Starting at the same time policewomen A and B chase thief T. They all run in the same direction at constant speeds. A runs twice as fast and B thrice as fast as T. If A and B catch up with T at the same time, B must have started

Correct option is C

Given:
Speed of Thief = x
Speed of Policewoman A = 2x
Speed of Policewoman B = 3x
Distance of A behind the thief = d
Distance of B behind the thief = D
Both catch the thief at the same time.

Formula Used:
​$$Time = \dfrac{\text{Distance}}{\text{Relative Speed}}$$​​

Solution:
Time taken by A:

​$$\dfrac{d}{2x - x} = \dfrac{d}{x}$$​​

Time taken by B:
​$$\dfrac{D}{3x - x} = \dfrac{D}{2x}$$​​

Since both times are equal:
​$$\dfrac{d}{x} = \dfrac{D}{2x}$$​​

Now, cancel x from both sides:
​$$d = \dfrac{D}{2}$$​​

Multiply both sides by 2:
2d = D

Conclusion:
D = 2d means B was twice as far behind the thief as A was.

Final Answer: (C) twice as far behind T as A did