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We are given a table showing the daily expenditure of 100 families, where the number of families in two expenditure ranges (30-39 and 50-59) is missing. The mode of the distribution is 43, and we need to find the missing frequencies.

### Data provided:
| Expenditure (₹) | 20 - 29 | 30 - 39 | 40 - 49 | 50 - 59 | 60 - 69 |
|-----------------|---------|---------|---------|---------|---------|
| Number of families | 14      | ?       | 27      | ?       | 15      |

- Total number of families = 100
- Mode = 43 (which falls in the class 40-49, making it the modal class)

### Step-by-step approach:

1. **Let the missing frequencies be:**
   - \( f_2 \) for the class 30-39
   - \( f_4 \) for the class 50-59

2. **Given total frequency:**
   \[
   14 + f_2 + 27 + f_4 + 15 = 100
   \]
   Simplifying:
   \[
   56 + f_2 + f_4 = 100
   \]
   \[
   f_2 + f_4 = 44 \tag{1}
   \]

3. **Using the mode formula:**

   The mode is given by:
   \[
   \text{Mode} = L + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) \times h
   \]
   Where:
   - \( L = 40 \) (lower limit of the modal class)
   - \( f_1 = 27 \) (frequency of the modal class)
   - \( f_0 = f_2 \) (frequency of the class preceding the modal class)
   - \( f_2 = f_4 \) (frequency of the class following the modal class)
   - \( h = 10 \) (class width)

   Substituting the values we have:
   \[
   43 = 40 + \left( \frac{27 - f_2}{2(27) - f_2 - f_4} \right) \times 10
   \]
   Simplifying:
   \[
   43 = 40 + \left( \frac{27 - f_2}{54 - f_2 - f_4} \right) \times 10
   \]
   \[
   3 = \left( \frac{27 - f_2}{54 - f_2 - f_4} \right) \times 10
   \]
   \[
   \frac{3}{10} = \frac{27 - f_2}{54 - f_2 - f_4}
   \]
   Cross-multiplying:
   \[
   3(54 - f_2 - f_4) = 10(27 - f_2)
   \]
   Expanding both sides:
   \[
   162 - 3f_2 - 3f_4 = 270 - 10f_2
   \]
   Rearranging:
   \[
   270 - 162 = 10f_2 - 3f_2 - 3f_4
   \]
   \[
   108 = 7f_2 - 3f_4 \tag{2}
   \]

4. **Solving equations (1) and (2):**
   We now have two equations:
   \[
   f_2 + f_4 = 44 \tag{1}
   \]
   \[
   7f_2 - 3f_4 = 108 \tag{2}
   \]
   Solving these will give us the values of \( f_2 \) and \( f_4 \).

Let's solve these equations to find the missing frequencies.The missing frequencies are:
- \( f_2 \) (for the class 30-39) = 24
- \( f_4 \) (for the class 50-59) = 20

Thus, the complete table of daily expenditure of the 100 families is:

| Expenditure (₹) | 20 - 29 | 30 - 39 | 40 - 49 | 50 - 59 | 60 - 69 |
|-----------------|---------|---------|---------|---------|---------|
| Number of families | 14      | 24      | 27      | 20      | 15      |

Let me know if you need further clarifications or details!

### Related questions:
1. How is the mode of a grouped data distribution calculated?
2. What is the significance of the modal class in data analysis?
3. Can the mode be used to describe skewed distributions effectively?
4. How do you calculate the mean from grouped data when frequencies are missing?
5. What are other central tendency measures that can be derived from this data?

### Tip:
When solving for missing frequencies in statistics, always cross-check your final values by ensuring the total frequency matches the given data, and use the mode formula for more complex problems.

The daily expenditure of 100 families is given with some missing frequencies. The mode of the distribution is 43. Find the missing frequencies.

Solve a grouped data problem involving the daily expenditure of 100 families, with missing frequencies in two ranges. Using the mode formula, find the missing frequencies and complete the table. Suitable for students in Grades 9-11.

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