Question

A man travels at 125% of his usual speed and reaches his office 10 minutes early.What is the usual time he takes to reach his office?

Answer 1

Answer:

The usual time is 50 min

Step-by-step explanation:

Let the normal speed of the man is v km/min

If the distance is d then the time taken

$t=\frac{d}{v}$

New speed is 125% of normal speed

Hence new speed

$v'=v\times\frac{125}{100}$

$\implies v'=1.25v$

Now time taken

$t'=t-10$

or $t-10=\frac{d}{1.25v}$

$\implies t-10=\frac{t}{1.25}$

$\implies 1.25t-12.5=t$

$\implies 0.25t=12.5$

$\implies t=\frac{12.5}{0.25}$ min

$\implies t=50$ min

Hope this helps.