## A candidate score 25% maeks and failed by 60 marks while the other candidate got 50% marks and got 40 marks more than the maximum marks requ

Question

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## Answers ( )

Answer:Let maximum mark be x and minimum mark be y.

25% of maximum mark = (25/100)x = x/4 = y-30 …………..(1)

50% of maximum mark = (50/100)x = x/2 = y+20 ………..(2)

subtract eqn.(1) from eqn.(2), we get x/4 = 50 or x = 200

substituting value of x in eqn.(1), we get y = 80

hence maximum mark is 200 and minimum mark is 80

Step-by-step explanation:please mark me as brainly

Answer:Minimum passing marks are 80.

Step-by-step explanation: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