If `r%` of students scored less than 33 marks, then find r, given `N=50`, class size=`10`, frequency of the rth. Percentile class `=12` and cumulative frequency of the rth percentile class `=22`, lower limit of rth percentile class `=30`

Answer

To solve the problem step by step, we will use the given data and apply the formula for calculating percentiles.

### Given Data:
- Total number of students, \( N = 50 \)
- Class size, \( H = 10 \)
- Frequency of the rth percentile class, \( f = 12 \)
- Cumulative frequency of the rth percentile class, \( CF = 22 \)
- Lower limit of the rth percentile class, \( L = 30 \)
- Minimum marks to be considered, \( X = 33 \)

### Step 1: Write the formula for the rth percentile
The formula for the rth percentile is given by:

\[ X = L + \frac{N \cdot \left(\frac{r}{100}\right) - CF}{f} \cdot H \]

### Step 2: Substitute the known values into the formula
Substituting the known values into the formula:

\[ 33 = 30 + \frac{50 \cdot \left(\frac{r}{100}\right) - 22}{12} \cdot 10 \]

### Step 3: Simplify the equation
First, subtract 30 from both sides:

\[ 33 - 30 = \frac{50 \cdot \left(\frac{r}{100}\right) - 22}{12} \cdot 10 \]

This simplifies to:

\[ 3 = \frac{50 \cdot \left(\frac{r}{100}\right) - 22}{12} \cdot 10 \]

### Step 4: Multiply both sides by 12 to eliminate the denominator
\[ 3 \cdot 12 = (50 \cdot \left(\frac{r}{100}\right) - 22) \cdot 10 \]

This gives:

\[ 36 = (50 \cdot \left(\frac{r}{100}\right) - 22) \cdot 10 \]

### Step 5: Divide both sides by 10
\[ \frac{36}{10} = 50 \cdot \left(\frac{r}{100}\right) - 22 \]

This simplifies to:

\[ 3.6 = 50 \cdot \left(\frac{r}{100}\right) - 22 \]

### Step 6: Add 22 to both sides
\[ 3.6 + 22 = 50 \cdot \left(\frac{r}{100}\right) \]

This gives:

\[ 25.6 = 50 \cdot \left(\frac{r}{100}\right) \]

### Step 7: Multiply both sides by \( \frac{100}{50} \)
\[ \frac{25.6 \cdot 100}{50} = r \]

This simplifies to:

\[ r = 51.2 \]

### Conclusion
Thus, the value of \( r \) is \( 51.2 \).

---