Given below are two statements :
Statement I : Ravi walks from his house at a speed of 5 km per hour and reaches the college 10 minutes late. If he increases the speed by 1 km per hour next day, he reaches the college 4 minutes  earlier than the scheduled time. If the college is P km far from his house, then P = 7.5 km.

Statement II: Amit runs $$2\frac{1}{3}$$ times as fast as Babita. If Amit gives Babita a start of 80 meters, then the winning post must be 140 meters far so that Amit and Babita might reach it at the same time.
In the light of the above statements, choose the correct answer from the options given below.

Statement I:

Let the actual time taken by Ravi to reach the college be 't'.

If Ravi travels at a speed of 5 km/hr, time taken is t + $$\frac{10}{60}$$

If Ravi travels at a speed of 6 km/hr, time taken is t - $$\frac{4}{60}$$

Distance = speed*time = P = $$5\left(t+\frac{10}{60}\right)=6\left(t-\frac{4}{60}\right)$$

$$5\left(t+\frac{10}{60}\right)=6\left(t-\frac{4}{60}\right)$$

$$5t+\frac{50}{60}=6t-\frac{24}{60}$$

$$t=\frac{74}{60}$$

Distance(P) = $$5\left(\frac{74}{60}+\frac{10}{60}\right)=5\left(\frac{84}{60}\right)=7$$ km

Therefore, statement I is incorrect.

Statement II:

Let the speed of Babita be 3x.

The speed of Amit is 7x.

B covers distance y. 

In the same time, A should cover 80+y.

$$\frac{y}{3x}=\ \frac{\ 80+y}{7x}$$

7y = 240 + 3y

4y = 240

y = 60

Total distance = 80 + 60 = 140m

Therefore, statement II is correct.

The answer is option D.