If the mode for the following frequency distribution is 22 and `10 gt y gt x, "them "y=`
`{:("Class-interval:", 0-10,10-20,20-30,30-40,40-50,"Total"),("Frequency:", 5,8,10,x,y,30):}`

Options

A. 2
B. 5
C. 3
D. 4

Explanation

To solve the problem, we need to find the values of \( x \) and \( y \) in the frequency distribution given that the mode is 22 and \( 10 > y > x \).

### Step-by-Step Solution:

1. **Identify the Class Intervals and Frequencies:**
- Class intervals: 0-10, 10-20, 20-30, 30-40, 40-50
- Frequencies: 5, 8, 10, \( x \), \( y \), Total = 30

2. **Determine the Modal Class:**
- The mode is given as 22, which falls in the class interval 20-30.
- Therefore, the modal class is 20-30.

3. **Identify Values for the Mode Formula:**
- The mode formula for grouped data is:
\[ \text{Mode} = l + \frac{f - f_1}{2f - f_1 - f_2} \times h \]
- Where:
- \( l \) = lower boundary of the modal class = 20
- \( f \) = frequency of the modal class = 10
- \( f_1 \) = frequency of the class preceding the modal class = 8
- \( f_2 \) = frequency of the class succeeding the modal class = \( x \)
- \( h \) = width of the class interval = 10

4. **Substituting Values into the Mode Formula:**
- Set the mode equal to 22:
\[ 22 = 20 + \frac{10 - 8}{2 \cdot 10 - 8 - x} \cdot 10 \]

5. **Simplify the Equation:**
- Rearranging gives:
\[ 22 - 20 = \frac{2}{20 - 8 - x} \cdot 10 \]
\[ 2 = \frac{20}{12 - x} \]
- Cross-multiplying gives:
\[ 2(12 - x) = 20 \]
\[ 24 - 2x = 20 \]
\[ 2x = 4 \quad \Rightarrow \quad x = 2 \]

6. **Finding the Value of \( y \):**
- Now substitute \( x = 2 \) back into the total frequency equation:
\[ 5 + 8 + 10 + x + y = 30 \]
\[ 5 + 8 + 10 + 2 + y = 30 \]
\[ 25 + y = 30 \quad \Rightarrow \quad y = 5 \]

7. **Final Values:**
- The values are \( x = 2 \) and \( y = 5 \).

### Summary of Results:
- \( x = 2 \)
- \( y = 5 \)