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

Options

A. 8
B. 7
C. 10
D. 12

Answer

To find the lower limit of the class in the given frequency distribution, we can follow these steps:

### Step 1: Understand the given information
We are given:
- Mid value of the class (M) = 10
- Width of the class (W) = 6

### Step 2: Set up the equations
1. The mid value of a class is calculated using the formula:
\[ M = \frac{\text{Lower Limit} + \text{Upper Limit}}{2} \]
Let's denote the lower limit as \( x \) and the upper limit as \( y \). Therefore, we can write:
\[ \frac{x + y}{2} = 10 \]
Multiplying both sides by 2 gives us:
\[ x + y = 20 \quad \text{(Equation 1)} \]

2. The width of the class is calculated using the formula:
\[ W = \text{Upper Limit} - \text{Lower Limit} \]
Thus, we can write:
\[ y - x = 6 \quad \text{(Equation 2)} \]

### Step 3: Solve the equations
From Equation 2, we can express \( y \) in terms of \( x \):
\[ y = x + 6 \quad \text{(Equation 3)} \]

### Step 4: Substitute Equation 3 into Equation 1
Now, substitute Equation 3 into Equation 1:
\[ x + (x + 6) = 20 \]
This simplifies to:
\[ 2x + 6 = 20 \]

### Step 5: Isolate \( x \)
Subtract 6 from both sides:
\[ 2x = 20 - 6 \]
\[ 2x = 14 \]
Now, divide by 2:
\[ x = \frac{14}{2} = 7 \]

### Step 6: Find \( y \) (Upper Limit)
Using Equation 3 to find \( y \):
\[ y = x + 6 = 7 + 6 = 13 \]

### Conclusion
The lower limit of the class is:
\[ \text{Lower Limit} = x = 7 \]