A student who gets 20% marks fails by 20 marks, but another student who gets 36% marks gets 44 marks more than minimum passing marks. Find the maximum number of marks and percentage necessary for passing.

Let the maximum number of marks be 'x' and minimum passing marks be 'y'.

Given, A student who gets 20% marks fails by 20 marks.

$$\Rightarrow \frac{20}{100}\times x = y - 20$$

$$\Rightarrow 20x = 100y - 2000$$.................(1)

Also given that, another student who gets 36% marks gets 44 marks more than minimum passing marks.

$$\Rightarrow \frac{36}{100}\times x = y + 44$$

$$\Rightarrow 36x = 100y + 4400$$....................(2)

(2) - (1) $$\Rightarrow$$ 16x = 6400

$$\Rightarrow$$ x = 400

From (1) or (2), we get y = 100

Hence, maximum number of marks = x = 400

Percentage necessary for passing = $$\frac{y}{x}\times 100$$ = 25%