By increasing the speed of a car by 10 km/hour, the time of journey for a distance of 72 km is reduced by 36 minutes. The original speed of the car was:

Given:
Distance = 72 km
Increase in speed = 10  km/h
Reduction in time} = 36  minutes = 0.6 hours
Let original speed = x  km/h
​$$T_1 = \frac{72}{x}\\ \ \\T_2 = \frac{72}{x+10}\\[6pt]T_1 - T_2 = 0.6\\[6pt]\frac{72}{x} - \frac{72}{x+10} = 0.6\\[6pt]72\left(\frac{x+10 - x}{x(x+10)}\right) = 0.6\\[6pt]72 \left(\frac{10}{x(x+10)}\right) = 0.6\\[6pt]\frac{720}{x(x+10)} = 0.6\\[6pt]720 = 0.6x(x+10)\\[6pt]1200 = x(x+10)\\[6pt]x^2 + 10x - 1200 = 0\\[6pt]x = 30$$ 

Correct answer is (B).
Alternate Method: