An ant moved for several seconds and covered 3 mm in the first second and 4 mm more in each successive second than in its predecessor. If the ant had covered 1 mm in the first second and 8 mm more in each successive second, then the difference between the path it would cover during the same time and the actual path would be more than 6 mm but less than 30 mm. Find the time for which the ant moved (in seconds)

Options

A. 5 s
B. 4 s
C. 6 s
D. 2 s

Answer

To solve the problem step by step, we need to analyze the movement of the ant in both scenarios and then find the time for which the ant moved.

### Step 1: Determine the distance covered by the ant in the first scenario.
In the first scenario:
- The ant covers 3 mm in the first second.
- In each successive second, it covers 4 mm more than the previous second.

Let’s calculate the distance covered in the first few seconds:
- **1st second:** 3 mm
- **2nd second:** 3 mm + 4 mm = 7 mm
- **3rd second:** 7 mm + 4 mm = 11 mm
- **4th second:** 11 mm + 4 mm = 15 mm

Now, let's find the total distance covered in `t` seconds:
- Total distance = 3 + 7 + 11 + 15 + ... (up to `t` seconds)

This forms an arithmetic series where:
- First term (a) = 3 mm
- Common difference (d) = 4 mm
- Number of terms (n) = t

The formula for the sum of the first `n` terms of an arithmetic series is:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]

So, the total distance covered in `t` seconds is:
\[ S_t = \frac{t}{2} \times (2 \times 3 + (t - 1) \times 4) \]
\[ S_t = \frac{t}{2} \times (6 + 4t - 4) \]
\[ S_t = \frac{t}{2} \times (4t + 2) \]
\[ S_t = t(2t + 1) \text{ mm} \]

### Step 2: Determine the distance covered by the ant in the second scenario.
In the second scenario:
- The ant covers 1 mm in the first second.
- In each successive second, it covers 8 mm more than the previous second.

Let’s calculate the distance covered in the first few seconds:
- **1st second:** 1 mm
- **2nd second:** 1 mm + 8 mm = 9 mm
- **3rd second:** 9 mm + 8 mm = 17 mm
- **4th second:** 17 mm + 8 mm = 25 mm

Now, let's find the total distance covered in `t` seconds:
- Total distance = 1 + 9 + 17 + 25 + ... (up to `t` seconds)

This also forms an arithmetic series where:
- First term (a) = 1 mm
- Common difference (d) = 8 mm
- Number of terms (n) = t

Using the same formula for the sum:
\[ S_t = \frac{t}{2} \times (2 \times 1 + (t - 1) \times 8) \]
\[ S_t = \frac{t}{2} \times (2 + 8t - 8) \]
\[ S_t = \frac{t}{2} \times (8t - 6) \]
\[ S_t = t(4t - 3) \text{ mm} \]

### Step 3: Set up the inequality based on the problem statement.
According to the problem, the difference between the two paths covered is more than 6 mm but less than 30 mm:
\[ 6 < |t(4t - 3) - t(2t + 1)| < 30 \]

This simplifies to:
\[ 6 < t |(4t - 3) - (2t + 1)| < 30 \]
\[ 6 < t |2t - 4| < 30 \]

### Step 4: Solve the inequality.
We can break this down into two cases based on the absolute value.

1. **Case 1:** \( 2t - 4 \geq 0 \) (i.e., \( t \geq 2 \))
- The inequality becomes:
\[ 6 < t(2t - 4) < 30 \]

2. **Case 2:** \( 2t - 4 < 0 \) (i.e., \( t < 2 \))
- The inequality becomes:
\[ 6 < t(4 - 2t) < 30 \]

### Step 5: Analyze the cases.
For **Case 1**:
- \( 6 < 2t^2 - 4t < 30 \)
- Solving \( 2t^2 - 4t - 6 > 0 \) gives us the roots and intervals.
- Solving \( 2t^2 - 4t - 30 < 0 \) gives us the roots and intervals.

For **Case 2**:
- \( 6 < 4t - 2t^2 < 30 \)
- Solving \( 2t^2 - 4t + 6 < 0 \) gives us the roots and intervals.
- Solving \( 2t^2 - 4t - 30 > 0 \) gives us the roots and intervals.

### Conclusion:
After analyzing both cases, we find that the only integer solution that satisfies the conditions of the problem is \( t = 4 \) seconds.

Thus, the time for which the ant moved is **4 seconds**.