A car after traveling 18 km from a point A developed some problem in the engine and the speed became $$\frac{4}{5}$$ th of its original speed. As a result, the car reached point B 45 minutes late. If the engine had developed the same problem after traveling 30 km from A, then it would have reached 36 minutes late. The original speed of the car (in km/h) is:
a) 25 kmph
b) 30 kmph
c) 20 kmph
d) 35 kmph

Answer: c
Explanation: He proceeds at $$\frac{4}{5}$$ S where S is his usual speed means $$\frac{1}{5}$$ decrease in speed which will lead to $$\frac{1}{4}$$ increase in time.
Now the main difference comes in those 12km (30 – 18) and the change in difference of time = (45 – 36) min = 9 min.
$$\frac{1}{4}$$ × T = 9 min where T is the time required to cover the distance of (30 – 18) = 12 km
T = 36 min = $$\frac{36}{60}$$ hours = 0.6 hours.
Speed of the car = $$\frac{12}{0.6}$$ = 20 kmph.