A candidate scores 25% marks and fails by 68 marks, while another candidate, who scores 50% marks, gets 42 marks more than the minimum required marks to pass the examination. Find the maximum marks for the examination.

Correct option is B

Given:

Candidate A: Scores 25% and fails by 68 marks.

Candidate B: Scores 50% and gets 42 marks more than the passing marks.

Solution:

Let the maximum marks be M

Let the passing marks be P

For Candidate A (25% marks, fails by 68):

25%  of  M = P - 68

0.25M = P - 68 ............ (Equation 1)

For Candidate B (50% marks, passes by 42):

50%  of  M = P + 42

0.50M = P + 42 ...........(Equation 2)

Subtracting Equation 1 from Equation 2 to eliminate P:

0.50M - 0.25M = (P + 42) - (P - 68)

0.25M = 110

M = $$\frac{110}{0.25}$$​ = 440

Thus, the maximum passing marks is 440. 
Alternate Solution:
Let maximum marks be = x
Marks scored by A = 25% of x = 0.25x
Marks scored by B = 50% of x = 0.50x
Passing marks = Marks scored by A + 68 = 0.25x + 68
Passing marks = Marks scored by B - 42 = 0.50x - 42
=> 0.25x + 68 = 0.50x - 42
=> 68 + 42 = 0.50x - 0.25x
=> 110 = 0.25x
=> x = 110 ÷ 0.25
=> x = 440