Every day a mongo seller sells half his stock `10%` of the stock overnight gets spoiled. If 1983 mangoes rotted over 3 nights then how many did he start with on the first day?

Options

A. 25000
B. 24000
C. 30000
D. 32000

Answer

To solve the problem, we need to work backwards from the information given about the mangoes that rotted over three nights. We will denote the initial number of mangoes on the first day as \( x \).

### Step 1: Understand the process
Each day, the seller sells half of his stock and 10% of the remaining stock gets spoiled overnight. We need to find out how many mangoes he started with on the first day given that a total of 1983 mangoes rotted over three nights.

### Step 2: Calculate the rotting process backwards
Let's denote the number of mangoes left at the end of each day as follows:
- Day 1: \( x \) (initial stock)
- Day 2: \( y \) (stock left after Day 1)
- Day 3: \( z \) (stock left after Day 2)

### Step 3: Calculate the number of mangoes spoiled each night
1. **Night 3**:
- Let’s denote the number of mangoes left after Day 2 as \( z \).
- 10% of \( z \) gets spoiled: \( 0.1z \)
- The total rotting after Night 3 is \( 0.1z \).

2. **Night 2**:
- Let’s denote the number of mangoes left after Day 1 as \( y \).
- After selling half, the seller has \( y = \frac{x}{2} - 0.1\left(\frac{x}{2}\right) = 0.9 \cdot \frac{x}{2} = \frac{0.9x}{2} \).
- The number of mangoes left after Day 2 is \( z = y - 0.1y = 0.9y = 0.9 \cdot \frac{0.9x}{2} = \frac{0.81x}{2} \).
- Therefore, the number of mangoes spoiled after Night 2 is \( 0.1z = 0.1 \cdot \frac{0.81x}{2} = \frac{0.081x}{2} \).

3. **Night 1**:
- The initial stock was \( x \), and after selling half, the seller has \( \frac{x}{2} \).
- The number of mangoes left after Day 1 is \( y = \frac{x}{2} - 0.1\left(\frac{x}{2}\right) = 0.9 \cdot \frac{x}{2} = \frac{0.9x}{2} \).
- The number of mangoes spoiled after Night 1 is \( 0.1y = 0.1 \cdot \frac{x}{2} = \frac{0.1x}{2} \).

### Step 4: Set up the equation for total rotting
The total number of mangoes that rotted over three nights is given as 1983:
\[ 0.1\left(\frac{x}{2}\right) + 0.1\left(\frac{0.9x}{2}\right) + 0.1\left(\frac{0.81x}{2}\right) = 1983 \]

### Step 5: Solve the equation
Combining the terms:
\[ 0.1\left(\frac{x}{2}\right) + 0.1\left(\frac{0.9x}{2}\right) + 0.1\left(\frac{0.81x}{2}\right) = 1983 \]
\[ \frac{0.1x}{2} + \frac{0.09x}{2} + \frac{0.081x}{2} = 1983 \]
\[ \frac{0.271x}{2} = 1983 \]
\[ 0.271x = 1983 \times 2 \]
\[ 0.271x = 3966 \]
\[ x = \frac{3966}{0.271} \approx 14600 \]

### Final Answer
The seller started with approximately **14600 mangoes** on the first day.

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