A girl, walking at $$\frac{5}{6}$$ of her usual speed, misses the bus by 7 minutes. How long does she usually take to reach the bus stop?

Let's assume the usual speed of girl is '6y'.

By the speed of 6y, she covers 'd' distance is 't' time.

d = 6yt    Eq.(i)

A girl, walking at $$\frac{5}{6}$$ of her usual speed, misses the bus by 7 minutes.

d = $$6y\times\frac{5}{6}(t+7)$$    Eq.(ii)

Eq.(i) = Eq.(ii)

So $$6yt = 6y\times\frac{5}{6}(t+7)$$

$$t = \frac{5}{6}(t+7)$$

6t = 5(t+7)

6t = 5t+35

6t - 5t = 35

t = 35 minutes

She usually takes 35 minutes to reach the bus stop.