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?

a man started 15 min late and by traveling at a speed which is 5/4th o

1.	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» $d = st_{1} = \frac{5s}{4} (t_{1} - \frac{35}{60})$ $\frac{4}{5} t_{1} =t_{1}- \frac{35}{60}$ $\frac{t_{1}}{5} =\frac{35}{60}$ $t_{1} = 5 * 35 = 175 mins$

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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?

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