A car left 42 minutes early than the scheduled time but in order to reach its destination 210 km away in time, it had to slow its usual speed by 10 km/hr. What is the usual speed of the car?

Distance = 210 km

Early departure time = 42 minutes =$$\frac{42}{60}$$​hours = 0.7 hours

Speed reduction = 10 km/hr

​$$\text{Time} = \frac{\text{Distance}}{\text{Speed}}$$​​

Let the usual speed of the car be x km/h.

Usual time taken:

​$$\text{Usual time} = \frac{210}{x}$$​​

New time taken after reducing speed by 10 km/h:

​$$\text{New time} = \frac{210}{x - 10}$$​​

According to the problem:

​$$\frac{210}{x - 10} - \frac{210}{x} = 0.7 \\\frac{210(x - x + 10)}{x(x - 10)} = 0.7 \\\frac{210 \times 10}{x(x - 10)} = 0.7 \\2100 = 0.7x(x - 10) \\3000 = x(x - 10) \\x^2 - 10x - 3000 = 0 \\x^2 – 60x + 50x – 3000 = 0 \\x(x – 60) + 50(x – 60) = 0 \\(x + 50) (x - 60) = 0$$​​

Now, x = -50 or 60 

As speed can’t be negative, So

x = 60 km/hr  

Thus, the usual speed of car be 60 km/hr