A car travels from point A to point B at a constant speed of V km/h. If the car increases its speed by 20%, it will reach B one hour earlier. if the car increases its speed by 25% after traveling at V km/h for 120km, it will reach b 48mins earlier. if the distance between the two towns is X km, find the value of x?

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The answer is 360km. May i know how to derive the answer? Thanks!
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You can derive by taking original time t and speed v and distance x ,use distance formula,and make two equations and solve them.
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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= xEquation 2: 1.2*(t-1)= xEquating the L.H.S. on both sides, we get t= 1.2t-1.2So,t= 6 hours. Case 2:...
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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= xEquation 2: 1.2*(t-1)= xEquating the L.H.S. on both sides, we get t= 1.2t-1.2So,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-0.8Às 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 getV= 60km/hr.Hence distance x= 60*6=360 km read less
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