While calculating mode the following observation are found. Lower limit of modal class =40, frequency of modal class =20, frequency of previous class =12 , frequency of preceding class =11 and width of the class =10, find mode .

Answer

To find the mode using the given data, we will use the formula for mode in a grouped frequency distribution:

\[ \text{Mode} = L + \frac{f_1 - f_0}{(2f_1 - f_0 - f_2)} \times h \]

Where:
- \( L \) = lower limit of the modal class
- \( f_1 \) = frequency of the modal class
- \( f_0 \) = frequency of the preceding class
- \( f_2 \) = frequency of the succeeding class
- \( h \) = width of the class

### Step-by-Step Solution:

1. **Identify the values from the problem:**
- Lower limit of modal class \( L = 40 \)
- Frequency of modal class \( f_1 = 20 \)
- Frequency of preceding class \( f_0 = 12 \)
- Frequency of succeeding class \( f_2 = 11 \)
- Width of the class \( h = 10 \)

2. **Substitute the values into the mode formula:**
\[ \text{Mode} = 40 + \frac{20 - 12}{(2 \times 20 - 12 - 11)} \times 10 \]

3. **Calculate the numerator:**
\[ 20 - 12 = 8 \]

4. **Calculate the denominator:**
\[ 2 \times 20 - 12 - 11 = 40 - 12 - 11 = 17 \]

5. **Now substitute these results back into the formula:**
\[ \text{Mode} = 40 + \frac{8}{17} \times 10 \]

6. **Calculate \( \frac{8}{17} \times 10 \):**
\[ \frac{8 \times 10}{17} = \frac{80}{17} \]

7. **Now, calculate \( 40 + \frac{80}{17} \):**
- First, convert 40 into a fraction with a denominator of 17:
\[ 40 = \frac{680}{17} \]
- Now add:
\[ \text{Mode} = \frac{680}{17} + \frac{80}{17} = \frac{760}{17} \]

8. **Finally, divide to find the mode:**
\[ \frac{760}{17} \approx 44.71 \]

### Final Answer:
The mode is approximately **44.71**.