QJEEM25Mad3c1: JEE Main 2025 Mathematics - Class 11 JEEM25Mad3c1 Solution: Marks obtains by all the students of class 12 are presented in…

The median for grouped data is given by:

$\text{Median} = L + \left(\frac{\frac{n}{2} - CF}{f}\right) \times h$

where

$L$ is the lower limit (or boundary) of the median class.

$CF$ is the cumulative frequency of all classes preceding the median class.

$f$ is the frequency of the median class.

$h$ is the class width.

$n$ is the total number of students.

Given:

Median $= 14$

Median class interval is $12-18$ , so $L = 12$ and the class width $h = 18 - 12 = 6$ .

Frequency of median class $f = 12$ .

Cumulative frequency below the median class $= 18$ (i.e., $CF = 18$ ).

Plugging these into the formula:

$14 = 12 + \left(\frac{\frac{n}{2} - 18}{12}\right) \times 6$

Step 1: Subtract 12 from both sides:

$2 = \left(\frac{\frac{n}{2} - 18}{12}\right) \times 6$

Step 2: Simplify the multiplication factor:

$\left(\frac{6}{12}\right) = \frac{1}{2}$

So the equation becomes:

$2 = \frac{1}{2}\left(\frac{n}{2} - 18\right)$

Step 3: Multiply both sides by 2 to remove the fraction:

$4 = \frac{n}{2} - 18$

Step 4: Solve for $\frac{n}{2}$ :

$\frac{n}{2} = 4 + 18 = 22$

Step 5: Multiply both sides by 2 to find $n$ :

$n = 44$

Thus, the total number of students is $44$ .