walking at 5 / 6 of its usual speed , a train is 10 minutes too late . what is usual time to cover the journey ?
"et speed be s and time to cover the journey be t s * t = d ( distance of journey ) - - - 1 now as per the question stem we have ( 5 / 6 ) s * ( t + 10 ) = d - - - - 2 equating value of d from ( 1 ) in ( 2 ) we get t = 50 mins or 5 / 6 hrs . answer : d"
divide(10, subtract(inverse(divide(5, 6)), const_1))