Understanding the Problem Setup

This problem involves calculating distances based on relative positions and movement speeds. We have three locations in a straight line: a police station (P), a bank (B), and a safe house (S). The bank is located between the police station and the safe house. We are given information about the distances between these locations and a chase scenario involving a thief and a policeman.

Defining Variables and Relationships

Let's define the distances:

• Let '$d_{BS}$' be the distance between the Bank (B) and the Safe House (S).
• Let '$d_{PB}$' be the distance between the Police Station (P) and the Bank (B).
• Let '$d_{PS}$' be the distance between the Police Station (P) and the Safe House (S).

Since the bank is between the police station and the safe house in a straight line, the total distance '$d_{PS}$' is the sum of '$d_{PB}$' and '$d_{BS}$':

$$ d_{PS} = d_{PB} + d_{BS} $$

We are told that the distance between the police station and the safe house is four times the distance between the bank and the safe house:

$$ d_{PS} = 4 \times d_{BS} $$

Now, we can combine these two equations:

$$ d_{PB} + d_{BS} = 4 \times d_{BS} $$

Solving for '$d_{PB}$', we get:

$$ d_{PB} = 4 \times d_{BS} - d_{BS} $$

$$ d_{PB} = 3 \times d_{BS} $$

This means the distance between the police station and the bank is three times the distance between the bank and the safe house. Our goal is to find '$d_{PB}$'.

Analyzing the Chase Scenario

The thief loots the bank at 9:15 A.M. and runs away. The policeman starts chasing instantly. The policeman catches the thief at the gate of the safe house after fifteen minutes.

• The time taken for the chase is 15 minutes. Let's convert this to hours:
  $t = 15 \text{ minutes} = \frac{15}{60} \text{ hours} = \frac{1}{4} \text{ hour}$.
• The policeman's speed is given as $v_p = 16$ km/h.
• The policeman catches the thief at the safe house. This implies the policeman traveled from the police station (P) to the safe house (S) in 15 minutes.
• The distance covered by the policeman is '$d_{PS}$'.

Calculating the Distances

Using the formula: distance = speed × time, we can find the distance the policeman traveled ('$d_{PS}$'):

$$ d_{PS} = v_p \times t $$

$$ d_{PS} = 16 \text{ km/h} \times \frac{1}{4} \text{ h} $$

$$ d_{PS} = 4 \text{ km} $$

Now we know the distance between the police station and the safe house is 4 km.

We can use the relationship '$d_{PS} = 4 \times d_{BS}$' to find the distance between the bank and the safe house ('$d_{BS}$'):

$$ 4 \text{ km} = 4 \times d_{BS} $$

Dividing both sides by 4:

$$ d_{BS} = \frac{4 \text{ km}}{4} $$

$$ d_{BS} = 1 \text{ km} $$

Finally, we need to find the distance between the bank and the police station ('$d_{PB}$'). We use the relationship we derived earlier: '$d_{PB} = 3 \times d_{BS}$':

$$ d_{PB} = 3 \times 1 \text{ km} $$

$$ d_{PB} = 3 \text{ km} $$

Conclusion

The distance between the bank and the police station is 3 km.