Question

One day, eesha started 30 minutes late from home and reached her office 50 minutes late, while driving 25% slower than her usual speed. how much time in minutes does eesha usually take to reach her office from home?

Answer 1

Let the speed which Eesha usually take to reach her office from home = s

Let the time taken by Eesha in normal days when she was not late = t

Since, Eesha was driving 25% slower than her usual speed.

Therefore, 25% of s

= $\frac{25}{100} \times s$

= $\frac{s}{4}$

Therefore, speed at which Eesha was driving when she was late = $s-\frac{s}{4}$

= $\frac{3s}{4}$

Since, Distance = Speed $speed \times time$

As speed and time are inversely related to each other.

When speed became $\frac{3s}{4}$  then Time taken should be  $\frac{4t}{3}$ .

So, Extra time taken = $\frac{4t}{3} - t$ = $\frac{t}{3}$

Since, Eesha took 20 min extra time due to slow driving.

Therefore, $\frac{t}{3} = 20$

t = 3$\times 20$ = 60

Therefore, Actual time taken by Eesha = 60 minutes.

Answer 2

Let time taken by eesha daily = x

and usual speed = y

(x+20)*(.75*y) = x*y extra time taken = 50-30 = 20min

.75x+15 = x

.25x = 15

x = 60 min