A student who gets 20% marks fails by 20 marks, but another student who gets 36% marks gets 44 marks more than minimum passing marks. Find the maximum number of marks and percentage necessary for passing.
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let the max. marks be x
20℅ of x + 20= 36℅ of x- 44
20/100 x+ 20= 36/100 x- 44
x/5 + 20 = 9/25 x- 44
x+100/5= 9x- 1100/25
25(x+100)=5(9x-1100)
25x+2500=45x-5500
2500+5500=45x-25x
8000=20x
x=8000/20
=400
max. marks=400
℅ required for pass=20℅of 400+20
=20/100×400 +20
=80+20
=100 marks
20℅ of x + 20= 36℅ of x- 44
20/100 x+ 20= 36/100 x- 44
x/5 + 20 = 9/25 x- 44
x+100/5= 9x- 1100/25
25(x+100)=5(9x-1100)
25x+2500=45x-5500
2500+5500=45x-25x
8000=20x
x=8000/20
=400
max. marks=400
℅ required for pass=20℅of 400+20
=20/100×400 +20
=80+20
=100 marks
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