In a 1000m race, B took 50% more time than C to finish the race, and B also took 25% more time than A. If all three started the race from the same point on a circular track of 1100m length, with A and B traveling clockwise and C traveling counterclockwise, what is the distance between the first meeting point of A and C and the meeting point of B and C?

We are told that  B took 50% more time than C to finish the race, and B also took 25% more time than A.

So, $$1.5*T_c = T_b$$, $$ T_b=1.25T_a$$.

From this $$T_a = \frac{6}{5}T_c$$

Let's find the speeds of all three. $$S_c =  \frac{1000}{T_c}$$, $$S_b = \frac{1000}{T_b} = \frac{1000}{1.5*T_c}$$, $$S_a=\frac{1000}{T_a}=\frac{1000}{\frac{6}{5}*T_c}=\frac{\left(5\cdot1000\right)}{6\cdot T_c}$$

The ratio of the speeds of A, B, C will be $$\frac{5}{6}:\frac{1}{1.5}:1=\frac{5}{6}:\frac{2}{3}:1=5:4:6$$.

The distance between the meeting points of A, C and B, C. 

The ratio of the speeds of A and C is 5:6. Time taken for them to meet = $$\frac{1100}{5x+6x}=\frac{1100}{11x}=\frac{100}{x}$$

 Distance C travelled = $$6x\cdot\frac{100}{x}$$ = 600m.

The ratio of the speeds of B and C is 4:6. Time taken for them to meet = $$\frac{1100}{4x+6x}=\frac{1100}{10x}=\frac{110}{x}$$

Distance C travelled = $$6x\cdot\frac{110}{x}$$ = 660m.

So, the distance between two meeting points = 60m