x= 60*6=360 km
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