A drives at the rate of 45 km/h and reaches its destination 4 minutes late. If speed is 60 km/h. A reaches 5 minutes early. The distance traveled by A is:

Let say, distance travel is X km.

Time taken at 45km/h speed is=$$\ \frac{\ X}{45}\ hr.$$

and time taken 60km/h speed is=$$\ \frac{\ X}{60}\ hr.$$

So, $$\ \frac{\ X}{45}-\ \frac{\ 4}{60}=\ \frac{\ X}{60}+\ \frac{\ 5}{60}$$

or,$$\ \frac{\ X}{45}-\ \frac{\ X}{60}=\ \frac{\ 4}{60}+\ \frac{\ 5}{60}$$

or,$$\ \frac{\ 4X-3X}{180}=\ \frac{\ 4+5}{60}$$

or,$$\ \frac{\ X}{3}=\ \frac{\ 9}{1}$$

or,$$\ \ X=\ 27.$$

So, C is correct choice.