Question

a candidate scored 25% marks in an examination and failed by 30 marks,while another candidate who scored 50% got 20 marks more than the minimum pass marks . find the maximum marks and the minimum pass marks?

Answer 1

Solution:-

Let the maximum marks be 'x'.
Then, according to the question.
(25 % of x) + 30 = (50 % of x) - 20
⇒ x/4 + 30 = x/2 - 20
⇒ (x/4) - (x/2) = - 20 - 30
⇒ (x - 2x)/4 = - 50
⇒ - x/4 = - 50
⇒ - x = - 200
⇒ x = 200
So, the maximum marks are 200.
Minimum passing marks are = x/4 + 30
= 200/4 + 30
= 50 + 30
= 80 marks
Minimum passing marks are 80.
Answer.

Answer 2

Answer:

Maximum mark 200 and minimum marks 80

Step-by-step explanation:

Let x and $$y$ be the passing marks and maximum marks respectively.

Therefore,

10025y=x−30

4y=x−30

⇒x=4y+30.....(1)

Also,

10050y=x+20

2y=x+20

⇒2y=4y+30+20(From (1))

⇒2y−4y=50

⇒2y=50

⇒y=50×2=100

Substituting the value of y in equation (1), we have

x=4200+30=50+30=80

Hence the minimum passing marks are 80 and the maximum marks are 200.

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