Question

a man started 15 min late and by traveling at a speed which is 5/4th of the usual speed reached his office 20 min early.what is the usual time of his journey?

Answer 1

$d = st_{1} = \frac{5s}{4} (t_{1} - \frac{35}{60})$

$\frac{4}{5} t_{1} =t_{1}- \frac{35}{60}$

$\frac{t_{1}}{5} =\frac{35}{60}$

$t_{1} = 5 * 35 = 175 mins$