A thief runs away from a police station with a uniform speed of 100 m/minute. After one minute a policeman runs behind the thief to catch him. He goes at speed of 100 m/minute in first minute and increases his speed 10 m each succeeding minute. After how many minutes, the policeman will catch the thief?

Options

A. 2 mins
B. 3 mins
C. 4 mins
D. 5 mins

Answer

To solve the problem step by step, we will analyze the situation of the thief and the policeman, and use the concept of arithmetic progressions (AP) to find out when the policeman catches the thief.

### Step-by-Step Solution:

1. **Understanding the Speeds**:
- The thief runs at a constant speed of 100 meters per minute.
- The policeman starts running 1 minute later. In the first minute, he runs at 100 meters per minute and increases his speed by 10 meters per minute each subsequent minute.

2. **Distance Traveled by the Thief**:
- In the first minute, the thief covers:
\[ \text{Distance}_{\text{thief}} = 100 \text{ m/min} \times 1 \text{ min} = 100 \text{ meters} \]
- After that, for every minute \( N \), the thief continues to run at 100 m/min. Therefore, after \( N \) minutes, the total distance traveled by the thief is:
\[ \text{Distance}_{\text{thief}} = 100 \text{ m/min} \times (N + 1) \text{ min} = 100(N + 1) \text{ meters} \]

3. **Distance Traveled by the Policeman**:
- In the first minute, the policeman runs 100 meters.
- For the next \( N \) minutes, his speed increases by 10 m/min each minute. The speeds for each minute are:
- 1st minute: 100 m/min
- 2nd minute: 110 m/min
- 3rd minute: 120 m/min
- ...
- \( N \)th minute: \( 100 + 10(N - 1) \) m/min

- The distance traveled by the policeman can be calculated using the formula for the sum of an arithmetic series:
\[ \text{Distance}_{\text{policeman}} = 100 + (100 + 10) + (100 + 20) + \ldots + (100 + 10(N - 1)) \]
- This can be simplified using the formula for the sum of the first \( N \) terms of an AP:
\[ S_N = \frac{N}{2} \times (2a + (N - 1)d) \]
where \( a = 100 \) and \( d = 10 \):
\[ \text{Distance}_{\text{policeman}} = \frac{N}{2} \times (2 \times 100 + (N - 1) \times 10) = \frac{N}{2} \times (200 + 10N - 10) = \frac{N}{2} \times (10N + 190) \]

4. **Setting Distances Equal**:
- To find when the policeman catches the thief, we set the distances equal:
\[ 100(N + 1) = \frac{N}{2} \times (10N + 190) \]

5. **Solving the Equation**:
- Multiply both sides by 2 to eliminate the fraction:
\[ 200(N + 1) = N(10N + 190) \]
- Expanding both sides:
\[ 200N + 200 = 10N^2 + 190N \]
- Rearranging gives:
\[ 10N^2 - 10N - 200 = 0 \]
- Dividing the entire equation by 10:
\[ N^2 - N - 20 = 0 \]

6. **Factoring the Quadratic**:
- Factor the quadratic:
\[ (N - 5)(N + 4) = 0 \]
- This gives us:
\[ N = 5 \quad \text{or} \quad N = -4 \]
- Since time cannot be negative, we take \( N = 5 \).

7. **Conclusion**:
- The policeman will catch the thief after **5 minutes**.