An engineering student has to secure $24\%$ marks to pass. He gets $61$ marks and fails by $29$ marks. What is the maximum marks for that exam?

Let the maximum marks for the exam be $M$.The student gets $61$ marks and fails by $29$ marks, which means the minimum passing marks are:
$$\text{Passing Marks}=61+29=90$$
According to the question, the passing marks are $24\%$ of the maximum marks:
$$24\%\text{ of }M=90$$
$$\frac{24}{100}\times M=90$$
$$M=\frac{90\times 100}{24}$$
$$M=\frac{9000}{24}=375\text{ marks}$$