one day eesha started 30 min late from home and reached her office 50 min late while driving 25 % slower than her usual speed . how much time in min does eesha usually take to reach her office from home ?
we know that speed is inversely proportional to time while she drives 25 % slower means she drove at 3434 ( s ) we know that d = s x t when speed became 3434 ( s ) then time taken should be 4343 ( t ) i . e , she has taken 4343 ( t ) - t extra to cover the distance . extra time = t 3 t 3 = 20 min ( as 20 min late due to slow driving ) actual time t = 60 minutes answer : b
divide(multiply(subtract(const_1, divide(25, const_100)), subtract(50, 30)), divide(25, const_100))