A candidate who gets 20% marks in the examination fails by 5 marks, but another candidate, who gets 30% marks, gets 20 more marks than the minimum pass marks. What is the necessary percentage required for passing?

Hint: The only thing that we need to focus on to solve this problem is the percentage calculation. Take all the percentages to be with respect to the maximum marks. The two unknown elements are maximum marks and passing marks.

Complete step-by-step answer:
We are letting the maximum marks to be x and passing marks to be y.
Now, it is given in the question that if a student scores 20% in the exam fails by 5 marks. We can mathematically represent it as:
20% of x = y – 5
$\Rightarrow \dfrac{20}{100}\times x=y-5$
On taking 5 to the other side of the equation, we get
$\dfrac{20}{100}\times x+5=y...............(i)$
It is also given in the question that a candidate who gets 30% marks, gets 20 more marks than the minimum pass marks. When we write this in the form of a mathematical equation, we get
30% of x = y +20
$\Rightarrow \dfrac{30}{100}\times x=y+20$
Now, let us substitute the value of y from equation (i).
$\dfrac{30}{100}\times x=\dfrac{20}{100}\times x+20+5$
$\Rightarrow \dfrac{30}{100}\times x-\dfrac{20}{100}\times x=25$
$\Rightarrow \dfrac{10}{100}\times x=25$
$\Rightarrow x=\dfrac{25\times 100}{10}$
$\therefore x=250$
So, the maximum marks in the exam is 250 marks. On substituting the value of x in equation (i), we get
 $\dfrac{20}{100}\times 250+5=y$
$\Rightarrow 50+5=y$
$\therefore y=55$
Now, let us try to calculate the percentage passing marks.
Percentage of marks required for passing = $\dfrac{\text{passing marks}}{\text{Maximum marks}}\times 100=\dfrac{y}{x}\times 100=\dfrac{55}{250}\times 100=22%$
Therefore, the maximum marks is 250 marks, and the passing percentage is 22%.

Note: Always remember the percentages for an exam are calculated with respect to maximum marks. It is preferable to let the exact marks be variables, however, if we want, we can let the percentages as well.