A Woman travelling at 130% of her usual speed reaches her office 12 minutes early. Her usual time to cover the journey is:

Let's assume the distance of her office is 'd' km.

Let's assume the usual speed of a woman is 'sp' km/h.

$$Usual\ time=\frac{d}{sp}$$    Eq.(i)

A Woman travelling at 130% of her usual speed reaches her office 12 minutes early.

$$Usual\ time\ -\ \frac{12}{60}=\frac{d}{130\%\ of\ sp}$$

$$Usual\ time\ -\ \frac{12}{60}=\frac{d}{1.3sp}$$    Eq.(ii)

Put Eq.(ii) in Eq.(i).

$$\frac{d}{sp} -\ \frac{12}{60}=\frac{d}{1.3sp}$$

$$\frac{d}{sp} -\frac{d}{1.3sp} = \frac{12}{60}$$

$$\frac{1.3d}{1.3sp} -\frac{d}{1.3sp} = \frac{1}{5}$$

$$\frac{0.3d}{1.3sp} = \frac{1}{5}$$

$$\frac{3d}{13sp}=\frac{1}{5}$$

$$\frac{d}{sp}=\frac{13}{5\times3}$$

$$\frac{d}{sp}=\frac{13}{15}$$

From Eq.(i), Her usual time to cover the journey = $$\frac{13}{15}$$ hours

= 0.87 hours