A car left 3 minutes early than the scheduled time and in order to reach the destination 126 km away in time, it has to slow its speed by 6 km/h from the usual. What is the usual speed (in km/hr) of the car?

Let usual speed of car = $$x$$ km/hr, speed on that day = $$(x-6)$$ km/hr

Let usual time = $$t$$ hours, time on that day = $$(t-\frac{3}{60})$$ hours

Using, time = distance/speed,

=> $$(\frac{126}{x-6})-(\frac{126}{x})=\frac{3}{60}$$

=> $$126(\frac{1}{x-6}-\frac{1}{x})=\frac{1}{20}$$

=> $$\frac{x-(x-6)}{x(x-6)}=\frac{1}{2520}$$

=> $$x^2-6x-15120=0$$

=> $$x^2-126x+120x-15120=0$$

=> $$x(x-126)+120(x-126)=0$$

=> $$(x-126)(x+120)=0$$

=> $$x=126,-120$$

$$\because x$$ cannot be negative, hence usual speed of car = 126 km/hr

=> Ans - (D)