Question

in an examination, a candidate A score 30% and fails by 40 marks while another
candidate B scores 40% and get 20 marks more than the minimum pass marks. Find
the maximum marks and minimum pass marks.​

Answer 1

Answer: Hey please mark my answer as brainliest

Step-by-step explanation:

Let the maximum mark be x

And minimum mark be y

30x/100=y -40

3x/10=y-40 (equation 1)

40x/100=y+20

2x/5=20+y

Y=(2x/5)-20. (Equation 2)

Substitute equation 2in equation 1

3x/10=(2x/5)-20-40

3x/10=(2x/5)-60

3x=(20x/5)-600

3x=4x-600

600=4x-3x

X=600

Therefore maximum mark is 600

Y=(2x/5)-20. From equation 2

=(1200/5)-20

=240-20

=220

Therefore pass mark is 220

Answer 2

Answes : The Answers are correct. Pls Mark as Brainliest

Step-by-step explanation:

Let Maximum Marks be = x

Let Minimum Marks be = y

A =  30/100 X x = y -40

   =  30x/100 = y -40

   = y = 30x/100 + 40

     

B = 40/100 X x = y +20

  = 40x/100 = y + 20

   = y = 40x/100 - 20

    30x/100 + 40 =  40x/100 - 20

     30x/100 - 40x/100  = -20 -40

     -10x/100 = -60

      10x/100 = 60 [Negative signs are removed]

      10x = 6000

      x = 600 [Maximum Marks are 600]

   Passing Marks = 600 x 30/100 + 40

                              = 220 [ Passing Marks]

[Maximum Marks are 600]

[ Passing Marks are 220]