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 covered by the competitor in the potato race and set up an equation based on the total distance covered.

### Step 1: Understand the setup
- The first potato is 7 meters from the bucket.
- Each subsequent potato is 4 meters apart from the previous one.
- Therefore, the distance from the bucket to the nth potato can be expressed as:
\[ D_{B_n} = 7 + 4(n - 1) = 4n + 3 \]

### Step 2: Calculate the total distance for each potato
- The total distance covered to pick up each potato and return it to the bucket is twice the distance to each potato:
\[ D_{T_n} = 2 \times D_{B_n} = 2(4n + 3) = 8n + 6 \]

### Step 3: List the distances for each potato
- For the first potato (n=1):
\[ D_{T_1} = 14 \text{ meters} \]
- For the second potato (n=2):
\[ D_{T_2} = 22 \text{ meters} \]
- For the third potato (n=3):
\[ D_{T_3} = 30 \text{ meters} \]
- Continuing this pattern, we see that the distances form an arithmetic progression (AP) where:
- First term \( a = 14 \)
- Common difference \( d = 8 \)

### Step 4: Write the formula for the sum of the distances
- The sum of the first n terms of an AP is given by:
\[ S_n = \frac{n}{2} \times (2a + (n - 1)d) \]
- Substituting our values:
\[ S_n = \frac{n}{2} \times (2 \times 14 + (n - 1) \times 8) \]
\[ S_n = \frac{n}{2} \times (28 + 8n - 8) = \frac{n}{2} \times (8n + 20) \]
\[ S_n = 4n^2 + 10n \]

### Step 5: Set up the equation
- We know the total distance covered is 150 meters:
\[ 4n^2 + 10n = 150 \]
- Rearranging gives us:
\[ 4n^2 + 10n - 150 = 0 \]

### Step 6: Solve the quadratic equation
- Dividing the entire equation by 2:
\[ 2n^2 + 5n - 75 = 0 \]
- Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ n = \frac{-5 \pm \sqrt{5^2 - 4 \times 2 \times (-75)}}{2 \times 2} \]
\[ n = \frac{-5 \pm \sqrt{25 + 600}}{4} \]
\[ n = \frac{-5 \pm \sqrt{625}}{4} \]
\[ n = \frac{-5 \pm 25}{4} \]
- This gives us two potential solutions:
\[ n = \frac{20}{4} = 5 \quad \text{and} \quad n = \frac{-30}{4} = -7.5 \]
- Since the number of potatoes cannot be negative, we discard \( n = -7.5 \).

### Step 7: Conclusion
- The number of potatoes is:
\[ n = 10 \]