Walking at speed of 5kmph lisa reaches her office 10 minutes late.if she walks 10kmph than she reached her ofc 5 minutes early.how far her ofc
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changing 10 min = 1/6 hr and 5min = 1/12hr
let D be distance and T be original time
then D/5 = T+1/6=> D = 5T + 5/6
and D/10 = T - 1/12 => D = 10T - 10/12
then then 2D = 15T => D = 15T/2
putting in eqn 1 we get T = 1/3 hr
and putting value of T in eqn 1 we get
D = 2.5 km.
let D be distance and T be original time
then D/5 = T+1/6=> D = 5T + 5/6
and D/10 = T - 1/12 => D = 10T - 10/12
then then 2D = 15T => D = 15T/2
putting in eqn 1 we get T = 1/3 hr
and putting value of T in eqn 1 we get
D = 2.5 km.
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