Umesh has a few chocolates. He wants to distribute these chocolates among his three sons and three daughters. He also wants to keep some chocolates with himself. He keeps one chocolate for himself and gives 1/5 of the remaining to his first son and keeps one and gives 1/5 of the remaining to his second son, and does the same with his last son. He then distributes the remaining chocolates among his three daughters in the ratio 23:46:55. What is the minimum number of chocolates that Umesh had initially?

Pick ONE option.

246

271

221

196

Let the initial number of chocolates Umesh has be denoted by \( x \).

1. Umesh keeps **1 chocolate** for himself. The remaining number of chocolates is \( x - 1 \).

2. He gives **1/5** of the remaining chocolates to his first son:
   \[
   \text{Remaining after first son} = \frac{4}{5} (x - 1)
   \]

3. Umesh keeps **1 chocolate** again and gives **1/5** of the remaining to his second son:
   \[
   \text{Remaining after Umesh keeps 1} = \frac{4}{5} \left(\frac{4}{5} (x - 1) - 1\right)
   \]

4. He keeps **1 chocolate** again and gives **1/5** of the remaining to his third son:
   \[
   \text{Remaining after Umesh keeps 1 again} = \frac{4}{5} \left(\frac{4}{5} \left(\frac{4}{5} (x - 1) - 1\right) - 1\right)
   \]

5. After the chocolates are distributed to his three sons, the remaining chocolates are divided among his three daughters in the ratio \( 23:46:55 \). Since the total of this ratio is \( 23 + 46 + 55 = 124 \), the remaining number of chocolates must be divisible by 124.

Let’s now calculate the number of chocolates step by step and set it divisible by 124:

### Step 1: Total chocolates after third son’s distribution

To simplify, we compute this iteratively.

\[
\text{Remaining after first son} = \frac{4}{5}(x - 1)
\]
\[
\text{Remaining after second son} = \frac{4}{5} \left(\frac{4}{5}(x - 1) - 1\right)
\]
\[
\text{Remaining after third son} = \frac{4}{5} \left(\frac{4}{5} \left(\frac{4}{5}(x - 1) - 1\right) - 1\right)
\]

Now, for simplicity, we will check the options by substituting values and seeing which results in a number divisible by 124.

### Option Check:

1. **For \( x = 246 \):**
   \[
   \text{Remaining after first son} = \frac{4}{5}(246 - 1) = \frac{4}{5}(245) = 196
   \]
   \[
   \text{Remaining after second son} = \frac{4}{5}(196 - 1) = \frac{4}{5}(195) = 156
   \]
   \[
   \text{Remaining after third son} = \frac{4}{5}(156 - 1) = \frac{4}{5}(155) = 124
   \]
   Since the remaining chocolates after distributing to the sons is exactly 124, which can be divided among the daughters in the ratio 23:46:55, **246** is the correct answer.

Thus, the minimum number of chocolates Umesh had initially is **246**.

Would you like more details or have any questions?

Here are 5 questions to expand the information:
1. How does changing the number of chocolates affect the distribution process?
2. Why is the remaining number of chocolates divisible by 124?
3. What happens if Umesh had fewer than 246 chocolates?
4. How would the distribution change if Umesh had 4 sons instead of 3?
5. What is the mathematical significance of the ratio 23:46:55?

**Tip:** When dealing with ratio-based distribution, ensure that the total amount left is divisible by the sum of the ratio parts.

Umesh has a few chocolates. He wants to distribute these chocolates among his three sons and three daughters. He also wants to keep some chocolates with himself. He keeps one chocolate for himself and gives 1/5 of the remaining to his first son and keeps one and gives 1/5 of the remaining to his second son, and does the same with his last son. He then distributes the remaining chocolates among his three daughters in the ratio 23:46:55. What is the minimum number of chocolates that Umesh had initially? Pick ONE option. 246, 271, 221, 196

Solve the problem of determining the minimum number of chocolates Umesh initially had based on fractional distribution among sons and proportional division among daughters in the ratio 23:46:55. This algebraic problem involves key concepts such as fractional operations and ratio analysis, suitable for Grades 8-10 students.

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