the beginning of day 1 of a month, Rajesh has 500 eggs. He sells 20% of the eggs by the end of the day and added b% of the eggs at the beginning of the next day and sold 20% of the eggs at the end of the next day. This pattern continued up to the end of the third day of the month when he is left with 1024 eggs. The value of b is equal to

To solve this problem, we need to determine the percentage \( b \) of the eggs Rajesh adds at the beginning of each day.

### Step-by-Step Solution

1. **Day 1:**
   - Rajesh starts with \(500\) eggs.
   - He sells \(20\%\) of the eggs by the end of the day.
   - Number of eggs sold on Day 1: 
     \[
     \frac{20}{100} \times 500 = 100
     \]
   - Eggs remaining after Day 1:
     \[
     500 - 100 = 400
     \]

2. **Day 2:**
   - At the beginning of Day 2, he adds \( b\% \) of the remaining eggs from Day 1.
   - The number of eggs added:
     \[
     \frac{b}{100} \times 400
     \]
   - Total eggs at the beginning of Day 2:
     \[
     400 + \frac{b}{100} \times 400 = 400 \left(1 + \frac{b}{100}\right)
     \]
   - He sells \(20\%\) of the eggs at the end of Day 2.
   - Eggs sold on Day 2:
     \[
     0.2 \times 400 \left(1 + \frac{b}{100}\right) = 80 \left(1 + \frac{b}{100}\right)
     \]
   - Eggs remaining after Day 2:
     \[
     400 \left(1 + \frac{b}{100}\right) - 80 \left(1 + \frac{b}{100}\right) = 320 \left(1 + \frac{b}{100}\right)
     \]

3. **Day 3:**
   - At the beginning of Day 3, he adds \( b\% \) of the remaining eggs from Day 2.
   - The number of eggs added:
     \[
     \frac{b}{100} \times 320 \left(1 + \frac{b}{100}\right)
     \]
   - Total eggs at the beginning of Day 3:
     \[
     320 \left(1 + \frac{b}{100}\right) + \frac{b}{100} \times 320 \left(1 + \frac{b}{100}\right) = 320 \left(1 + \frac{b}{100}\right) \left(1 + \frac{b}{100}\right)
     \]
   - He sells \(20\%\) of the eggs at the end of Day 3.
   - Eggs sold on Day 3:
     \[
     0.2 \times 320 \left(1 + \frac{b}{100}\right)^2 = 64 \left(1 + \frac{b}{100}\right)^2
     \]
   - Eggs remaining after Day 3:
     \[
     320 \left(1 + \frac{b}{100}\right)^2 - 64 \left(1 + \frac{b}{100}\right)^2 = 256 \left(1 + \frac{b}{100}\right)^2
     \]

   We know that after Day 3, Rajesh has 1024 eggs. Therefore, we equate the remaining eggs to 1024:
   \[
   256 \left(1 + \frac{b}{100}\right)^2 = 1024
   \]

4. **Solve for \( b \):**
   \[
   \left(1 + \frac{b}{100}\right)^2 = \frac{1024}{256} = 4
   \]
   \[
   1 + \frac{b}{100} = \sqrt{4} = 2
   \]
   \[
   \frac{b}{100} = 2 - 1 = 1
   \]
   \[
   b = 100
   \]

So, the value of \( b \) is **100**.

### Conclusion

Rajesh adds 100% of the eggs at the beginning of each day.

Would you like more details or have any questions? Here are some related questions you might find interesting:

1. What if the percentage sold each day was different?
2. How would the solution change if the initial number of eggs was different?
3. What happens if Rajesh decides to add a fixed number of eggs each day instead of a percentage?
4. How would the solution be affected if the percentage added each day was less than 100%?
5. Can this method be used to solve similar problems with different conditions?

**Tip:** When dealing with percentage problems, it’s helpful to express percentages as decimals to simplify calculations.

Learn how to solve Rajesh's egg problem involving percentage sales and exponential growth over three days. Follow a step-by-step solution to find the percentage of eggs Rajesh adds daily. Suitable for Grades 7-9.

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