In a frequency distribution, the mid-value of a class is 10 and width of each class is 6. The upper limit of the class is

Options

A. 13
B. 7
C. 8
D. 12

Answer

To find the upper limit of the class in a frequency distribution where the mid-value is given, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Mid-value of the class = 10
- Width of each class = 6

2. **Understand the Concept of Mid-value**:
- The mid-value of a class is the average of the lower limit and the upper limit of that class.
- It can be expressed as:
\[ \text{Mid-value} = \frac{\text{Lower limit} + \text{Upper limit}}{2} \]

3. **Express the Upper Limit in Terms of Lower Limit**:
- Let the lower limit be \( L \) and the upper limit be \( U \).
- From the mid-value formula, we have:
\[ 10 = \frac{L + U}{2} \]
- Rearranging gives:
\[ L + U = 20 \quad \text{(Equation 1)} \]

4. **Relate the Width of the Class to Lower and Upper Limits**:
- The width of the class is defined as:
\[ \text{Width} = U - L \]
- Given that the width is 6, we can express this as:
\[ U - L = 6 \quad \text{(Equation 2)} \]

5. **Solve the Equations**:
- Now we have two equations:
1. \( L + U = 20 \)
2. \( U - L = 6 \)

- We can add these two equations:
\[ (L + U) + (U - L) = 20 + 6 \]
\[ 2U = 26 \]
\[ U = 13 \]

6. **Find the Lower Limit**:
- Substitute \( U = 13 \) back into Equation 1:
\[ L + 13 = 20 \]
\[ L = 20 - 13 = 7 \]

7. **Conclusion**:
- The upper limit of the class is \( U = 13 \).

### Final Answer:
The upper limit of the class is **13**.