The student got 32% marks and failed by 12 marks. Another student got 42% marks with 28 marks more than the minimum marks required to pass. Find the maximum marks and pass percent require to pass.
Answers
Given:
The student got 32% marks and failed by 12 marks.
Another student got 42% marks with 28 marks more than the minimum marks required to clear.
To Find:
The maximum marks and clearing percent required.
Solution:
Let the marks be x and marks required to clear be y.
Now, The student got 32% of x and failed by 12 marks. Another student got 42% of x and got 28% marks more than the minimum marks required to clear. So,
⇒ 32% of x +12 = 42% of x-28
⇒ 12 + 28 = 42% 0f x-32% of x
⇒ 42x/100-32x/100=40
⇒ 10x/100 = 40
⇒ x/10 = 40
⇒ x = 40×100
⇒ x = 400
So, the maximum marks is 400.
Minimum clearing marks = 32% of x+12
= 32/100×400+12 [substituting the value of x]
= 140.
So, minimum clearing marks = 140
Now, the clearing percentage = minimum clearing marks/maximum marks × 100
= 140/400×100
= 35%
So, the clearing percentage = 35%
Therefore, the maximum marks are 400 and the passing percentage is 35%.