Build the two cumulative-frequency tables, find the class where cumulative frequency first reaches N/2 = 25, then apply the median formula in that class. …

Step 1 — 'less than' cumulative frequency table (cumulate from the top):

Step 2 — 'more than' cumulative frequency table (cumulate from the bottom):

Step 3 — locate the median class. Total $N = 50$, so $\dfrac{N}{2} = 25$. In the 'less than' table, the cumulative frequency first reaches 25 exactly at 'less than 30' — so the two ogives cross at $x = 30$, giving a graphically-read median of 30 marks. This also matches: the 'more than' table shows cumulative frequency 25 at 'more than or equal to 30' — the same crossing point, confirming the two ogives genuinely intersect at (30, 25).

Step 4 — confirm with the median formula. The median class is $20–30$ (it is the class whose 'less than' cumulative frequency first reaches/exceeds 25, and the cumulative frequency just before it, 13, corresponds to the previous class 10–20):

$L=20$, $\dfrac{N}{2}=25$, $c.f.=13$ (cumulative frequency of the class before the median class), $f=12$ (frequency of the median class 20–30), $h=10$ (class width). …

An equally valid alternative is to locate the median class using the 'more than' cumulative frequency table instead — the class whose 'more than' cumulative frequency first drops to or below 25 is again 20–30 (more-than-or-equal-to-30 …

A common mistake is using the frequency of the wrong class as $f$ in the formula (e.g., using the cumulative frequency instead of the class frequency), or mis-identifying the median class when the cumulative frequency lands exactly on N/2, as it does here — the rule is to take the c …