Paulson usually leaves his home everyday at 8:15 am and travels at a certain speed to reach office on time. On

Question

Paulson usually leaves his home everyday at 8:15 am and travels at a certain speed to reach office on time. One day he travelled at 3/5 of the usual speed and hence arrived 45 minutes late. How much time does Paulson usually take to reach his office?

Accepted Answer

Since the distance from home to office is constant, speed and time are inversely proportional.
Let the usual speed be

S

and the usual time be

T

.
New speed

= \frac{3}{5} S

Since

Speed \propto \frac{1}{Time}

, the new time taken will be

\frac{5}{3}

of the usual time (

T

).

New Time = \frac{5}{3} T

Given that he arrives 45 minutes late:

New Time - Usual Time = 45 minutes

\frac{5}{3} T - T = 45

\frac{2}{3} T = 45

T = \frac{45 \times 3}{2} = \frac{135}{2} minutes

Convert the time into hours:

T = \frac{135}{2 \times 60} hours = \frac{135}{120} hours = \frac{9}{8} hours

T = 1 \frac{1}{8} hours

Paulson usually leaves his home everyday at 8:15 am and travels at a certain speed to reach office on time. One day he travelled at $\frac{3}{5}$ of the usual speed and hence arrived 45 minutes late. How much time does Paulson usually take to reach his office?

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Paulson usually leaves his home everyday at 8:15 am and travels at a certain speed to reach office on time. One day he travelled at 53​ of the usual speed and hence arrived 45 minutes late. How much time does Paulson usually take to reach his office?