If a student walks at a speed 30% more than the usual speed, he reaches 15 min earlier to his destination. How much time (in minutes) does he take to reach his destination normally?

If a student walks at a speed 30% more than the usual speed, he reaches 15 min earlier to his destination.

Let's assume the usual speed of student is 10y and the time is taken to reach the destination is 'T'.

Speed after increase = 130% of 10y = 13y

Time after the increase in the speed = (T-15)

As we know that the distance is the same in both cases.

$$10y \times T = 13y \times (T-15)$$

$$10T = 13T-13\times15$$

$$13T-10T = 13\times15$$

$$3T = 13\times15$$

$$T = 13\times5$$

= 65 minutes