An engineering student has to secure $36\%$ marks to pass. He gets $53$ marks and fails by $37$ marks. Find the maximum marks.

Let the maximum marks be $M$.The passing marks required are $36\%$ of $M$.The student gets $53$ marks and fails by $37$ marks, so the passing marks are:
$$\text{Passing Marks}=53+37=90$$
Therefore,
$$36\%\text{ of }M=90$$
$$\frac{36}{100}\times M=90$$
$$M=\frac{90\times 100}{36}$$
$$M=\frac{5\times 100}{2}=250$$
Hence, the maximum marks are $250$.