In a potato race, a bucket is placed at the starting point, which is 7 meter from the first potato. The other potatoes are placed 4 m a part in a straight line from the first one. There are n potatoes in the line. Each competitor starts from the bucket, picks up the nearest potato, runs back with it, drops in the bucket, runs back to pick up the next potato, runs to the bucket and drops it and this process continues till all the potatoes are picked up and dropped in the bucket. Each competitor ran a total of 150 m. The number of potatoes is.`"____________"`

Answer

To solve the problem step by step, we will analyze the distances each competitor runs to collect the potatoes and then set up an equation based on the total distance run.

### Step 1: Understand the setup
- The first potato is 7 meters from the bucket.
- Each subsequent potato is placed 4 meters apart from the previous one.
- The distances to the potatoes can be represented as follows:
- Distance to the 1st potato (P1) = 7 m
- Distance to the 2nd potato (P2) = 7 + 4 = 11 m
- Distance to the 3rd potato (P3) = 7 + 4 + 4 = 15 m
- Distance to the nth potato (Pn) = 7 + 4(n - 1)

### Step 2: Calculate the total distance for each potato
Each competitor runs to pick up a potato and returns to the bucket. Therefore, the total distance for each potato is:
- For P1: 7 m (to P1) + 7 m (back) = 14 m
- For P2: 11 m (to P2) + 11 m (back) = 22 m
- For P3: 15 m (to P3) + 15 m (back) = 30 m
- For Pn: (7 + 4(n - 1)) + (7 + 4(n - 1)) = 2(7 + 4(n - 1))

### Step 3: Generalize the distance formula
The distance for the nth potato can be generalized as:
\[ D_n = 2(7 + 4(n - 1)) = 2(4n + 3) = 8n + 6 \]

### Step 4: Calculate the total distance for n potatoes
The total distance \( D \) run by the competitor for n potatoes is the sum of the distances for each potato:
\[ D = D_1 + D_2 + D_3 + ... + D_n \]
\[ D = 14 + 22 + 30 + ... + (8n + 6) \]

### Step 5: Identify the sequence
This forms an arithmetic sequence where:
- First term \( a = 14 \)
- Common difference \( d = 8 \)
- Number of terms \( n \)

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

### Step 6: Set up the equation
We know the total distance run is 150 m:
\[ S_n = 150 \]
Substituting the values:
\[ \frac{n}{2} \times (2 \times 14 + (n - 1) \times 8) = 150 \]
\[ \frac{n}{2} \times (28 + 8n - 8) = 150 \]
\[ \frac{n}{2} \times (8n + 20) = 150 \]

### Step 7: Simplify the equation
Multiply both sides by 2:
\[ n(8n + 20) = 300 \]
\[ 8n^2 + 20n - 300 = 0 \]

### Step 8: Solve the quadratic equation
Divide the entire equation by 4:
\[ 2n^2 + 5n - 75 = 0 \]

### Step 9: Factor the quadratic
Factoring:
\[ (2n - 10)(n + 15) = 0 \]
Setting each factor to zero gives:
1. \( 2n - 10 = 0 \) → \( n = 5 \)
2. \( n + 15 = 0 \) → \( n = -15 \) (not valid since n cannot be negative)

### Final Answer
Thus, the number of potatoes \( n \) is:
\[ n = 5 \]