The distance of the college and the home of Rajeev is 80 km. One day he was late by 1 hour than the normal time to leave for the college, so he increased his speed by 4 km/h and thus he reached to college at the normal time. What is the changed (or increased) speed of Rajeev?

Let normal speed = v km/h, time = t hours. Distance = vt = 80. Late by 1 hour, new speed = v+4, time = t-1. Distance = (v+4)(t-1) = 80. Since vt = 80, we have vt + 4t - v - 4 = vt. 4t - v - 4 = 0. Also t = 80/v. Substitute: 4(80/v) - v - 4 = 0. 320/v - v - 4 = 0. Multiply by v: 320 - v² - 4v = 0. v² + 4v - 320 = 0. (v+20)(v-16) = 0. v = 16 (positive). New speed = v+4 = 20 km/h.